Pandas (Extended)¶

The Pandas module is Python's fundamental data analytics library and it provides high-performance, easy-to-use data structures and tools for data analysis. Many Excel features are available in Pandas, including creating pivot tables, computing columns based on other columns, etc. Pandas also facilitates grouping rows by column values and joining tables as in SQL. Pandas is a very comprehensive and mature module that can be used for advanced data analytics, so this tutorial presents a basic overview of Pandas' capabilities.

Table of Contents¶

Series
DataFrame
Combining DataFrames
- SQL like joins
- Concatenation
Categories
Saving and loading
- Saving
- Loading
Exercises
- Possible solutions

Let's import Pandas with the usual convention as pd.

# let's suppress warnings, as they can get annoying sometimes
import warnings
warnings.filterwarnings("ignore")

import pandas as pd

Series¶

A Series is a one-dimensional array-like object containing a sequence of values and an associated array of data labels, called its index.

Creating a Series¶

The simplest Series is formed from only an array of data.

obj = pd.Series([2,-1,3,5])
obj

0    2
1   -1
2    3
3    5
dtype: int64

The string representation of a Series displayed interactively shows the index on the left and the values on the right. You can get the array representation and index object of the Series via its values and index attributes, respectively.

obj.values

array([ 2, -1,  3,  5], dtype=int64)

obj.index # Like range(4)

RangeIndex(start=0, stop=4, step=1)

list(obj.index) # if you want the index as a list of values

[0, 1, 2, 3]

We may want to create a Series with an index identifying each data point with a label.

obj2 = pd.Series([2,-1,3,5], index=['a', 'b', 'c', 'd'])
obj2

a    2
b   -1
c    3
d    5
dtype: int64

list(obj2.index)

['a', 'b', 'c', 'd']

A Series can have a name.

obj_withName = pd.Series([83, 68], index=["bob", "alice"], name="weights")
obj_withName

bob      83
alice    68
Name: weights, dtype: int64

You can also create a Series object from a dict. The keys will be used as index labels.

weightdata = {"john": 86, "michael": 68, "alice": 68, "bob": 83}
obj3 = pd.Series(weightdata)
obj3

john       86
michael    68
alice      68
bob        83
dtype: int64

Selecting and filtering in Series¶

Compared with NumPy arrays, you can use labels in the index when selecting single values or a set of values.

obj2

a    2
b   -1
c    3
d    5
dtype: int64

obj2['a']

2

obj2[['b','c','d']] 
# Here ['b', 'c', 'd'] is interpreted as a list of indices, even though it contains strings instead of integers.

b   -1
c    3
d    5
dtype: int64

You can still access the items by integer location, like in a regular array. By default, the rank of the item in the Series starts at 0.

obj2[0]

2

obj2[[1,3]]

b   -1
d    5
dtype: int64

obj2[obj2 < 0]

b   -1
dtype: int64

Slicing with labels behaves differently than normal Python slicing in that the endpoint is inclusive.

obj2['b':'c']

b   -1
c    3
dtype: int64

To make it clear when you are accessing by label or by integer location, it is recommended to always use the loc attribute when accessing by label, and the iloc attribute when accessing by integer location.

obj2.loc["a"]

2

obj2.iloc[0]

2

Slicing a Series also slices the index labels.

obj2.iloc[1:3]

b   -1
c    3
dtype: int64

This can lead to unexpected results when using the default numeric labels, so be careful:

surprise = pd.Series([1000, 1001, 1002, 1003])
surprise

0    1000
1    1001
2    1002
3    1003
dtype: int64

surprise_slice = surprise[2:]
surprise_slice

2    1002
3    1003
dtype: int64

Oh look! The first element has index label 2. The element with index label 0 is absent from the slice.

But remember that you can access elements by integer location using the iloc attribute. This illustrates another reason why it's always better to use loc and iloc to access Series objects.

surprise_slice.iloc[0]

1002

Operations on Series¶

Series objects behave much like one-dimensional ndarrays, and you can often pass them as parameters to NumPy functions.

import numpy as np
np.exp(obj2)

a      7.389056
b      0.367879
c     20.085537
d    148.413159
dtype: float64

Arithmetic operations on Series are also possible, and they apply elementwise, just like for ndarrays.

obj2 + [1000,2000,3000,4000]

a    1002
b    1999
c    3003
d    4005
dtype: int64

Similar to NumPy, if you add a single number to a Series, that number is added to all items in the Series. This is called broadcasting.

obj2 + 1000

a    1002
b     999
c    1003
d    1005
dtype: int64

The same is true for all binary operations and even conditional operations.

Plotting a Series¶

pandas makes it easy to plot Series data using matplotlib. Just import matplotlib and call the plot() method:

%matplotlib inline
import matplotlib.pyplot as plt
%config InlineBackend.figure_format = 'retina'
plt.style.use("ggplot")
###
temperatures = [4.4,5.1,6.1,6.2,6.1,6.1,5.7,5.2,4.7,4.1,3.9,3.5]
s7 = pd.Series(temperatures, name="Temperature")
s7.plot()
plt.show();

DataFrame¶

A DataFrame represents a rectangular table of data and contains an ordered collection of columns, each of which can be a different value type (numeric, string, boolean, etc.). You can see DataFrames as dict of Series.

Creating a DataFrame¶

There are many ways to construct a DataFrame, though one of the most common is from a dict of equal-length lists or NumPy arrays.

data = {'state': ['Ohio', 'Ohio', 'Ohio', 'Nevada', 'Nevada', 'Nevada'],
'year': [2000, 2001, 2002, 2001, 2002, 2003],
'pop': [1.5, 1.7, 3.6, 2.4, 2.9, 3.2]}
df = pd.DataFrame(data)
df

For large DataFrames, the head method selects only the first five rows.

df.head()

You can also create a DataFrame by passing a dict of Series objects.

people_dict = {
    "weight": pd.Series([68, 83, 112], index=["alice", "bob", "charles"]),
    "birthyear": pd.Series([1984, 1985, 1992], index=["bob", "alice", "charles"], name="year"),
    "children": pd.Series([0, 3], index=["charles", "bob"]),
    "hobby": pd.Series(["Biking", "Dancing"], index=["alice", "bob"]),
}
people = pd.DataFrame(people_dict)
people

A few things to note:

The Series were automatically aligned based on their index.
Missing values are represented as NaN.
Series names are ignored (the name "year" was dropped).

If you pass a list of columns and/or index row labels to the DataFrame constructor, it will guarantee that these columns and/or rows will exist, in that order, and no other column/row will exist.

df2 = pd.DataFrame(
        people_dict,
        columns=["birthyear", "weight", "height"],
        index=["bob", "alice", "eugene"]
     )
df2

Another convenient way to create a DataFrame is to pass all the values to the constructor as an ndarray, or a list of lists, and specify the column names and row index labels separately:

values = [
            [1985, np.nan, "Biking",   68],
            [1984, 3,      "Dancing",  83],
            [1992, 0,      np.nan,    112]
         ]
df3 = pd.DataFrame(
        values,
        columns=["birthyear", "children", "hobby", "weight"],
        index=["alice", "bob", "charles"]
     )
df3

You can access columns pretty much as you would expect. They are returned as Series objects.

people["birthyear"]

alice      1985
bob        1984
charles    1992
Name: birthyear, dtype: int64

You can also get multiple columns at once.

people[["birthyear", "hobby"]]

To specify missing values, you can either use np.nan or NumPy's masked arrays:

masked_array = np.ma.asarray(values, dtype=np.object)
masked_array[(0, 2), (1, 2)] = np.ma.masked
df3 = pd.DataFrame(
        masked_array,
        columns=["birthyear", "children", "hobby", "weight"],
        index=["alice", "bob", "charles"]
     )
df3

Instead of an ndarray, you can also pass a DataFrame object:

df4 = pd.DataFrame(
         df3,
         columns=["hobby", "children"],
         index=["alice", "bob"]
     )
df4

Selecting and filtering in DataFrame¶

Let's go back to the people:

people

The loc attribute lets you access rows instead of columns. The result is a Series object in which the DataFrame's column names are mapped to row index labels.

people.loc["charles"]

weight        112
birthyear    1992
children        0
hobby         NaN
Name: charles, dtype: object

You can also access rows by integer location using the iloc attribute.

people.iloc[2]

weight        112
birthyear    1992
children        0
hobby         NaN
Name: charles, dtype: object

You can also get a slice of rows, and this returns a DataFrame object.

people.iloc[1:3]

Finally, you can pass a boolean array to get the matching rows.

people[np.array([True, False, True])]

This is most useful when combined with boolean expressions.

people[people["birthyear"] < 1990]

Transposing¶

You can swap columns and indices using the T attribute.

people.T

Adding and removing columns¶

You can generally treat DataFrame objects like dictionaries of Series, so the following work fine.

people

people["age"] = 2018 - people["birthyear"]  # adds a new column "age"
people["over 30"] = people["age"] > 30      # adds another column "over 30"
birthyears = people.pop("birthyear")
del people["children"]

people

birthyears

alice      1985
bob        1984
charles    1992
Name: birthyear, dtype: int64

When you add a new colum, it must have the same number of rows. Missing rows are filled with NaN, and extra rows are ignored.

people["pets"] = pd.Series({"bob": 0, "charles": 5, "eugene":1})  # alice is missing, eugene is ignored
people

When adding a new column, it is added at the end (on the right) by default. You can also insert a column anywhere else using the insert() method.

people.insert(1, "height", [172, 181, 185])
people

Assigning new columns¶

You can also create new columns by calling the assign() method. Note that this returns a new DataFrame object, the original is not modified.

people.assign(
    body_mass_index = people["weight"] / (people["height"] / 100) ** 2,
    has_pets = people["pets"] > 0
)

Note that you cannot access columns created within the same assignment.

try:
    people.assign(
        body_mass_index = people["weight"] / (people["height"] / 100) ** 2,
        overweight = people["body_mass_index"] > 25
    )
except KeyError as e:
    print("Key error:", e)

Key error: 'body_mass_index'

The solution is to split this assignment in two consecutive assignments.

df5 = people.assign(body_mass_index = people["weight"] / (people["height"] / 100) ** 2)
df5.assign(overweight = df5["body_mass_index"] > 25)

Having to create a temporary variable df5 is not very convenient. You may want to just chain the assigment calls, but it does not work because the people object is not actually modified by the first assignment.

try:
    (people
         .assign(body_mass_index = people["weight"] / (people["height"] / 100) ** 2)
         .assign(overweight = people["body_mass_index"] > 25)
    )
except KeyError as e:
    print("Key error:", e)

Key error: 'body_mass_index'

But fear not, there is a simple solution. You can pass a function to the assign() method (typically a lambda function), and this function will be called with the DataFrame as a parameter.

(people
     .assign(body_mass_index = lambda df: df["weight"] / (df["height"] / 100) ** 2)
     .assign(overweight = lambda df: df["body_mass_index"] > 25)
)

Problem solved!

Evaluating an expression¶

A great feature supported by pandas is expression evaluation. This relies on the numexpr library which must be installed.

people.eval("weight / (height/100) ** 2 > 25")

alice      False
bob         True
charles     True
dtype: bool

Assignment expressions are also supported. Let's set inplace=True to directly modify the DataFrame rather than getting a modified copy:

people.eval("body_mass_index = weight / (height/100) ** 2", inplace=True)
people

You can use a local or global variable in an expression by prefixing it with '@'.

overweight_threshold = 30
people.eval("overweight = body_mass_index > @overweight_threshold", inplace=True)
people

Querying a DataFrame¶

The query() method lets you filter a DataFrame based on a query expression.

people.query("age > 30 and pets == 0")

Sorting a DataFrame¶

You can sort a DataFrame by calling its sort_index method. By default it sorts the rows by their index label, in ascending order, but let's reverse the order.

people.sort_index(ascending=False)

Note that sort_index returned a sorted copy of the DataFrame. To modify people directly, we can set the inplace argument to True. Also, we can sort the columns instead of the rows by setting axis=1.

people.sort_index(axis=1, inplace=True)
people

To sort the DataFrame by the values instead of the labels, we can use sort_values and specify the column to sort by.

people.sort_values(by="age", inplace=True)
people

Operations on DataFrame¶

Although DataFrames do not try to mimick NumPy arrays, there are a few similarities. Let's create a DataFrame to demonstrate this:

grades_array = np.array([[8,8,9],[10,9,9],[4, 8, 2], [9, 10, 10]])
grades = pd.DataFrame(grades_array, columns=["sep", "oct", "nov"], index=["alice","bob","charles","darwin"])
grades

You can apply NumPy mathematical functions on a DataFrame. The function is applied to all values.

np.sqrt(grades)

Similarly, adding a single value to a DataFrame will add that value to all elements in the DataFrame. This is called broadcasting.

grades + 1

Of course, the same is true for all other binary operations, including arithmetic (*,/,**...) and conditional (>, ==...) operations.

grades >= 5

Aggregation operations, such as computing the max, the sum or the mean of a DataFrame, apply to each column, and you get back a Series object.

grades.mean()

sep    7.75
oct    8.75
nov    7.50
dtype: float64

The all method is also an aggregation operation: it checks whether all values are True or not. Let's see during which months all students got a grade greater than 5.

(grades > 5).all()

sep    False
oct     True
nov    False
dtype: bool

Most of these functions take an optional axis parameter which lets you specify along which axis of the DataFrame you want the operation executed. The default is axis=0, meaning that the operation is executed vertically (on each column). You can set axis=1 to execute the operation horizontally (on each row). For example, let's find out which students had all grades greater than 5:

(grades > 5).all(axis = 1)

alice       True
bob         True
charles    False
darwin      True
dtype: bool

The any method returns True if any value is True. Let's see who got at least one grade 10:

(grades == 10).any(axis = 1)

alice      False
bob         True
charles    False
darwin      True
dtype: bool

If you add a Series object to a DataFrame (or execute any other binary operation), pandas attempts to broadcast the operation to all rows in the DataFrame. This only works if the Series has the same size as the DataFrames rows. For example, let's substract the mean of the DataFrame (a Series object) from the DataFrame:

grades - grades.mean()  # equivalent to: grades - [7.75, 8.75, 7.50]

We substracted 7.75 from all September grades, 8.75 from October grades and 7.50 from November grades. It is equivalent to substracting this DataFrame.

pd.DataFrame([[7.75, 8.75, 7.50]]*4, index=grades.index, columns=grades.columns)

If you want to substract the global mean from every grade, here is one way to do it:

grades - grades.values.mean() # substracts the global mean (8.00) from all grades

Automatic alignment for DataFrames¶

Similar to Series, when operating on multiple DataFrames, pandas automatically aligns them by row index label, but also by column names. Let's create a DataFrame with bonus points for each person from October to December:

bonus_array = np.array([[0,np.nan,2],[np.nan,1,0],[0, 1, 0], [3, 3, 0]])
bonus_points = pd.DataFrame(bonus_array, columns=["oct", "nov", "dec"], index=["bob","colin", "darwin", "charles"])
bonus_points

grades + bonus_points

Looks like the addition worked in some cases but way too many elements are now empty. That's because when aligning the DataFrames, some columns and rows were only present on one side, and thus they were considered missing on the other side (NaN). Then adding NaN to a number results in NaN, hence the result.

Plotting a DataFrame¶

Just like for Series, pandas makes it easy to draw nice graphs based on a DataFrame.

For example, it is easy to create a bar plot from a DataFrame's data by calling its plot method.

people.plot(kind = "bar", y = ["body_mass_index"])
plt.show();

You can pass extra arguments supported by matplotlib's functions. For example, we can create scatterplot and pass it a list of sizes using the s argument of matplotlib's scatter() function.

people.plot(kind = "scatter", x = "height", y = "weight", s=[40, 120, 200])
plt.show();

Again, there are way too many options to list here: the best option is to scroll through the Visualization page in pandas documentation and find the plot you are interested in.

Handling missing data¶

Dealing with missing data is a frequent task when working with real life data. pandas offers a few tools to handle missing data.

The isnull and notnull functions in pandas can be used to detect missing data.

pd.isnull(grades)

pd.notnull(grades)

It's a bit unfair that we're setting grades to zero in September, though. Perhaps we should decide that missing grades are missing grades, but missing bonus points should be replaced by zeros.

grades

fixed_bonus_points = bonus_points.fillna(0)
fixed_bonus_points.insert(0, "sep", 0)
fixed_bonus_points.loc["alice"] = 0
grades + fixed_bonus_points

That's much better: although we made up some data, we have not been too unfair.

Another way to handle missing data is to interpolate. Let's look at the bonus_points DataFrame again:

bonus_points

Now let's call the interpolate method. By default, it interpolates vertically (axis=0), so let's tell it to interpolate horizontally (axis=1).

bonus_points.interpolate(axis=1)

Bob had 0 bonus points in October, and 2 in December. When we interpolate for November, we get the mean: 1 bonus point. Colin had 1 bonus point in November, but we do not know how many bonus points he had in September, so we cannot interpolate, this is why there is still a missing value in October after interpolation. To fix this, we can set the September bonus points to 0 before interpolation.

better_bonus_points = bonus_points.copy()
better_bonus_points.insert(0, "sep", 0)
better_bonus_points.loc["alice"] = 0
better_bonus_points = better_bonus_points.interpolate(axis=1)
better_bonus_points

Great, now we have reasonable bonus points everywhere. Let's find out the final grades:

grades + better_bonus_points

It is slightly annoying that the September column ends up on the right. This is because the DataFrames we are adding do not have the exact same columns (the grades DataFrame is missing the "dec" column), so to make things predictable, pandas orders the final columns alphabetically. To fix this, we can simply add the missing column before adding.

grades["dec"] = np.nan
final_grades = grades + better_bonus_points
final_grades

There's not much we can do about December and Colin: it's bad enough that we are making up bonus points, but we can't reasonably make up grades (well I guess some teachers probably do). So let's call the dropna() method to get rid of rows that are full of NaNs:

final_grades_clean = final_grades.dropna(how="all")
final_grades_clean

Now let's remove columns that are full of NaNs by setting the axis argument to 1:

final_grades_clean = final_grades_clean.dropna(axis=1, how="all")
final_grades_clean

Overview functions¶

When dealing with large DataFrames, it is useful to get a quick overview of its content. pandas offers a few functions for this. First, let's create a large DataFrame with a mix of numeric values, missing values and text values. Notice how Jupyter displays only the corners of the DataFrame:

much_data = np.fromfunction(lambda x,y: (x+y*y)%17*11, (10000, 26))
large_df = pd.DataFrame(much_data, columns=list("ABCDEFGHIJKLMNOPQRSTUVWXYZ"))
large_df[large_df % 16 == 0] = np.nan
large_df.insert(3,"some_text", "Blabla")
large_df.head(10)

The head() method returns the top 5 rows.

large_df.head()

Of course there's also a tail() function to view the bottom 5 rows. You can pass the number of rows you want.

large_df.tail(n=2)

The info() method prints out a summary of each columns contents.

large_df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 10000 entries, 0 to 9999
Data columns (total 27 columns):
A            8823 non-null float64
B            8824 non-null float64
C            8824 non-null float64
some_text    10000 non-null object
D            8824 non-null float64
E            8822 non-null float64
F            8824 non-null float64
G            8824 non-null float64
H            8822 non-null float64
I            8823 non-null float64
J            8823 non-null float64
K            8822 non-null float64
L            8824 non-null float64
M            8824 non-null float64
N            8822 non-null float64
O            8824 non-null float64
P            8824 non-null float64
Q            8824 non-null float64
R            8823 non-null float64
S            8824 non-null float64
T            8824 non-null float64
U            8824 non-null float64
V            8822 non-null float64
W            8824 non-null float64
X            8824 non-null float64
Y            8822 non-null float64
Z            8823 non-null float64
dtypes: float64(26), object(1)
memory usage: 2.1+ MB

Finally, the describe() method gives a nice overview of the main aggregated values over each column:

count: number of non-null (not NaN) values
mean: mean of non-null values
std: standard deviation of non-null values
min: minimum of non-null values
25%, 50%, 75%: 25th, 50th and 75th percentile of non-null values
max: maximum of non-null values

large_df.describe()

Combining DataFrames¶

SQL-like joins¶

One powerful feature of pandas is it's ability to perform SQL-like joins on DataFrames. Various types of joins are supported: inner joins, left/right outer joins and full joins. To illustrate this, let's start by creating a couple simple DataFrames:

city_loc = pd.DataFrame(
    [
        ["CA", "San Francisco", 37.781334, -122.416728],
        ["NY", "New York", 40.705649, -74.008344],
        ["FL", "Miami", 25.791100, -80.320733],
        ["OH", "Cleveland", 41.473508, -81.739791],
        ["UT", "Salt Lake City", 40.755851, -111.896657]
    ], columns=["state", "city", "lat", "lng"])
city_loc

city_pop = pd.DataFrame(
    [
        [808976, "San Francisco", "California"],
        [8363710, "New York", "New-York"],
        [413201, "Miami", "Florida"],
        [2242193, "Houston", "Texas"]
    ], index=[3,4,5,6], columns=["population", "city", "state"])
city_pop

Now let's join these DataFrames using the merge() function:

pd.merge(left=city_loc, right=city_pop, on="city")

Note that both DataFrames have a column named state, so in the result they got renamed to state_x and state_y.

Also, note that Cleveland, Salt Lake City and Houston were dropped because they don't exist in both DataFrames. This is the equivalent of a SQL INNER JOIN. If you want a FULL OUTER JOIN, where no city gets dropped and NaN values are added, you must specify how="outer".

all_cities = pd.merge(left=city_loc, right=city_pop, on="city", how="outer")
all_cities

Of course LEFT OUTER JOIN is also available by setting how="left": only the cities present in the left DataFrame end up in the result. Similarly, with how="right" only cities in the right DataFrame appear in the result. For example:

pd.merge(left=city_loc, right=city_pop, on="city", how="right")

If the key to join on is actually in one (or both) DataFrame's index, you must use left_index=True and/or right_index=True. If the key column names differ, you must use left_on and right_on. For example:

city_pop2 = city_pop.copy()
city_pop2.columns = ["population", "name", "state"]
pd.merge(left=city_loc, right=city_pop2, left_on="city", right_on="name")

Concatenation¶

Rather than joining DataFrames, we may just want to concatenate them using concat() method.

result_concat = pd.concat([city_loc, city_pop], sort = False)
result_concat

Note that this operation aligned the data horizontally (by columns) but not vertically (by rows). In this example, we end up with multiple rows having the same index (eg. 3). pandas handles this rather gracefully.

result_concat.loc[3]

Or you can tell pandas to just ignore the index.

pd.concat([city_loc, city_pop], ignore_index=True, sort = False)

Notice that when a column does not exist in a DataFrame, it acts as if it was filled with NaN values. If we set join="inner", then only columns that exist in both DataFrames are returned.

pd.concat([city_loc, city_pop], join="inner")

You can concatenate DataFrames horizontally instead of vertically by setting axis=1.

pd.concat([city_loc, city_pop], axis=1)

In this case it really does not make much sense because the indices do not align well (eg. Cleveland and San Francisco end up on the same row, because they shared the index label 3). So let's reindex the DataFrames by city name before concatenating:

pd.concat([city_loc.set_index("city"), city_pop.set_index("city")], axis=1, sort = False)

This looks a lot like a FULL OUTER JOIN, except that the state columns were not renamed to state_x and state_y, and the city column is now the index.

The append() method is a useful shorthand for concatenating DataFrames vertically:

city_loc.append(city_pop, sort = False)

Categories¶

As always in pandas, the append() method does not actually modify city_loc: it works on a copy and returns the modified copy.

It is quite frequent to have values that represent categories, for example 1 for female and 2 for male, or "A" for Good, "B" for Average, "C" for Bad. These categorical values can be hard to read and cumbersome to handle, but fortunately pandas makes it easy. To illustrate this, let's take the city_pop DataFrame we created earlier, and add a column that represents a category:

city_econ = city_pop.copy()
city_econ["econ_code"] = [17, 17, 34, 20]
city_econ

Right now the econ_code column is full of apparently meaningless codes. Let's fix that. First, we will create a new categorical column based on the econ_codes:

city_econ["economy"] = city_econ["econ_code"].astype('category')
city_econ["economy"].cat.categories

Int64Index([17, 20, 34], dtype='int64')

Now we can give each category a meaningful name:

city_econ["economy"].cat.categories = ["Finance", "Energy", "Tourism"]
city_econ

Note that categorical values are sorted according to their categorical order, not their alphabetical order:

city_econ.sort_values(by="economy", ascending=False)

Saving and loading¶

pandas can save DataFrames to various backends, including file formats such as CSV, Excel, JSON, HTML and HDF5, or to a SQL database. Let's create a DataFrame to demonstrate this:

my_df = pd.DataFrame(
    [["Biking", 68.5, 1985, np.nan], ["Dancing", 83.1, 1984, 3]], 
    columns=["hobby","weight","birthyear","children"],
    index=["alice", "bob"]
)
my_df

Saving¶

Let's save it as a CSV file.

my_df.to_csv("my_df.csv")

Loading¶

Now let's load our CSV file back into a DataFrame:

my_df_loaded = pd.read_csv("my_df.csv", index_col=0)
my_df_loaded

As you might guess, there are similar read_json, read_html, read_excel functions as well. We can also read data straight from the Internet. For example, let's load all New York State Zip codes from data.ny.gov:

ny_zip = None
try:
    csv_url = "https://data.ny.gov/api/views/juva-r6g2/rows.csv"
    ny_zip = pd.read_csv(csv_url, index_col=0)
    ny_zip = ny_zip.head()
except IOError as e:
    print(e)
ny_zip

There are more options available, in particular regarding datetime format. Check out the documentation for more details.

Exercises¶

Create a Series using the following number: 3.14, 2.718, 1.618 with the following labels "pi", "euler's number", "golden ratio". Then filter values that are only greater than 2.

Create the following DataFrame (But use name as the index):

name	age	state	num_children	num_pets
john	23	iowa	2	0
mary	78	dc	2	4
peter	22	california	0	0
jeff	19	texas	1	5
bill	45	washington	2	0
lisa	33	dc	1	0

Then, create a bar plot that shows age for each person in name.

Add another person as a new row to the previous DataFrame with the following values (HINT: use pd.concat):

name: mike, age: 0, state: new york, num_children: 1, num_pets: 0.

Since this new person has a child, his age cannot be zero. Replace it with the median age of all other people in the DataFrame.

Possible solutions¶

Indexing and selecting in Series

s = pd.Series([3.14, 2.718, 1.618], index = ["pi", "euler's number", "golden ratio"])

s[s > 2]

Creating and plotting a DataFrame

df1 = pd.DataFrame(data={'age':[23,78,22,19,45,33],'state':['iowa','dc','california','texas','washington','dc'],'num_children':[2,2,0,1,2,1],'num_pets':[0,4,0,5,0,0]},index=['john','mary','peter','jeff','bill','lisa'])

df1.plot(kind = "bar", y = "age")

Adding new row and handling missing data

df2 = pd.DataFrame(data={'age':[0],'state':['new york'],'num_children':[1],'num_pets':[0]},index=['mike'])
df3 = pd.concat([df1,df2]) 
df2['age'] = df1['age'].median() # Calculate and add the median age of all other people to 'mike'
df3 = pd.concat([df1,df2]) # Update the DataFrame

References¶

www.featureranking.com

	sep	oct	nov
alice	2.828427	2.828427	3.000000
bob	3.162278	3.000000	3.000000
charles	2.000000	2.828427	1.414214
darwin	3.000000	3.162278	3.162278

	sep	oct	nov
alice	0.25	-0.75	1.5
bob	2.25	0.25	1.5
charles	-3.75	-0.75	-5.5
darwin	1.25	1.25	2.5

	A	B	C	some_text	D	E	F	G	H	I	...	Q	R	S	T	U	V	W	X	Y	Z
0	NaN	11.0	44.0	Blabla	99.0	NaN	88.0	22.0	165.0	143.0	...	11.0	NaN	11.0	44.0	99.0	NaN	88.0	22.0	165.0	143.0
1	11.0	22.0	55.0	Blabla	110.0	NaN	99.0	33.0	NaN	154.0	...	22.0	11.0	22.0	55.0	110.0	NaN	99.0	33.0	NaN	154.0
2	22.0	33.0	66.0	Blabla	121.0	11.0	110.0	44.0	NaN	165.0	...	33.0	22.0	33.0	66.0	121.0	11.0	110.0	44.0	NaN	165.0
3	33.0	44.0	77.0	Blabla	132.0	22.0	121.0	55.0	11.0	NaN	...	44.0	33.0	44.0	77.0	132.0	22.0	121.0	55.0	11.0	NaN
4	44.0	55.0	88.0	Blabla	143.0	33.0	132.0	66.0	22.0	NaN	...	55.0	44.0	55.0	88.0	143.0	33.0	132.0	66.0	22.0	NaN
5	55.0	66.0	99.0	Blabla	154.0	44.0	143.0	77.0	33.0	11.0	...	66.0	55.0	66.0	99.0	154.0	44.0	143.0	77.0	33.0	11.0
6	66.0	77.0	110.0	Blabla	165.0	55.0	154.0	88.0	44.0	22.0	...	77.0	66.0	77.0	110.0	165.0	55.0	154.0	88.0	44.0	22.0
7	77.0	88.0	121.0	Blabla	NaN	66.0	165.0	99.0	55.0	33.0	...	88.0	77.0	88.0	121.0	NaN	66.0	165.0	99.0	55.0	33.0
8	88.0	99.0	132.0	Blabla	NaN	77.0	NaN	110.0	66.0	44.0	...	99.0	88.0	99.0	132.0	NaN	77.0	NaN	110.0	66.0	44.0
9	99.0	110.0	143.0	Blabla	11.0	88.0	NaN	121.0	77.0	55.0	...	110.0	99.0	110.0	143.0	11.0	88.0	NaN	121.0	77.0	55.0

	A	B	C	some_text	D	E	F	G	H	I	...	Q	R	S	T	U	V	W	X	Y	Z
0	NaN	11.0	44.0	Blabla	99.0	NaN	88.0	22.0	165.0	143.0	...	11.0	NaN	11.0	44.0	99.0	NaN	88.0	22.0	165.0	143.0
1	11.0	22.0	55.0	Blabla	110.0	NaN	99.0	33.0	NaN	154.0	...	22.0	11.0	22.0	55.0	110.0	NaN	99.0	33.0	NaN	154.0
2	22.0	33.0	66.0	Blabla	121.0	11.0	110.0	44.0	NaN	165.0	...	33.0	22.0	33.0	66.0	121.0	11.0	110.0	44.0	NaN	165.0
3	33.0	44.0	77.0	Blabla	132.0	22.0	121.0	55.0	11.0	NaN	...	44.0	33.0	44.0	77.0	132.0	22.0	121.0	55.0	11.0	NaN
4	44.0	55.0	88.0	Blabla	143.0	33.0	132.0	66.0	22.0	NaN	...	55.0	44.0	55.0	88.0	143.0	33.0	132.0	66.0	22.0	NaN

	A	B	C	some_text	D	E	F	G	H	I	...	Q	R	S	T	U	V	W	X	Y	Z
9998	22.0	33.0	66.0	Blabla	121.0	11.0	110.0	44.0	NaN	165.0	...	33.0	22.0	33.0	66.0	121.0	11.0	110.0	44.0	NaN	165.0
9999	33.0	44.0	77.0	Blabla	132.0	22.0	121.0	55.0	11.0	NaN	...	44.0	33.0	44.0	77.0	132.0	22.0	121.0	55.0	11.0	NaN

	state	year	pop
0	Ohio	2000	1.5
1	Ohio	2001	1.7
2	Ohio	2002	3.6
3	Nevada	2001	2.4
4	Nevada	2002	2.9
5	Nevada	2003	3.2

	state	year	pop
0	Ohio	2000	1.5
1	Ohio	2001	1.7
2	Ohio	2002	3.6
3	Nevada	2001	2.4
4	Nevada	2002	2.9

	sep	oct	nov
alice	0.0	0.0	1.0
bob	2.0	1.0	1.0
charles	-4.0	0.0	-6.0
darwin	1.0	2.0	2.0

	oct	nov	dec
bob	0.0	NaN	2.0
colin	NaN	1.0	0.0
darwin	0.0	1.0	0.0
charles	3.0	3.0	0.0