用列表中的子字符串替换“熊猫"列中的字符串
问题描述：
我有一个DF:
DF
camp, value
asd_abcd_gr_yxz_aaaa, 5
efgh_kr_ijk, 10
hjssaasd_kr_adsad, 15
asdas_kr_asd, 2
asd_fr_asda_bb_bbbbbbb, 12
adklasdj_gr_asdsad, 3
更长的时间.
与列表[_gr_, _kr_, _fr_, etc..]中的元素进行比较之后，我希望结果为
After comparing with elements in list [_gr_, _kr_, _fr_, etc..] I want the result to be
DF
camp, value
gr, 8
kr, 27
fr, 12
最好尽可能短而不会循环通过DF.该列表比_gr_, _kr_, _fr_
preferably as short as possible without looping through the DF. The list is much longer than _gr_, _kr_, _fr_
提前谢谢！
答
您可以尝试 loc :
You can try str.contains with loc:
print df
camp value
0 abcd_gr_yxz 5
1 efgh_kr_ijk 10
2 hjssaasd_kr_adsad 15
3 asdas_kr_asd 2
4 asd_fr_asda 12
5 adklasdj_gr_asdsad 3
ABR = ['_gr_', '_kr_', '_fr_']
for x in ABR:
df.loc[df['camp'].str.contains(x), 'camp'] = x
print df
camp value
0 _gr_ 5
1 _kr_ 10
2 _kr_ 15
3 _kr_ 2
4 _fr_ 12
5 _gr_ 3
print df.groupby('camp')['value'].sum().reset_index()
camp value
0 _fr_ 12
1 _gr_ 8
2 _kr_ 27
或 str.extract 和 str.strip :
ABR = ['_gr_', '_kr_', '_fr_']
s = '(' + '|'.join(ABR) + ')'
print s
(_gr_|_kr_|_fr_)
df['camp'] = df['camp'].str.extract(s, expand=False)
df = df.groupby('camp', as_index=False)['value'].sum()
df['camp'] = df['camp'].str.strip('_')
print df
camp value
0 fr 12
1 gr 8
2 kr 27