我正在尝试制作三个 csv 文件,并仅创建一个合并有用数据的文件。
现在,我陷入了这个问题:
我有两列( SUFFIX 和 COD_METEL ),其中有150万行,我需要详细说明并创建包含结果的另一列。
SUFFIX COD_METEL
0 CBR CBR8901027
1 CBR CBR8901028
2 CBR CBR8904001
3 CBR CBR8904002
4 CBR CBR8904008
5 CBR CBR8904027
6 CBR CBR8904039
7 THO THO96666290
8 THO THO96666294
9 THO THO96666298
10 THO THO96666302
11 THO THO96666322
12 THO THO96666326
13 ZV ZV111900NI
14 ZV ZV111910NI
15 ZX ZX2021.AC
16 ZX ZX2021.AC
17 ZX ZX6066.AC
18 ZX ZX6111.AC
19 ZX ZX6111.AC
20 ZX ZX6380.AC
21 ZX ZX9030
22 ZX ZX9030
23 ZX ZX9030
24 ZZ ZZ00012565
在这里,我需要将 SUFFIX 值“减去”到 COD_METEL ,如下所示:
df["RESULT"] = df["COD_METEL"] - df["SUFFIX"]
SUFFIX COD_METEL RESULT
0 CBR CBR8901027 8901027
1 CBR CBR8901028 8901028
2 CBR CBR8904001 8904001
我知道无法使用“-”运算符,因此,我想问您一些技巧来解决此问题,并快速替换所有值。
我已经尝试做一些测试:
replaceList = list(set(df["SUFFIX"]))
for to_replace in replaceList:
df["RESULT"] = df["COD_METEL"].str.replace(to_replace,"")
答案 0 :(得分:1)
如果没有缺失值,您可以尝试list comprehension
:
df['new'] = [j.replace(i, '') for i, j in zip(df['SUFFIX'], df['COD_METEL'])]
print (df)
SUFFIX COD_METEL new
0 CBR CBR8901027 8901027
1 CBR CBR8901028 8901028
2 CBR CBR8904001 8904001
3 CBR CBR8904002 8904002
4 CBR CBR8904008 8904008
5 CBR CBR8904027 8904027
6 CBR CBR8904039 8904039
7 THO THO96666290 96666290
8 THO THO96666294 96666294
9 THO THO96666298 96666298
10 THO THO96666302 96666302
11 THO THO96666322 96666322
12 THO THO96666326 96666326
13 ZV ZV111900NI 111900NI
14 ZV ZV111910NI 111910NI
15 ZX ZX2021.AC 2021.AC
16 ZX ZX2021.AC 2021.AC
17 ZX ZX6066.AC 6066.AC
18 ZX ZX6111.AC 6111.AC
19 ZX ZX6111.AC 6111.AC
20 ZX ZX6380.AC 6380.AC
21 ZX ZX9030 9030
22 ZX ZX9030 9030
23 ZX ZX9030 9030
24 ZZ ZZ00012565 00012565
性能:
#[250000 rows x 2 columns]
df = pd.concat([df] * 10000, ignore_index=True)
#print (df)
In [289]: %timeit df['RESULT'] = df.apply(lambda x: x['COD_METEL'].replace(x['SUFFIX'], ''), axis=1)
5.05 s ± 347 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [290]: %timeit df['new'] = [j.replace(i, '') for i, j in zip(df['SUFFIX'], df['COD_METEL'])]
98.7 ms ± 8.8 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
答案 1 :(得分:1)
另一种方法是:
df['RESULT'] = df.apply(lambda x: x['COD_METEL'].replace(x['SUFFIX'], ''), axis=1)
df
SUFFIX COD_METEL RESULT
0 CBR CBR8901027 8901027
1 CBR CBR8901028 8901028
2 CBR CBR8904001 8904001
3 CBR CBR8904002 8904002
4 CBR CBR8904008 8904008
5 CBR CBR8904027 8904027
6 CBR CBR8904039 8904039
7 THO THO96666290 96666290
8 THO THO96666294 96666294
9 THO THO96666298 96666298
10 THO THO96666302 96666302
11 THO THO96666322 96666322
12 THO THO96666326 96666326
13 ZV ZV111900NI 111900NI
14 ZV ZV111910NI 111910NI
15 ZX ZX2021.AC 2021.AC
16 ZX ZX2021.AC 2021.AC
17 ZX ZX6066.AC 6066.AC
18 ZX ZX6111.AC 6111.AC
19 ZX ZX6111.AC 6111.AC
20 ZX ZX6380.AC 6380.AC
21 ZX ZX9030 9030
22 ZX ZX9030 9030
23 ZX ZX9030 9030
24 ZZ ZZ00012565 00012565