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python - pandas: sort by the order of appearance

问题描述

Suppose we have a dataframe:

df = pd.DataFrame(pd.np.zeros((15,10,)), dtype=int, \
    index=[['a']*5+['b']*5+['c']*5, list(range(15))])
df.index.names=['index0', 'index1']
pd.np.random.seed(0)
for i, v in df.iterrows():
    v.loc[pd.np.random.randint(10)] = 1

df

               0  1  2  3  4  5  6  7  8  9
index0 index1                              
a      0       0  0  0  0  0  1  0  0  0  0
       1       1  0  0  0  0  0  0  0  0  0
       2       0  0  0  1  0  0  0  0  0  0
       3       0  0  0  1  0  0  0  0  0  0
       4       0  0  0  0  0  0  0  1  0  0
b      5       0  0  0  0  0  0  0  0  0  1
       6       0  0  0  1  0  0  0  0  0  0
       7       0  0  0  0  0  1  0  0  0  0
       8       0  0  1  0  0  0  0  0  0  0
       9       0  0  0  0  1  0  0  0  0  0
c      10      0  0  0  0  0  0  0  1  0  0
       11      0  0  0  0  0  0  1  0  0  0
       12      0  0  0  0  0  0  0  0  1  0
       13      0  0  0  0  0  0  0  0  1  0
       14      0  1  0  0  0  0  0  0  0  0

How to sort first the rows in the blocks a, b and c by the order of appearance of "1" and then sort also a, b and c?

Expected output:

               0  1  2  3  4  5  6  7  8  9
index0 index1                              
a      1       1  0  0  0  0  0  0  0  0  0
       2       0  0  0  1  0  0  0  0  0  0
       3       0  0  0  1  0  0  0  0  0  0
       0       0  0  0  0  0  1  0  0  0  0
       4       0  0  0  0  0  0  0  1  0  0
c      14      0  1  0  0  0  0  0  0  0  0
       11      0  0  0  0  0  0  1  0  0  0
       10      0  0  0  0  0  0  0  1  0  0
       12      0  0  0  0  0  0  0  0  1  0
       13      0  0  0  0  0  0  0  0  1  0
b      8       0  0  1  0  0  0  0  0  0  0
       6       0  0  0  1  0  0  0  0  0  0
       9       0  0  0  0  1  0  0  0  0  0
       7       0  0  0  0  0  1  0  0  0  0
       5       0  0  0  0  0  0  0  0  0  1

Edit: The values can be other than "1", in reality these are different text values.

标签： pythonpandassorting

解决方案

One approach is to use pandas.DataFrame.groupby with idxmax and sort_values:

import pandas as pd

l = (d.loc[d.idxmax(1).sort_values().index] for _, d in df.groupby('index0'))
new_df = pd.concat(sorted(l, key= lambda x:list(x.sum()), reverse=True))
print(new_df)

Output:

               0  1  2  3  4  5  6  7  8  9
index0 index1                              
a      1       1  0  0  0  0  0  0  0  0  0
       2       0  0  0  1  0  0  0  0  0  0
       3       0  0  0  1  0  0  0  0  0  0
       0       0  0  0  0  0  1  0  0  0  0
       4       0  0  0  0  0  0  0  1  0  0
c      14      0  1  0  0  0  0  0  0  0  0
       11      0  0  0  0  0  0  1  0  0  0
       10      0  0  0  0  0  0  0  1  0  0
       12      0  0  0  0  0  0  0  0  1  0
       13      0  0  0  0  0  0  0  0  1  0
b      8       0  0  1  0  0  0  0  0  0  0
       6       0  0  0  1  0  0  0  0  0  0
       9       0  0  0  0  1  0  0  0  0  0
       7       0  0  0  0  0  1  0  0  0  0
       5       0  0  0  0  0  0  0  0  0  1

In case the 1s are texts and rest is same, try using pandas.Dataframe.ne

tmp = df.ne(0)
# same operation
new_df = df.loc[new_tmp.index]

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