您可以在子集中使用flag_1、flag_2和flag_3的子集中的{a1}，然后使用^{}通过列表理解找到位置。在
但是所有0值的位置不是0，而是flag_1。所以用numpy.where来纠正它。在#get min value of columns 'flag_1','flag_2','flag_3'
print df[['flag_1','flag_2','flag_3']].idxmax(axis=1)
0 flag_1
1 flag_1
2 flag_1
3 flag_2
4 flag_1
5 flag_3
6 flag_1
7 flag_1
8 flag_1
9 flag_2
dtype: object
#get position of flag
print df.columns.get_loc('flag_1')
1
#get positions all flags
flag = [df.columns.get_loc(k) for k in df[['flag_1','flag_2','flag_3']].idxmax(axis=1)]
print flag
[1, 1, 1, 2, 1, 3, 1, 1, 1, 2]
#alternative solution for positions of flags - last digit has to be number
print [int(x[-1]) for x in df[['flag_1','flag_2','flag_3']].idxmax(axis=1)]
[1, 1, 1, 2, 1, 3, 1, 1, 1, 2]
^{pr2}$
编辑：
您还可以使用文本flag动态检查列：#get columns where first value before _ is text 'flag'
cols = [x for x in df.columns if x.split('_')[0] == 'flag']
print cols
['flag_1', 'flag_2', 'flag_3']
#get min value of columns 'flag_1','flag_2','flag_3'
print df[cols].idxmax(axis=1)
0 flag_1
1 flag_1
2 flag_1
3 flag_2
4 flag_1
5 flag_3
6 flag_1
7 flag_1
8 flag_1
9 flag_2
dtype: object
#get positions of flag
print df.columns.get_loc('flag_1')
1
#get positions all flags
flag = [df.columns.get_loc(k) for k in df[cols].idxmax(axis=1)]
print flag
[1, 1, 1, 2, 1, 3, 1, 1, 1, 2]
#alternative solution for positions of flags - last digit has to be number
print [int(x[-1]) for x in df[cols].idxmax(axis=1)]
[1, 1, 1, 2, 1, 3, 1, 1, 1, 2]#if all values in 'flag_1','flag_2','flag_3' are 0, get 0 else flag
df['new'] = np.where((df[cols].sum(axis=1)) == 0, 0, flag)
print df
cat flag_1 flag_2 flag_3 pop state year new
0 value1 1 0 0 1.5 Ohio 2000 1
1 value3 1 1 0 1.7 Ohio 2001 1
2 value2 1 1 0 3.6 Ohio 2002 1
3 value11 0 1 0 2.4 Nevada 2001 2
4 value5 0 0 0 2.9 Nevada 2002 0
5 value9 0 0 1 11.1 New York 2003 3
6 value13 0 0 0 23.4 New York 2004 0
7 value10 1 1 0 0.1 California 2009 1
8 value7 0 0 0 0.3 California 2010 0
9 value14 0 1 1 1.1 California 2009 2