我有这个数据框架,并且只需要保留2列(这里是numA和numB )具有倒数值的那些行。
gpm = pd.DataFrame(data={
'id':[1,2,3,4,5,6,7,8,9],
'time':[150315,150315,150315,150315,150315,150315,150315,150315,150315],
'numA':['A','D','C','B','A','C','A','E','D'],
'numB':['B','C','B','A','B','D','B','A','A'],
'antA':['MSPDV','VIELU','RMPC1','MJCIH','PALT2','M2PV3','MACIF','MACIF','VIELU'],
'antB':['BPDV8','0GRI3','SSFDJ','SSFDJ','SSFDJ','CCPG1','0GRI3','SSFDJ','SSFDJ']
})我只想要列numA和numB互换的行。也就是说,保留出现对(A,B),(B,A)和(C,D),(D,C)的所有线。
目前,我的解决方案包括创建一个包含所有唯一标识符的列表,并遍历每一行,查看实际的合作伙伴是否在合作伙伴列表中
它非常慢..。(也许是不正确的!)
## here's my code
parties = {}
nums = gpm['numA']+gpm['numB']
for i in nums.unique():
parties[i] = gpm['numB'][gpm['numA'] == i]
parties[i] = gpm['numA'][gpm['numB'] == i]
new_d = gpm.iloc[[0]]
for i in np.arange(1,gpm.shape[0]):
numa = gpm.iloc[i]['numA']
if gpm.iloc[i]['numB'] in parties[numa]:
new_d.append(gpm.iloc[[i]])有没有什么精明的程序员可以帮你加快速度?要解析的实际文件是大约15 is的csv。
谢谢
发布于 2018-11-14 22:25:34
在您的示例中,我假设包含id=3、8和9的行(C,B)、(E,A)和(D,A)是不需要的?如果是这样,下面是通过比较numA和numB中的值来选择特定可接受组合的标准方法:
In [5]: gpm[((gpm['numA'] == 'A') & (gpm['numB'] == 'B')) |
...: ((gpm['numA'] == 'B') & (gpm['numB'] == 'A')) |
...: ((gpm['numA'] == 'C') & (gpm['numB'] == 'D')) |
...: ((gpm['numA'] == 'D') & (gpm['numB'] == 'C'))
...: ]
Out[5]:
id time numA numB antA antB
0 1 150315 A B MSPDV BPDV8
1 2 150315 D C VIELU 0GRI3
3 4 150315 B A MJCIH SSFDJ
4 5 150315 A B PALT2 SSFDJ
5 6 150315 C D M2PV3 CCPG1
6 7 150315 A B MACIF 0GRI3(将结果赋给new_d)
https://stackoverflow.com/questions/53300569
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