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[英]Python Openpyxl: How to take values from a column, append it to a string and then make the new string part of an array?
[英]Split and take part of string from column values and make new column from that in pandas python
我有一个这样的字符串作为我的df中一列的值。
ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232
ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232
如何使用此列的一部分获取新列。 我需要的部分是
74
89
string.split()
允许您根据分隔符(此处为/
和-
)将字符串分解为部件列表。
s = 'ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232'
print s.split('/')[2].split('-')[1]
# 74
使用pandas.apply()
将其应用于您的列
df['b'] = df['a'].apply(lambda s:s.split('/')[2].split('-')[1])
print (df)
产量
a b
0 ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232 74
1 ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232 89
nb:使用@ A-Za-z的解决方案,它比我的快。
如果这是df
val
0 ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232
1 ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232
您可以使用str.extract
df['num_val'] = df.val.str.extract('LNFFF-(\d+)/', expand = False)
你得到
val num_val
0 ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232 74
1 ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232 89
假设您的数据框名为df且列col:
df['sub_col'] = pd.Series([s[21:23] for s in df['col'].values], index=df.index)
看来你需要str.extract
:
df = pd.DataFrame({'a': ['ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232',
'ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232']})
print (df)
a
0 ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232
1 ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232
df['new'] = df['a'].str.extract('LNFFF-(\d+)', expand=False)
#if necessary convert to ints
df['new'] = df['new'].astype(int)
print (df)
a new
0 ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232 74
1 ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232 89
df['new'] = df['a'].str.split('/').str[2].str.extract('(\d+)', expand=False)
print (df)
a new
0 ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232 74
1 ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232 89
df['new'] = df['a'].str.split('/').str[2].str.split('-').str[1]
print (df)
a new
0 ttt-OPP/MKKL-7/LNFFF-74/OOOP-71/AAD-1/RRR-232 74
1 ttt-OPP/MKKL-7/LNFFF-89/OOOP-71/AAD-1/RRR-232 89
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