groupby().apply(lambda x:x.copy()) raise error.

[ruoyu0088]

df = pd.DataFrame({"g":[1, 2, 2, 2], "a":[1, 2, 3, 4], "b":[5, 6, 7, 8]})
df.groupby("g").apply(lambda x:x.copy())

raise

ValueError: Shape of passed values is (3, 4), indices imply (3, 2)

but lambda x:x or lambda x:x[:] works.

[ruoyu0088]

I debuged this and found the different between df.copy() and df[:]:

import pandas as pd
import numpy as np

def f1(x):
    return x.copy()

def f2(x):
    return x[:]

df = pd.DataFrame({"g":[1, 2, 2, 2], "a":[1, 2, 3, 4], "b":[5, 6, 7, 8]})

print f1(df).index._id is df.index._id  #True
print f2(df).index._id is df.index._id  #False

although df.copy().index is not df.index, the _id attribute is the same.

(df1, df2), _ = pd.lib.apply_frame_axis0(df, f1, [1, 2], np.array([0, 1], np.int64), np.array([1, 4], np.int64))
print df1.index._id is df2.index._id #True

the apply() calls apply_frame_axis0() to do the job, the index of the result DataFrames share the same _id object. This will cause concat() raise the ValueError: Shape of passed values is (3, 4), indices imply (3, 2) error.

[jreback]

you mentioned this in #9867 but we'll take this as a bug-report.

Yes these should do the same.

The entire point of the inference is to guess whether the user is mutating the input or not. IMHO this should be banned, but that ship has sailed.

[jorisvandenbossche]

So this doesn't raise an error anymore on master, but you still get an inconsistent output:

In [111]: df = pd.DataFrame({"g":[1, 2, 1, 2], "a":[1, 2, 3, 4], "b":[5, 6, 7, 8]})

In [112]: df.groupby("g").apply(lambda x:x)
Out[112]: 
   a  b  g
0  1  5  1
1  2  6  2
2  3  7  1
3  4  8  2

In [113]: df.groupby("g").apply(lambda x:x.copy())
Out[113]: 
     a  b  g
g           
1 0  1  5  1
  2  3  7  1
2 1  2  6  2
  3  4  8  2

[mroeschke]

close in favor of #14927?

[jorisvandenbossche]

Yep