有没有办法遍历数据框中的字符串列表？

Question

I wrote the following code. I want to replace the number "1" with "0" whenever it appear twice or more for a particular universal_id and the number "1" that is left should be in the row where days are the lowest. The below code does the work but I want to iterate over more then one universal_id. Column "e" is ok for 'efra" I want this to do for other ID's and other columns.

pdf1 = pd.DataFrame(
    [[1, 0,1, 0,1, 60, 'fdaf'],
     [1, 1,0, 0,1, 350, 'fdaf'],
     [1, 1,0, 0,1, 420, 'erfa'],
     [0, 1,0, 0,1, 410, 'erfa']],
    columns=['A', 'B', 'c', 'd', 'e', 'days','universal_id'])

pdf1['A'] = np.where(pdf1['days']==pdf1['days'].min(),1,0)
zet = pdf1.loc[pdf1['e'].isin([1]) & 
pdf1['universal_id'].str.contains('erfa')]
zet['e'] = np.where(zet['days']==zet['days'].min(),1,0)
pdf1.loc[zet.index, :] = zet[:]
pdf1

Output:

    A   B   c   d   e   days    universal_id
 0  1   0   1   0   1   60     fdaf
 1  0   1   0   0   1   350    fdaf
 2  0   1   0   0   0   420    erfa
 3  0   1   0   0   1   410    erfa

Answer 1

You can use:

df2 = pdf1.sort_values(by='days')

m1 = df2['A'].eq(1)
m2 = df2[['A', 'universal_id']].duplicated()

pdf1.loc[m1&m2, 'A'] = 0

output:

   A  B  c  d  e  days universal_id
0  1  0  1  0  1    60         fdaf
1  0  1  0  0  1   350         fdaf
2  1  1  0  0  1   420         erfa
3  0  1  0  0  1   410         erfa

for e, f you want to follow the same logic:

m1 = df2['A'].eq(1)
m3 = df2[['e', 'universal_id']].duplicated()

pdf1.loc[m1&m3, 'e'] = 0

output:

   A  B  c  d  e  days universal_id
0  1  0  1  0  1    60         fdaf
1  0  1  0  0  0   350         fdaf
2  1  1  0  0  0   420         erfa
3  0  1  0  0  1   410         erfa