我有一个数据框，其布局如下，不，包括“flag_common”：
cat flag_1 flag_2 flag_3 pop state year flag_common
value1 1 0 0 1.5 Ohio 2000 1
value3 1 1 0 1.7 Ohio 2001 1
value2 1 1 0 3.6 Ohio 2002 1
value11 0 1 0 2.4 Nevada 2001 2
value5 0 0 0 2.9 Nevada 2002 0
value9 0 0 1 11.1 New York 2003 3
value13 0 0 0 23.4 New York 2004 0
value10 1 1 0 0.1 California 2009 1
value7 0 0 0 0.3 California 2010 0
value14 0 1 1 1.1 California 2009 2
列“flag_common”应该通过查看二进制标志并插入值1-3来设置，具体取决于标志是1还是0.当两个标志对同一行设置为1时，标志与最小数字被插入“flag_common”。这必须是动态的，能够将flag_1处理为“flag_n”。
我有点使用行迭代方法和for循环来解决它，但是我的数据非常大并且变得非常慢，所以我希望有一种“pythonic”方式来编写这个矢量化的方法。 / p>
数据框的代码如下：
df = pd.DataFrame({'state': ['Ohio', 'Ohio', 'Ohio', 'Nevada', 'Nevada', 'New York', 'New York', 'California', 'California', 'California'],
'year' : [2000, 2001, 2002, 2001, 2002, 2003, 2004, 2009, 2010, 2009],
'pop' : [1.5, 1.7, 3.6, 2.4, 2.9, 11.1, 23.4, 0.1, 0.3, 1.1],
'cat' : ['value1', 'value3', 'value2', 'value11', 'value5', 'value9', 'value13', 'value10', 'value7', 'value14'],
'flag_1' : [1, 1,1,0,0,0,0,1,0,0],
'flag_2' : [0, 1,1,1,0,0,0,1,0,1],
'flag_3' : [0, 0, 0, 0,0,1,0,0,0, 1]
})
感谢我提前提出任何想法和建议！
答案 0 :(得分：1)
您可以在columns，flag_1和flag_2列的子集中使用flag_3 0个idxmax，然后通过列表理解{{3}查找位置}。
但所有0值的排名不是flag_1，而是#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]
。因此，请使用get_loc进行更正。
#if all values in 'flag_1','flag_2','flag_3' are 0, get 0 else flag
df['new'] = np.where((df[['flag_1','flag_2','flag_3']].sum(axis=1)) == 0, 0, flag)
print df
cat flag_1 flag_2 flag_3 pop state year flag_common new
0 value1 1 0 0 1.5 Ohio 2000 1 1
1 value3 1 1 0 1.7 Ohio 2001 1 1
2 value2 1 1 0 3.6 Ohio 2002 1 1
3 value11 0 1 0 2.4 Nevada 2001 2 2
4 value5 0 0 0 2.9 Nevada 2002 0 0
5 value9 0 0 1 11.1 New York 2003 3 3
6 value13 0 0 0 23.4 New York 2004 0 0
7 value10 1 1 0 0.1 California 2009 1 1
8 value7 0 0 0 0.3 California 2010 0 0
9 value14 0 1 1 1.1 California 2009 2 2
flag编辑：
您还可以动态检查包含文字#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]
的列：
#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
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