Pandas df 通过过滤行设置动态列值

Question

这是我的数据集：

import pandas as pd
data = { 
    'xProductNumber': ['0000',
              '3505',
              '1056',
              '3501'], 
    'xy_0000': [1,
            0,
            0,
            0], 
    'xy_3613': [0,
            0,
            0,
            0],
    'xy_3505': [0,
            1,
            0,
            0],
    'xy_3671': [0,
            0,
            0,
            1],
    'xy_1056': [1,
            0,
            1,
            0],
    'xy_3070': [1,
            0,
            0,
            0],   
}

C =('0000', '3505', '1056', '1182')

df = pd.DataFrame(data)

我想做这样的事情：

df.loc[df.apply(lambda x: (x.xProductNumber in C) and (eval('x.xy_'+str(x.xProductNumber)) == 1), axis=1 ) ,  'xy_' +str(df['xProductNumber'])] = 11 

这将动态更新正确的列 - 而不是添加新列。

输出结果应该是：

xProductNumber	xy_0000	xy_3613	xy_3505	xy_3671	xy_1056	xy_3070
0000	11	0	0	0	1	1
3505	0	0	11	0	0	0
1056	0	0	0	0	11	0
3501	0	0	0	1	0	0

我尝试了很多组合，但无济于事。 任何帮助将不胜感激！

Answer 1

您可以创建一个 MultiIndex，并临时堆叠您的 df 以选择您要应用操作的区域。不需要显式循环：

# index where we want to modify df
idx = pd.MultiIndex.from_tuples([(i, f'xy_{i}') for i in C])

# or, alternatively:
a = pd.Index(C)
idx = pd.MultiIndex.from_arrays([a, 'xy_' + a])


# temporary stacked df, to make the operation
tmp = df.set_index('xProductNumber').stack()

# do the operation
tmp.loc[idx.intersection(tmp.index)] = 11

# unstack back to df
df = tmp.unstack().reset_index()

现在：

>>> df
  xProductNumber  xy_0000  xy_3613  xy_3505  xy_3671  xy_1056  xy_3070
0           0000       11        0        0        0        1        1
1           3505        0        0       11        0        0        0
2           1056        0        0        0        0       11        0
3           3501        0        0        0        1        0        0

为了理解起见，值得看看tmp和idx：

>>> tmp.head(10)
xProductNumber         
0000            xy_0000    11
                xy_3613     0
                xy_3505     0
                xy_3671     0
                xy_1056     1
                xy_3070     1
3505            xy_0000     0
                xy_3613     0
                xy_3505    11
                xy_3671     0
dtype: int64

>>> idx
MultiIndex([('0000', 'xy_0000'),
            ('3505', 'xy_3505'),
            ('1056', 'xy_1056'),
            ('1182', 'xy_1182')],
           )

Answer 2

不确定矢量化方法，但我们可以利用 stack 和 unstack 一次处理每一行，这正是 apply 所做的。

stacked = df.stack()

for n in range(len(df)):
    stack = stacked[n]
    pnum = stack.iloc[0]
    if pnum not in C:
        continue
    key = f"xy_{pnum}"
    try:
        val = stack.loc[key]
    except KeyError:
        continue
    if val == 1:
        stack.loc[key] = 11

stacked.unstack()

  xProductNumber xy_0000 xy_3613 xy_3505 xy_3671 xy_1056 xy_3070
0           0000      11       0       0       0       1       1
1           3505       0       0      11       0       0       0
2           1056       0       0       0       0      11       0
3           3501       0       0       0       1       0       0

Answer 3

解决此问题的另一种方法是重命名列并广播以进行比较：

a , b = df.iloc[:,1:], df['xProductNumber']
rename_dict = dict(zip(a.columns,a.columns.str.split("_").str[1]))
a = a.rename(columns=rename_dict) #renamed columns of a as per b

#Now compare tp create a mask and assign 11
m = pd.DataFrame([a.columns]*len(a),columns=a.columns) == b.to_numpy()[:,None]
out = df.assign(**a.mask(m,11).set_axis([*rename_dict.keys()],axis=1))

---

print(out)


  xProductNumber  xy_0000  xy_3613  xy_3505  xy_3671  xy_1056  xy_3070
0           0000       11        0        0        0        1        1
1           3505        0        0       11        0        0        0
2           1056        0        0        0        0       11        0
3           3501        0        0        0        1        0        0