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2018/11/8 100-pandas-puzzles-with-solutions
http://localhost:8889/notebooks/100-pandas-puzzles-with-solutions.ipynb 1/26
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 np2018/11/8 100-pandas-puzzles-with-solutions
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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__2018/11/8 100-pandas-puzzles-with-solutions
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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()2018/11/8 100-pandas-puzzles-with-solutions
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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)2018/11/8 100-pandas-puzzles-with-solutions
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In[5]:
6. Return the first 3 rows of the DataFrame df .
In?[6]:

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)2018/11/8 100-pandas-puzzles-with-solutions
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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']]2018/11/8 100-pandas-puzzles-with-solutions
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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)]2018/11/8 100-pandas-puzzles-with-solutions
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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()2018/11/8 100-pandas-puzzles-with-solutions
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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])2018/11/8 100-pandas-puzzles-with-solutions
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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')2018/11/8 100-pandas-puzzles-with-solutions
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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)2018/11/8 100-pandas-puzzles-with-solutions
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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)2018/11/8 100-pandas-puzzles-with-solutions
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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] += 12018/11/8 100-pandas-puzzles-with-solutions
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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: unutbu2018/11/8 100-pandas-puzzles-with-solutions
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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)2018/11/8 100-pandas-puzzles-with-solutions
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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(!)', ' (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')2018/11/8 100-pandas-puzzles-with-solutions
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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 spaces2018/11/8 100-pandas-puzzles-with-solutions
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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.nlevels2018/11/8 100-pandas-puzzles-with-solutions
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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()2018/11/8 100-pandas-puzzles-with-solutions
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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
In[]:
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)2018/11/8 100-pandas-puzzles-with-solutions
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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()2018/11/8 100-pandas-puzzles-with-solutions
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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")2018/11/8 100-pandas-puzzles-with-solutions
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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))2018/11/8 100-pandas-puzzles-with-solutions
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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.2018/11/8 100-pandas-puzzles-with-solutions
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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 r

辅导pandas-puzzles、讲解Python程序设计、辅导Cython留学生、讲解Python编程 辅导留学生 Statistics统计、

辅导pandas-puzzles、讲解Python程序设计、辅导Cython留学生、讲解Python编程辅导留学生 Statistics统计、