Pandas groupby - 将不同的功能应用于每组中一半的记录

Question

I have something like the following dataframe, where I have non-unique combinations of street address ranges and street names.

import pandas as pd
df=pd.DataFrame()
df['BlockRange']=['100-150','100-150','100-150','100-150','200-300','200-300','300-400','300-400','300-400']
df['Street']=['Main','Main','Main','Main','Spruce','Spruce','2nd','2nd','2nd']
df
  BlockRange  Street
0    100-150    Main
1    100-150    Main
2    100-150    Main
3    100-150    Main
4    200-300  Spruce
5    200-300  Spruce
6    300-400     2nd
7    300-400     2nd
8    300-400     2nd

Within each of the 3 'groups' - (100-150, Main), (200-300, Spruce), and (300-400, 2nd) - I want half of the records in each group to get a block number equal to the midpoint of the block range and half of the records to get a block number equal to the midpoint of the block range plus 1 (as to put it on the other side of the street).

I know this should be able to be done using groupby transform, but I can't figure out how to do so (I'm having trouble applying a function to the groupby key, 'BlockRange').

I'm able to get the result I'm looking for only by looping through each unique group, which will take a while when run on my full dataset. See below for my current solution and the end result I'm looking for:

groups=df.groupby(['BlockRange','Street'])

#Write function that calculates the mid point of the block range
def get_mid(x):
    block_nums=[int(y) for y in x.split('-')]
    return sum(block_nums)/len(block_nums)

final=pd.DataFrame()
for groupkey,group in groups:
    block_mid=get_mid(groupkey[0])
    halfway_point=len(group)/2
    group['Block']=0
    group.iloc[:halfway_point]['Block']=block_mid
    group.iloc[halfway_point:]['Block']=block_mid+1
    final=final.append(group)

final
  BlockRange  Street  Block
0    100-150    Main    125
1    100-150    Main    125
2    100-150    Main    126
3    100-150    Main    126
4    200-300  Spruce    250
5    200-300  Spruce    251
6    300-400     2nd    350
7    300-400     2nd    351
8    300-400     2nd    351

Any suggestions as to how I can do this more efficiently? Perhaps using groupby transform?

Answer 1

You can use apply with custom function f :

def f(x):
    df = pd.DataFrame([y.split('-') for y in x['BlockRange'].tolist()])
    df = df.astype(int)
    block_nums = df.sum(axis=1) / 2
    x['Block'] = block_nums[0]
    halfway_point=len(x)/2
    x.iloc[halfway_point:, 2] = block_nums[0] + 1
    return x

print df.groupby(['BlockRange','Street']).apply(f)

  BlockRange  Street  Block
0    100-150    Main    125
1    100-150    Main    125
2    100-150    Main    126
3    100-150    Main    126
4    200-300  Spruce    250
5    200-300  Spruce    251
6    300-400     2nd    350
7    300-400     2nd    351
8    300-400     2nd    351  

Timings:

In [32]: %timeit orig(df)
__main__:26: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
__main__:27: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
__main__:28: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
1 loops, best of 3: 290 ms per loop

In [33]: %timeit new(df)
100 loops, best of 3: 10.2 ms per loop  

Testing:

print df
df1 = df.copy()

def orig(df):
    groups=df.groupby(['BlockRange','Street'])

    #Write function that calculates the mid point of the block range
    def get_mid(x):
        block_nums=[int(y) for y in x.split('-')]
        return sum(block_nums)/len(block_nums)
    final=pd.DataFrame()

    for groupkey,group in groups:
        block_mid=get_mid(groupkey[0])
        halfway_point=len(group)/2
        group['Block']=0
        group.iloc[:halfway_point]['Block']=block_mid
        group.iloc[halfway_point:]['Block']=block_mid+1
        final=final.append(group)
    return final    

def new(df):
    def f(x):
        df = pd.DataFrame([y.split('-') for y in x['BlockRange'].tolist() ])
        df = df.astype(int)
        block_nums = df.sum(axis=1) / 2
        x['Block'] = block_nums[0]
        halfway_point=len(x)/2
        x.iloc[halfway_point:, 2] = block_nums[0] + 1
        return x

    return df.groupby(['BlockRange','Street']).apply(f)

print orig(df)
print new(df1)   

Answer 2

For comparison, note that you can do this without apply :

ss = df["BlockRange"].str.split("-")
midnum = (ss.str[1].astype(float) + ss.str[0].astype(float))//2
grouped = df.groupby(["BlockRange", "Street"])
df["Block"] = midnum + (grouped.cumcount()>= grouped["Street"].transform(len) // 2)

which gives me

>>> df
  BlockRange  Street  Block
0    100-150    Main    125
1    100-150    Main    125
2    100-150    Main    126
3    100-150    Main    126
4    200-300  Spruce    250
5    200-300  Spruce    251
6    300-400     2nd    350
7    300-400     2nd    351
8    300-400     2nd    351

This works because cumcount and transform(len) give us the pieces we need:

>>> grouped.cumcount()
0    0
1    1
2    2
3    3
4    0
5    1
6    0
7    1
8    2
dtype: int64
>>> grouped.transform(len)
   Block
0      4
1      4
2      4
3      4
4      2
5      2
6      3
7      3
8      3