Another solution to the distance to zero problem. This uses a group by. I think this one is more clear than the other two, but that could certainly just be to me.

Really enjoyed your puzzles, so I tried my hand at making some. Let me know if there's anything I can change, or if you believe there is some other pandas functionality that's deserving of puzzles as I'm happy to come up with more.

Hi Alex,

As a quest to better learn pandas I created a series of exercises in a different form than yours. I would like to know if you might to be interested to contribute in any way to my repo or if I can use the your exercises.

Thanks

Implemented the fix suggested by @Arten013 in the issue #17

This consists in a single, small change of the solutions notebook : in the second solution of exercice 29, replace x = (df1['X'] != 0).cumsum() y = x != x.shift() by y = df['X'] != 0

The solution below (Q29-2) output wrong answer when I input dataframe whose value starts with zero.

x = (df['X'] != 0).cumsum()
y = x != x.shift()
df['Y'] = y.groupby((y != y.shift()).cumsum()).cumsum()

In this code, Series y has to have True where its value is not zero and False otherwise. However, the first value of y become True in any case.

e.g.

df1 = pd.DataFrame({'X': [0, 2, 0, 3]})
df2 = pd.DataFrame({'X': [1, 2, 0, 3]})

x = (df1['X'] != 0).cumsum()
y = x != x.shift()
print(y[0])

x = (df2['X'] != 0).cumsum()
y = x != x.shift()
print(y[0])

outputs

True
True

This bug can be fixed by replacing first two lines into y = df['X'] != 0

Here's the code to compare the results between the solution 1 , solution 2 and modified solution2.

import pandas as pd
import numpy as np
df = pd.DataFrame({'X': [0, 2, 0, 3, 4, 2, 5, 0, 3, 4]})

def solution1(df):
    izero = np.r_[-1, (df['X'] == 0).nonzero()[0]] # indices of zeros
    idx = np.arange(len(df))
    return pd.Series(idx - izero[np.searchsorted(izero - 1, idx) - 1])

def solution2(df):
    x = (df['X'] != 0).cumsum()
    y = x != x.shift()
    return y.groupby((y != y.shift()).cumsum()).cumsum()

def solution2_modified(df):
    y = df['X'] != 0
    return y.groupby((y != y.shift()).cumsum()).cumsum()

check_df = pd.concat([df, solution1(df), solution2(df), solution2_modified(df)], axis=1)
check_df.columns = ['input_df', 'solution1', 'solution2', 'solution2_modified']
display(check_df)

|input_df|solution1|solution2|solution2_modified| |-------:|--------:|--------:|-----------------:| | 0| 0| 1| 0| | 2| 1| 2| 1| | 0| 0| 0| 0| | 3| 1| 1| 1| | 4| 2| 2| 2| | 2| 3| 3| 3| | 5| 4| 4| 4| | 0| 0| 0| 0| | 3| 1| 1| 1| | 4| 2| 2| 2|

I executed these code with Python 3.6.7 & pandas 0.24.0.

It would be nice if these exercises would have a license, so one knows under which conditions one can make use of them.

I don't have any particular license in mind myself, and of course that's not my call to make, tough in the name of reducing license proliferation I would suggest to use the same license as pandas itself uses: https://github.com/pandas-dev/pandas/blob/master/LICENSE .

27. A DataFrame has a column of groups 'grps' and and column of numbers 'vals'. For each group, find the sum of the three greatest values.

The solution starts with sorting the 'vals' column - this is not needed. The nlargest method selects the three greatest values irrespective of the order of element.

Suggestion: delete the sorting, the solution is provided by just by the second line of code.

Thanks for the project. When i working with question#21 using pandas1.2.4. It needs to fillna first.

df['age'] = df['age'].fillna(0)
df.pivot_table(index='animal', columns='visits', values='age', aggfunc='mean')

Question 20.1 added to convert age and visits column to float and int respectively. Having object datatype for age and visits causing issue in question no 21. creating pivot table

Added an alternate version of q20, it is a little longer than just .replace but it showcases how you could use lambda and apply in that situation. Let me know what you think.