熊猫将值列表更改为列

debugcn 发表于 Dev

吉米

我有这样的df，我想将值列表更改为column

```

    uid   device
0   000 [1.0, 3.0]
1   001 [3.0]
2   003 [nan]
3   004 [2.0, 3.0]
4   005 [1.0]
5   006 [1.0]
6   006 [nan]
7   007 [2.0]
```

应该

```

    uid  device      NA  just_1  just_2or3  Both
0   000 [1.0, 3.0]   0     0         0        1
1   001 [3.0]        0     0         1        0
2   003 [nan]        1     0         0        0
3   004 [2.0, 3.0]   0     0        "1"       0
4   005 [1.0]        0     1         0        0
5   006 [1.0]        0     1         0        0
6   006 [nan]        1     0         0        0
7   007 [2.0]        0     1         1        0
8   008 [1.0, 2.0]   0     0         0        1

```

我想更改为虚拟变量，如果设备仅1.0，则将对应的列值设置为1，如果2.0、3.0，[2.0,3.0]，则将just_2or3设置为1。

仅当列表中的1.0（例如[1.0,3.0]，[1.0,2.0]）都设置为1时

我怎样才能做到这一点？谢谢你

耶斯列尔

您可以将自定义函数f与列表推导结合使用，最后将boolean值强制转换为intby astype：

df = pd.DataFrame({'uid':['000','001','002','003','004','005','006','007'],
                   'device':[[1.0,3.0],[3.0],[np.nan],[2.0,3.0],
                             [1.0],[1.0],[np.nan],[2.0]]})

print (df)
       device  uid
0  [1.0, 3.0]  000
1       [3.0]  001
2       [nan]  002
3  [2.0, 3.0]  003
4       [1.0]  004
5       [1.0]  005
6       [nan]  006
7       [2.0]  007

def f(x):
    #print (x)
    NA = [np.nan in x][0]
    just_1  = [1 in x and not(2 in x or 3 in x)][0]
    both = [1 in x and (2 in x or 3 in x)][0]
    just_2or3 = [1 not in x and (2 in x or 3 in x)][0]
    return pd.Series([NA, just_1, just_2or3, both], 
                     index=['NA','just_1','just_2or3', 'both'])

print (df.set_index('uid').device.apply(f).astype(int).reset_index())
   uid  NA  just_1  just_2or3  both
0  000   0       0          0     1
1  001   0       0          1     0
2  002   1       0          0     0
3  003   0       0          1     0
4  004   0       1          0     0
5  005   0       1          0     0
6  006   1       0          0     0
7  007   0       0          1     0

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