代写pandas-puzzles、代做Python程序设计、代写Cython留学生、代做Python编程

日期：2018-12-13 10:10

2018/11/8 100-pandas-puzzles-with-solutions

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100 pandas puzzles

Inspired by 100 Numpy exerises (https://github.com/rougier/numpy-100), here are 100* short puzzles for

testing your knowledge of pandas' (http://pandas.pydata.org/) power.

Since pandas is a large library with many different specialist features and functions, these excercises focus

mainly on the fundamentals of manipulating data (indexing, grouping, aggregating, cleaning), making use of the

core DataFrame and Series objects.

Many of the excerises here are stright-forward in that the solutions require no more than a few lines of code (in

pandas or NumPy... don't go using pure Python or Cython!). Choosing the right methods and following best

practices is the underlying goal.

The exercises are loosely divided in sections. Each section has a difficulty rating; these ratings are subjective,

of course, but should be a seen as a rough guide as to how inventive the required solution is.

If you're just starting out with pandas and you are looking for some other resources, the official documentation

is very extensive. In particular, some good places get a broader overview of pandas are...

10 minutes to pandas (http://pandas.pydata.org/pandas-docs/stable/10min.html)

pandas basics (http://pandas.pydata.org/pandas-docs/stable/basics.html)

tutorials (http://pandas.pydata.org/pandas-docs/stable/tutorials.html)

cookbook and idioms (http://pandas.pydata.org/pandas-docs/stable/cookbook.html#cookbook)

Enjoy the puzzles!

* the list of exercises is not yet complete! Pull requests or suggestions for additional exercises, corrections and

improvements are welcomed.

Importing pandas

Getting started and checking your pandas setup

Difficulty: easy

1. Import pandas under the name pd .

In[1]:

2. Print the version of pandas that has been imported.

import pandas as pd

import numpy as np

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In[2]:

3. Print out all the version information of the libraries that are required by the pandas library.

Out[2]:

'0.23.4'

pd.__version__

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In[3]:

DataFrame basics

INSTALLED VERSIONS

------------------

commit: None

python: 3.6.6.final.0

python-bits: 64

OS: Darwin

OS-release: 17.7.0

machine: x86_64

processor: i386

byteorder: little

LC_ALL: None

LANG: zh_CN.UTF-8

LOCALE: zh_CN.UTF-8

pandas: 0.23.4

pytest: None

pip: 18.1

setuptools: 39.1.0

Cython: None

numpy: 1.14.5

scipy: 1.1.0

pyarrow: None

xarray: None

IPython: 6.5.0

sphinx: None

patsy: None

dateutil: 2.7.3

pytz: 2018.5

blosc: None

bottleneck: None

tables: None

numexpr: None

feather: None

matplotlib: 2.2.2

openpyxl: None

xlrd: None

xlwt: None

xlsxwriter: None

lxml: None

bs4: None

html5lib: 1.0.1

sqlalchemy: None

pymysql: None

psycopg2: None

jinja2: 2.10

s3fs: None

fastparquet: None

pandas_gbq: None

pandas_datareader: None

pd.show_versions()

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A few of the fundamental routines for selecting, sorting, adding and aggregating

data in DataFrames

Difficulty: easy

Note: remember to import numpy using:

import numpy as np

Consider the following Python dictionary data and Python list labels :

data = {'animal': ['cat', 'cat', 'snake', 'dog', 'dog', 'cat', 'snake', 'ca

t', 'dog', 'dog'],

'age': [2.5, 3, 0.5, np.nan, 5, 2, 4.5, np.nan, 7, 3],

'visits': [1, 3, 2, 3, 2, 3, 1, 1, 2, 1],

'priority': ['yes', 'yes', 'no', 'yes', 'no', 'no', 'no', 'yes', 'n

o', 'no']}

labels = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

(This is just some meaningless data I made up with the theme of animals and trips to a vet.)

4. Create a DataFrame df from this dictionary data which has the index labels .

In[2]:

5. Display a summary of the basic information about this DataFrame and its data.

data = {'animal': ['cat', 'cat', 'snake', 'dog', 'dog', 'cat', 'snake', 'cat', 'dog

'age': [2.5, 3, 0.5, np.nan, 5, 2, 4.5, np.nan, 7, 3],

'visits': [1, 3, 2, 3, 2, 3, 1, 1, 2, 1],

'priority': ['yes', 'yes', 'no', 'yes', 'no', 'no', 'no', 'yes', 'no', 'no']

labels = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

df = pd.DataFrame(data, index=labels)

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In[5]:

6. Return the first 3 rows of the DataFrame df .

In?[6]:

<class 'pandas.core.frame.DataFrame'>

Index: 10 entries, a to j

Data columns (total 4 columns):

animal 10 non-null object

age 8 non-null float64

visits 10 non-null int64

priority 10 non-null object

dtypes: float64(1), int64(1), object(2)

memory usage: 400.0+ bytes

Out[5]:

age visits

count 8.000000 10.000000

mean 3.437500 1.900000

std 2.007797 0.875595

min 0.500000 1.000000

25% 2.375000 1.000000

50% 3.000000 2.000000

75% 4.625000 2.750000

max 7.000000 3.000000

Out[6]:

animal age visits priority

a cat 2.5 1 yes

b cat 3.0 3 yes

c snake 0.5 2 no

df.info()

# ...or...

df.describe()

df.iloc[:3]

# or equivalently

df.head(3)

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7. Select just the 'animal' and 'age' columns from the DataFrame df .

In?[7]:

8. Select the data in rows [3, 4, 8] and in columns ['animal', 'age'] .

In?[3]:

9. Select only the rows where the number of visits is greater than 3.

Out[7]:

animal age

a cat 2.5

b cat 3.0

c snake 0.5

d dog NaN

e dog 5.0

f cat 2.0

g snake 4.5

h cat NaN

i dog 7.0

j dog 3.0

Out[3]:

animal age

d dog NaN

e dog 5.0

i dog 7.0

df.loc[:, ['animal', 'age']]

# or

df[['animal', 'age']]

df.loc[df.index[[3, 4, 8]], ['animal', 'age']]

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In[4]:

10. Select the rows where the age is missing, i.e. is NaN .

In[5]:

11. Select the rows where the animal is a cat and the age is less than 3.

In[6]:

12. Select the rows the age is between 2 and 4 (inclusive).

Out[4]:

animal age visits priority

Out[5]:

animal age visits priority

d dog NaN 3 yes

h cat NaN 1 yes

Out[6]:

animal age visits priority

a cat 2.5 1 yes

f cat 2.0 3 no

df[df['visits'] > 3]

df[df['age'].isnull()]

df[(df['animal'] == 'cat') & (df['age'] < 3)]

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In[7]:

13. Change the age in row 'f' to 1.5.

In[]:

14. Calculate the sum of all visits (the total number of visits).

In[]:

15. Calculate the mean age for each different animal in df .

In[8]:

16. Append a new row 'k' to df with your choice of values for each column. Then delete that row to return the

original DataFrame.

Out[7]:

animal age visits priority

a cat 2.5 1 yes

b cat 3.0 3 yes

f cat 2.0 3 no

j dog 3.0 1 no

Out[8]:

animal

cat 2.5

dog 5.0

snake 2.5

Name: age, dtype: float64

df[df['age'].between(2, 4)]

df.loc['f', 'age'] = 1.5

df['visits'].sum()

df.groupby('animal')['age'].mean()

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In[]:

17. Count the number of each type of animal in df .

In[9]:

18. Sort df first by the values in the 'age' in decending order, then by the value in the 'visit' column in

ascending order.

In[10]:

19. The 'priority' column contains the values 'yes' and 'no'. Replace this column with a column of boolean

values: 'yes' should be True and 'no' should be False .

Out[9]:

cat 4

dog 4

snake 2

Name: animal, dtype: int64

Out[10]:

animal age visits priority

i dog 7.0 2 no

e dog 5.0 2 no

g snake 4.5 1 no

j dog 3.0 1 no

b cat 3.0 3 yes

a cat 2.5 1 yes

f cat 2.0 3 no

c snake 0.5 2 no

h cat NaN 1 yes

d dog NaN 3 yes

df.loc['k'] = [5.5, 'dog', 'no', 2]

# and then deleting the new row...

df = df.drop('k')

df['animal'].value_counts()

df.sort_values(by=['age', 'visits'], ascending=[False, True])

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In?[?]:

20. In the 'animal' column, change the 'snake' entries to 'python'.

In?[14]:

21. For each animal type and each number of visits, find the mean age. In other words, each row is an animal,

each column is a number of visits and the values are the mean ages (hint: use a pivot table).

In[15]:

DataFrames: beyond the basics

Slightly trickier: you may need to combine two or more methods to get the right

answer

Difficulty: medium

The previous section was tour through some basic but essential DataFrame operations. Below are some ways

that you might need to cut your data, but for which there is no single "out of the box" method.

animal age visits priority

a cat 2.5 1 yes

b cat 3.0 3 yes

c python 0.5 2 no

d dog NaN 3 yes

e dog 5.0 2 no

f cat 2.0 3 no

g python 4.5 1 no

h cat NaN 1 yes

i dog 7.0 2 no

j dog 3.0 1 no

Out[15]:

visits 1 2 3

animal

cat 2.5 NaN 2.5

dog 3.0 6.0 NaN

python 4.5 0.5 NaN

df['priority'] = df['priority'].map({'yes': True, 'no': False})

df['animal'] = df['animal'].replace('snake', 'python')

print(df)

df.pivot_table(index='animal', columns='visits', values='age', aggfunc='mean')

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22. You have a DataFrame df with a column 'A' of integers. For example:

df = pd.DataFrame({'A': [1, 2, 2, 3, 4, 5, 5, 5, 6, 7, 7]})

How do you filter out rows which contain the same integer as the row immediately above?

In[16]:

23. Given a DataFrame of numeric values, say

df = pd.DataFrame(np.random.random(size=(5, 3))) # a 5x3 frame of float valu

es

how do you subtract the row mean from each element in the row?

In[]:

24. Suppose you have DataFrame with 10 columns of real numbers, for example:

df = pd.DataFrame(np.random.random(size=(5, 10)), columns=list('abcdefghij'

))

Which column of numbers has the smallest sum? (Find that column's label.)

Out[16]:

A

0 1

1 2

3 3

4 4

5 5

8 6

9 7

df = pd.DataFrame({'A': [1, 2, 2, 3, 4, 5, 5, 5, 6, 7, 7]})

df.loc[df['A'].shift() != df['A']]

# Alternatively, we could use drop_duplicates() here. Note

# that this removes *all* duplicates though, so it won't

# work as desired if A is [1, 1, 2, 2, 1, 1] for example.

df.drop_duplicates(subset='A')

df.sub(df.mean(axis=1), axis=0)

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In[17]:

25. How do you count how many unique rows a DataFrame has (i.e. ignore all rows that are duplicates)?

In[]:

The next three puzzles are slightly harder...

26. You have a DataFrame that consists of 10 columns of floating--point numbers. Suppose that exactly 5

entries in each row are NaN values. For each row of the DataFrame, find the column which contains the third

NaN value.

(You should return a Series of column labels.)

In[]:

27. A DataFrame has a column of groups 'grps' and and column of numbers 'vals'. For example:

df = pd.DataFrame({'grps': list('aaabbcaabcccbbc'),

'vals': [12,345,3,1,45,14,4,52,54,23,235,21,57,3,87]})

For each group, find the sum of the three greatest values.

In?[?]:

28. A DataFrame has two integer columns 'A' and 'B'. The values in 'A' are between 1 and 100 (inclusive). For

each group of 10 consecutive integers in 'A' (i.e. (0, 10] , (10, 20] , ...), calculate the sum of the

corresponding values in column 'B'.

Out[17]:

'A'

df.sum().idxmin()

len(df) - df.duplicated(keep=False).sum()

# or perhaps more simply...

len(df.drop_duplicates(keep=False))

(df.isnull().cumsum(axis=1) == 3).idxmax(axis=1)

df.groupby('grp')['vals'].nlargest(3).sum(level=0)

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In[]:

DataFrames: harder problems

These might require a bit of thinking outside the box...

...but all are solvable using just the usual pandas/NumPy methods (and so avoid using explicit for loops).

Difficulty: hard

29. Consider a DataFrame df where there is an integer column 'X':

df = pd.DataFrame({'X': [7, 2, 0, 3, 4, 2, 5, 0, 3, 4]})

For each value, count the difference back to the previous zero (or the start of the Series, whichever is closer).

These values should therefore be [1, 2, 0, 1, 2, 3, 4, 0, 1, 2] . Make this a new column 'Y'.

In[]:

Here's an alternative approach based on a cookbook recipe (http://pandas.pydata.org/pandasdocs/stable/cookbook.html#grouping):

In[]:

And another approach using a groupby:

In[]:

30. Consider a DataFrame containing rows and columns of purely numerical data. Create a list of the rowdf.groupby(pd.cut(df['A'],

np.arange(0, 101, 10)))['B'].sum()

izero = np.r_[-1, (df['X'] == 0).nonzero()[0]] # indices of zeros

idx = np.arange(len(df))

df['Y'] = idx - izero[np.searchsorted(izero - 1, idx) - 1]

# http://stackoverflow.com/questions/30730981/how-to-count-distance-to-the-previous-

# credit: Behzad Nouri

x = (df['X'] != 0).cumsum()

y = x != x.shift()

df['Y'] = y.groupby((y != y.shift()).cumsum()).cumsum()

df['Y'] = df.groupby((df['X'] == 0).cumsum()).cumcount()

# We're off by one before we reach the first zero.

first_zero_idx = (df['X'] == 0).idxmax()

df['Y'].iloc[0:first_zero_idx] += 1

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column index locations of the 3 largest values.

In[]:

31. Given a DataFrame with a column of group IDs, 'grps', and a column of corresponding integer values,

'vals', replace any negative values in 'vals' with the group mean.

In[]:

32. Implement a rolling mean over groups with window size 3, which ignores NaN value. For example consider

the following DataFrame:

>>> df = pd.DataFrame({'group': list('aabbabbbabab'),

'value': [1, 2, 3, np.nan, 2, 3,

np.nan, 1, 7, 3, np.nan, 8]})

>>> df

group value

0 a 1.0

1 a 2.0

2 b 3.0

3 b NaN

4 a 2.0

5 b 3.0

6 b NaN

7 b 1.0

8 a 7.0

9 b 3.0

10 a NaN

11 b 8.0

The goal is to compute the Series:

df.unstack().sort_values()[-3:].index.tolist()

# http://stackoverflow.com/questions/14941261/index-and-column-for-the-max-value-in-

# credit: DSM

def replace(group):

mask = group<0

group[mask] = group[~mask].mean()

return group

df.groupby(['grps'])['vals'].transform(replace)

# http://stackoverflow.com/questions/14760757/replacing-values-with-groupby-means/

# credit: unutbu

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0 1.000000

1 1.500000

2 3.000000

3 3.000000

4 1.666667

5 3.000000

6 3.000000

7 2.000000

8 3.666667

9 2.000000

10 4.500000

11 4.000000

E.g. the first window of size three for group 'b' has values 3.0, NaN and 3.0 and occurs at row index 5. Instead

of being NaN the value in the new column at this row index should be 3.0 (just the two non-NaN values are

used to compute the mean (3+3)/2)

In[]:

Series and DatetimeIndex

Exercises for creating and manipulating Series with datetime data

Difficulty: easy/medium

pandas is fantastic for working with dates and times. These puzzles explore some of this functionality.

33. Create a DatetimeIndex that contains each business day of 2015 and use it to index a Series of random

numbers. Let's call this Series s .

In?[?]:

34. Find the sum of the values in s for every Wednesday.

g1 = df.groupby(['group'])['value'] # group values

g2 = df.fillna(0).groupby(['group'])['value'] # fillna, then group values

s = g2.rolling(3, min_periods=1).sum() / g1.rolling(3, min_periods=1).count() # comp

s.reset_index(level=0, drop=True).sort_index() # drop/sort index

# http://stackoverflow.com/questions/36988123/pandas-groupby-and-rolling-apply-ignor

dti = pd.date_range(start='2015-01-01', end='2015-12-31', freq='B')

s = pd.Series(np.random.rand(len(dti)), index=dti)

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In[]:

35. For each calendar month in s , find the mean of values.

In[]:

36. For each group of four consecutive calendar months in s , find the date on which the highest value

occurred.

In[]:

37. Create a DateTimeIndex consisting of the third Thursday in each month for the years 2015 and 2016.

In]:

Cleaning Data

Making a DataFrame easier to work with

Difficulty: easy/medium

It happens all the time: someone gives you data containing malformed strings, Python, lists and missing data.

How do you tidy it up so you can get on with the analysis?

Take this monstrosity as the DataFrame to use in the following puzzles:

df = pd.DataFrame({'From_To': ['LoNDon_paris', 'MAdrid_miLAN', 'londON_Stock

hOlm',

'Budapest_PaRis', 'Brussels_londOn'],

'FlightNumber': [10045, np.nan, 10065, np.nan, 10085],

'RecentDelays': [[23, 47], [], [24, 43, 87], [13], [67, 32]],

'Airline': ['KLM(!)', '<Air France> (12)', '(British Airw

ays. )',

'12. Air France', '"Swiss Air"']})

(It's some flight data I made up; it's not meant to be accurate in any way.)

38. Some values in the the FlightNumber column are missing. These numbers are meant to increase by 10 with

s[s.index.weekday == 2].sum()

s.resample('M').mean()

s.groupby(pd.TimeGrouper('4M')).idxmax()

pd.date_range('2015-01-01', '2016-12-31', freq='WOM-3THU')

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each row so 10055 and 10075 need to be put in place. Fill in these missing numbers and make the column an

integer column (instead of a float column).

In[]:

39. The From_To column would be better as two separate columns! Split each string on the underscore

delimiter _ to give a new temporary DataFrame with the correct values. Assign the correct column names to

this temporary DataFrame.

In[]:

40. Notice how the capitalisation of the city names is all mixed up in this temporary DataFrame. Standardise

the strings so that only the first letter is uppercase (e.g. "londON" should become "London".)

In[]:

41. Delete the From_To column from df and attach the temporary DataFrame from the previous questions.

In[]:

42. In the Airline column, you can see some extra puctuation and symbols have appeared around the airline

names. Pull out just the airline name. E.g. '(British Airways. )' should become 'British

Airways' .

In[]:

43. In the RecentDelays column, the values have been entered into the DataFrame as a list. We would like each

first value in its own column, each second value in its own column, and so on. If there isn't an Nth value, the

value should be NaN.

Expand the Series of lists into a DataFrame named delays , rename the columns delay_1 , delay_2 ,

etc. and replace the unwanted RecentDelays column in df with delays .

df['FlightNumber'] = df['FlightNumber'].interpolate().astype(int)

temp = df.From_To.str.split('_', expand=True)

temp.columns = ['From', 'To']

temp['From'] = temp['From'].str.capitalize()

temp['To'] = temp['To'].str.capitalize()

df = df.drop('From_To', axis=1)

df = df.join(temp)

df['Airline'] = df['Airline'].str.extract('([a-zA-Z\s]+)', expand=False).str.strip()

# note: using .strip() gets rid of any leading/trailing spaces

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In[]:

The DataFrame should look much better now.

Using MultiIndexes

Go beyond flat DataFrames with additional index levels

Difficulty: medium

Previous exercises have seen us analysing data from DataFrames equipped with a single index level. However,

pandas also gives you the possibilty of indexing your data using multiple levels. This is very much like adding

new dimensions to a Series or a DataFrame. For example, a Series is 1D, but by using a MultiIndex with 2

levels we gain of much the same functionality as a 2D DataFrame.

The set of puzzles below explores how you might use multiple index levels to enhance data analysis.

To warm up, we'll look make a Series with two index levels.

44. Given the lists letters = ['A', 'B', 'C'] and numbers = list(range(10)) , construct a

MultiIndex object from the product of the two lists. Use it to index a Series of random numbers. Call this Series

s .

In[]:

45. Check the index of s is lexicographically sorted (this is a necessary proprty for indexing to work correctly

with a MultiIndex).

In[]:

# there are several ways to do this, but the following approach is possibly the simp

delays = df['RecentDelays'].apply(pd.Series)

delays.columns = ['delay_{}'.format(n) for n in range(1, len(delays.columns)+1)]

df = df.drop('RecentDelays', axis=1).join(delays)

letters = ['A', 'B', 'C']

numbers = list(range(10))

mi = pd.MultiIndex.from_product([letters, numbers])

s = pd.Series(np.random.rand(30), index=mi)

s.index.is_lexsorted()

# or more verbosely...

s.index.lexsort_depth == s.index.nlevels

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46. Select the labels 1 , 3 and 6 from the second level of the MultiIndexed Series.

In[]:

47. Slice the Series s ; slice up to label 'B' for the first level and from label 5 onwards for the second level.

In[]:

48. Sum the values in s for each label in the first level (you should have Series giving you a total for labels A,

B and C).

In[]:

49. Suppose that sum() (and other methods) did not accept a level keyword argument. How else could

you perform the equivalent of s.sum(level=1) ?

In[]:

50. Exchange the levels of the MultiIndex so we have an index of the form (letters, numbers). Is this new Series

properly lexsorted? If not, sort it.

In[]:

Minesweeper

s.loc[:, [1, 3, 6]]

s.loc[pd.IndexSlice[:'B', 5:]]

# or equivalently without IndexSlice...

s.loc[slice(None, 'B'), slice(5, None)]

s.sum(level=0)

# One way is to use .unstack()...

# This method should convince you that s is essentially

# just a regular DataFrame in disguise!

s.unstack().sum(axis=0)

new_s = s.swaplevel(0, 1)

# check

new_s.index.is_lexsorted()

# sort

new_s = new_s.sort_index()

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Generate the numbers for safe squares in a Minesweeper grid

Difficulty: medium to hard

If you've ever used an older version of Windows, there's a good chance you've played with [Minesweeper]

(https://en.wikipedia.org/wiki/Minesweeper_(video_game)

(https://en.wikipedia.org/wiki/Minesweeper_(video_game)). If you're not familiar with the game, imagine a grid

of squares: some of these squares conceal a mine. If you click on a mine, you lose instantly. If you click on a

safe square, you reveal a number telling you how many mines are found in the squares that are immediately

adjacent. The aim of the game is to uncover all squares in the grid that do not contain a mine.

In this section, we'll make a DataFrame that contains the necessary data for a game of Minesweeper:

coordinates of the squares, whether the square contains a mine and the number of mines found on adjacent

squares.

51. Let's suppose we're playing Minesweeper on a 5 by 4 grid, i.e.

X = 5

Y = 4

To begin, generate a DataFrame df with two columns, 'x' and 'y' containing every coordinate for this

grid. That is, the DataFrame should start:

x y

0 0 0

1 0 1

2 0 2

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52. For this DataFrame df , create a new column of zeros (safe) and ones (mine). The probability of a mine

occuring at each location should be 0.4.

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53. Now create a new column for this DataFrame called 'adjacent' . This column should contain the

number of mines found on adjacent squares in the grid.

(E.g. for the first row, which is the entry for the coordinate (0, 0) , count how many mines are found on the

coordinates (0, 1) , (1, 0) and (1, 1) .)

p = pd.tools.util.cartesian_product([np.arange(X), np.arange(Y)])

df = pd.DataFrame(np.asarray(p).T, columns=['x', 'y'])

# One way is to draw samples from a binomial distribution.

df['mine'] = np.random.binomial(1, 0.4, X*Y)

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54. For rows of the DataFrame that contain a mine, set the value in the 'adjacent' column to NaN.

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55. Finally, convert the DataFrame to grid of the adjacent mine counts: columns are the x coordinate, rows

are the y coordinate.

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Plotting

Visualize trends and patterns in data

Difficulty: medium

To really get a good understanding of the data contained in your DataFrame, it is often essential to create plots:

if you're lucky, trends and anomalies will jump right out at you. This functionality is baked into pandas and the

puzzles below explore some of what's possible with the library.

# Here is one way to solve using merges.

# It's not necessary the optimal way, just

# the solution I thought of first...

df['adjacent'] = \

df.merge(df + [ 1, 1, 0], on=['x', 'y'], how='left')\

.merge(df + [ 1, -1, 0], on=['x', 'y'], how='left')\

.merge(df + [-1, 1, 0], on=['x', 'y'], how='left')\

.merge(df + [-1, -1, 0], on=['x', 'y'], how='left')\

.merge(df + [ 1, 0, 0], on=['x', 'y'], how='left')\

.merge(df + [-1, 0, 0], on=['x', 'y'], how='left')\

.merge(df + [ 0, 1, 0], on=['x', 'y'], how='left')\

.merge(df + [ 0, -1, 0], on=['x', 'y'], how='left')\

.iloc[:, 3:]\

.sum(axis=1)

# An alternative solution is to pivot the DataFrame

# to form the "actual" grid of mines and use convolution.

# See https://github.com/jakevdp/matplotlib_pydata2013/blob/master/examples/mineswee

from scipy.signal import convolve2d

mine_grid = df.pivot_table(columns='x', index='y', values='mine')

counts = convolve2d(mine_grid.astype(complex), np.ones((3, 3)), mode='same').real.as

df['adjacent'] = (counts - mine_grid).ravel('F')

df.loc[df['mine'] == 1, 'adjacent'] = np.nan

df.drop('mine', axis=1)\

.set_index(['y', 'x']).unstack()

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56. Pandas is highly integrated with the plotting library matplotlib, and makes plotting DataFrames very userfriendly!

Plotting in a notebook environment usually makes use of the following boilerplate:

import matplotlib.pyplot as plt

%matplotlib inline

plt.style.use('ggplot')

matplotlib is the plotting library which pandas' plotting functionality is built upon, and it is usually aliased to

plt .

%matplotlib inline tells the notebook to show plots inline, instead of creating them in a separate

window.

plt.style.use('ggplot') is a style theme that most people find agreeable, based upon the styling of

R's ggplot package.

For starters, make a scatter plot of this random data, but use black X's instead of the default markers.

df = pd.DataFrame({"xs":[1,5,2,8,1], "ys":[4,2,1,9,6]})

Consult the documentation (https://pandas.pydata.org/pandasdocs/stable/generated/pandas.DataFrame.plot.html)

if you get stuck!

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57. Columns in your DataFrame can also be used to modify colors and sizes. Bill has been keeping track of his

performance at work over time, as well as how good he was feeling that day, and whether he had a cup of

coffee in the morning. Make a plot which incorporates all four features of this DataFrame.

(Hint: If you're having trouble seeing the plot, try multiplying the Series which you choose to represent size by

10 or more)

The chart doesn't have to be pretty: this isn't a course in data viz!

df = pd.DataFrame({"productivity":[5,2,3,1,4,5,6,7,8,3,4,8,9],

"hours_in" :[1,9,6,5,3,9,2,9,1,7,4,2,2],

"happiness" :[2,1,3,2,3,1,2,3,1,2,2,1,3],

"caffienated" :[0,0,1,1,0,0,0,0,1,1,0,1,0]})

import matplotlib.pyplot as plt

%matplotlib inline

plt.style.use('ggplot')

df = pd.DataFrame({"xs":[1,5,2,8,1], "ys":[4,2,1,9,6]})

df.plot.scatter("xs", "ys", color = "black", marker = "x")

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58. What if we want to plot multiple things? Pandas allows you to pass in a matplotlib Axis object for plots, and

plots will also return an Axis object.

Make a bar plot of monthly revenue with a line plot of monthly advertising spending (numbers in millions)

df = pd.DataFrame({"revenue":[57,68,63,71,72,90,80,62,59,51,47,52],

"advertising":[2.1,1.9,2.7,3.0,3.6,3.2,2.7,2.4,1.8,1.6,1.

3,1.9],

"month":range(12)

})

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Now we're finally ready to create a candlestick chart, which is a very common tool used to analyze stock price

data. A candlestick chart shows the opening, closing, highest, and lowest price for a stock during a time

window. The color of the "candle" (the thick part of the bar) is green if the stock closed above its opening price,

or red if below.

df = pd.DataFrame({"productivity":[5,2,3,1,4,5,6,7,8,3,4,8,9],

"hours_in" :[1,9,6,5,3,9,2,9,1,7,4,2,2],

"happiness" :[2,1,3,2,3,1,2,3,1,2,2,1,3],

"caffienated" :[0,0,1,1,0,0,0,0,1,1,0,1,0]})

df.plot.scatter("hours_in", "productivity", s = df.happiness * 30, c = df.caffienate

df = pd.DataFrame({"revenue":[57,68,63,71,72,90,80,62,59,51,47,52],

"advertising":[2.1,1.9,2.7,3.0,3.6,3.2,2.7,2.4,1.8,1.6,1.3,1.9],

"month":range(12)

})

ax = df.plot.bar("month", "revenue", color = "green")

df.plot.line("month", "advertising", secondary_y = True, ax = ax)

ax.set_xlim((-1,12))

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This was initially designed to be a pandas plotting challenge, but it just so happens that this type of plot is just

not feasible using pandas' methods. If you are unfamiliar with matplotlib, we have provided a function that will

plot the chart for you so long as you can use pandas to get the data into the correct format.

Your first step should be to get the data in the correct format using pandas' time-series grouping function. We

would like each candle to represent an hour's worth of data. You can write your own aggregation function

which returns the open/high/low/close, but pandas has a built-in which also does this.

The below cell contains helper functions. Call day_stock_data() to generate a DataFrame containing the

prices a hypothetical stock sold for, and the time the sale occurred. Call plot_candlestick(df) on your

properly aggregated and formatted stock data to print the candlestick chart.

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59. Generate a day's worth of random stock data, and aggregate / reformat it so that it has hourly summaries

of the opening, highest, lowest, and closing prices

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#This function is designed to create semi-interesting random stock price data

import numpy as np

def float_to_time(x):

return str(int(x)) + ":" + str(int(x%1 * 60)).zfill(2) + ":" + str(int(x*60 % 1

def day_stock_data():

#NYSE is open from 9:30 to 4:00

time = 9.5

price = 100

results = [(float_to_time(time), price)]

while time < 16:

elapsed = np.random.exponential(.001)

time += elapsed

if time > 16:

break

price_diff = np.random.uniform(.999, 1.001)

price *= price_diff

results.append((float_to_time(time), price))

df = pd.DataFrame(results, columns = ['time','price'])

df.time = pd.to_datetime(df.time)

return df

def plot_candlestick(agg):

fig, ax = plt.subplots()

for time in agg.index:

ax.plot([time.hour] * 2, agg.loc[time, ["high","low"]].values, color = "blac

ax.plot([time.hour] * 2, agg.loc[time, ["open","close"]].values, color = agg

ax.set_xlim((8,16))

ax.set_ylabel("Price")

ax.set_xlabel("Hour")

ax.set_title("OHLC of Stock Value During Trading Day")

plt.show()

df = day_stock_data()

df.head()

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60. Now that you have your properly-formatted data, try to plot it yourself as a candlestick chart. Use the

plot_candlestick(df) function above, or matplotlib's plot documentation

(https://matplotlib.org/api/_as_gen/matplotlib.axes.Axes.plot.html) if you get stuck.

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More exercises to follow soon...