Pandas groupby - 对每组中的一半记录应用不同的函数

Question

我有类似以下数据框的内容，其中我有街道地址范围和街道名称的非唯一组合。

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

在 3 个“组”中的每一个中 - (100-150, Main)、(200-300, Spruce) 和 (300-400, 2nd) - 我希望每个组中的一半记录获得一个块number 等于块范围的中点和一半的记录，得到一个等于块范围的中点加 1 的块号（将其放在街道的另一侧）。

我知道这应该可以使用 groupby 转换来完成，但我不知道该怎么做（我无法将函数应用于 groupby 键，'BlockRange'）。

我只能通过遍历每个唯一组来获得我正在寻找的结果，这在我的完整数据集上运行时需要一段时间。有关我当前的解决方案和我正在寻找的最终结果，请参见下文：

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

关于如何更有效地做到这一点有什么建议吗？也许使用 groupby 转换？

Answer 1

您可以将apply与自定义函数f一起使用：

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  

时间安排：

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  

测试：

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

为了比较，请注意您可以在没有apply 的情况下执行此操作：

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)

这给了我

>>> 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

之所以有效，是因为 cumcount 和 transform(len) 为我们提供了我们需要的部分：

>>> 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