我有类似下面的数据框,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
在每个3’组’ – (100-150,Main),(200-300,Spruce)和(300-400,2nd)中 – 我希望每组中的一半记录得到一个等于块范围的中点和一半的记录使得块编号等于块范围的中点加1(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,但我无法弄清楚如何这样做(我在将函数应用于groupby键时遇到了麻烦,’BlockRange’).
我只能通过循环遍历每个唯一的组来获得我正在寻找的结果,这将在我的完整数据集上运行时需要一段时间.请参阅下面的我当前的解决方案和我正在寻找的最终结果：
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的任何建议？也许使用groupby转换？
解决方法:
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
时序：
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)
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