Pandas change the dataframe shape column values to row - python-3.x

Question

I have a dataframe which is looking as below
df
school1 game1 game2 game3
school2 game1
school3 game2 game3
school4 game2
output
game1 school1 school2
game2 school1 school4 school3
game3 school1 school3
can any one suggest me how can I get the out put i am new to pandas please help me
Thank you

Answer 1

Here is one way that relies on the melt() method to first make a long table out of the original and then on the pivot() method to transform it to the new wide format:
import pandas as pd
import numpy as np
# Code that creates your input dataframe (replace column names as needed)
df = pd.DataFrame(
{'school':['school1', 'school2', 'school3', 'school4'],
'g1':['game1', 'game1', 'game2', None],
'g2':['game2', None, 'game3', 'game2'],
'g3':['game3', None, None, None],
}
)
# Convert to long format (one row per school-game)
long_df = df.set_index('school').melt(ignore_index=False).reset_index()
# Remove null (non-existing) school-game combinations
# Also, convert index to column for next step
long_df = long_df[long_df.value.notnull()].reset_index(drop=True).reset_index()
# Convert to dataframe with one row per game ID
by_game_df = long_df.pivot(index='value',columns='index',values='school')
At this point, the dataframe will look like this:
index value 0 1 2 3 4 5 6
0 game1 school1 school2 NaN NaN NaN NaN NaN
1 game2 NaN NaN school3 school1 NaN school4 NaN
2 game3 NaN NaN NaN NaN school3 NaN school1
You can perform these additional steps to shift non-null school values to the left and to remove columns with only NaN's remaining:
# per https://stackoverflow.com/a/65596853:
idx = pd.isnull(by_game_df.values).argsort(axis=1)
squeezed_df = pd.DataFrame(
by_game_df.values[np.arange(by_game_df.shape[0])[:,None], idx],
index=by_game_df.index,
columns=by_game_df.columns
)
result = squeezed_df.dropna(axis=1, how='all')
result
# index value 0 1 2
# 0 game1 school1 school2 NaN
# 1 game2 school3 school1 school4
# 2 game3 school3 school1 NaN

Answer 2

Or with a Series of lists and a much maligned loop:
d = {'School': ['s1','s2','s3','s4'], 'c1': ['g1','g1','g2',np.nan], 'c2': ['g2',np.nan,'g3','g2'], 'c3': ['g3',np.nan,np.nan,np.nan]}
df = pd.DataFrame(d)
df
School c1 c2 c3
0 s1 g1 g2 g3
1 s2 g1 NaN NaN
2 s3 g2 g3 NaN
3 s4 NaN g2 NaN
gg = pd.Series(dtype=object)
def add_gs(game, sch):
if type(game) is str:
if game in gg.keys():
gg[game] += [sch]
else:
gg[game] = [sch]
cols = df.filter(regex='c[0-9]').columns
for i in range(len(df)):
for col in cols:
add_gs(df.loc[i,col],df.loc[i,'School'])
gg
g1 [s1, s2]
g2 [s1, s3, s4]
g3 [s1, s3]

Answer 3

A solution that relies on defaultdict() for reshaping the data:
from collections import defaultdict
import pandas as pd
# Code that creates your input dataframe (replace column names as needed)
df = pd.DataFrame(
{'school':['school1', 'school2', 'school3', 'school4'],
'g1':['game1', 'game1', 'game2', None],
'g2':['game2', None, 'game3', 'game2'],
'g3':['game3', None, None, None],
}
)
# convert df to dictionary
d = df.set_index('school').to_dict(orient='index')
# reshape the dictionary
def_d = defaultdict(list)
for k, v in d.items():
for i in v.values():
if i is not None:
def_d[i].append(k)
d_rs = dict(def_d)
# prepare dictionary for converting back to dataframe
dict_for_df = {
k: pd.Series(
v + [None] * (len(max(d_rs.values(), key=lambda x: len(x))) - len(v))
) for k, v in d_rs.items()
}
# convert dictionary to dataframe
final_df = pd.DataFrame.from_dict(dict_for_df, orient='index')
}
# 0 1 2
# game1 school1 school2 None
# game2 school1 school3 school4
# game3 school1 school3 None