我试图替换pd.Dataframe中所有字符串的一部分,但它不起作用.
我的数据示例:
HLAA0101
HLAA0201
HLAA0202
HLAA0203
HLAA0205
我想要获得什么:
A0101
A0201
A0202
A0203
A0205
我的代码:
mhc = train_csv.mhc
for i in mhc:
i[0:2].replace(‘HLA’, ‘ ‘)
print(mhc)
但它不起作用.
最佳答案 选项1:
df[‘mhc’] = df[‘mhc’].str[3:]
选项2:
df[‘mhc’] = df[‘mhc’].str.replace(r’^HLA’,”)
选项3:
df[‘mhc’] = df[‘mhc’].str.extract(r’HLA(.*)’, expand=False)
选项4 :(注意:有时列表理解比字符串/对象dtypes的内部向量化方法更快)
df[‘mhc’] = [s[3:] for s in df[‘mhc’]]
所有选项都会产生相同的结果:
In [26]: df
Out[26]:
mhc
0 A0101
1 A0201
2 A0202
3 A0203
4 A0205
时间为50.000行DF:
In [29]: df = pd.concat([df] * 10**4, ignore_index=True)
In [30]: %timeit df[‘mhc’].str[3:]
35.9 ms ± 3.18 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [31]: %timeit df[‘mhc’].str.replace(r’^HLA’,”)
162 ms ± 3.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [32]: %timeit df[‘mhc’].str.extract(r’HLA(.*)’, expand=False)
164 ms ± 4.87 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [33]: %timeit [s[3:] for s in df[‘mhc’]]
14.6 ms ± 18.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [34]: df.shape
Out[34]: (50000, 1)
结论:列表理解方法获胜.
Original: https://blog.csdn.net/weixin_30945845/article/details/112960922
Author: 1360785665
Title: pythonforin替换字符_如何在pandas.Dataframe中替换字符串的一部分?
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