Question

我想基于以下条件创建一个具有数值的新列：

一个。如果性别是男性和pet1 = pet2，points = 5

湾如果性别是女性和（pet1是＆＃39; cat＆＃39;或pet1 =＆＃39; dog＆＃39;），points = 5

℃。所有其他组合，points = 0

    gender    pet1      pet2
0   male      dog       dog
1   male      cat       cat
2   male      dog       cat
3   female    cat       squirrel
4   female    dog       dog
5   female    squirrel  cat
6   squirrel  dog       cat

我希望最终结果如下：

    gender    pet1      pet2      points
0   male      dog       dog       5
1   male      cat       cat       5
2   male      dog       cat       0
3   female    cat       squirrel  5
4   female    dog       dog       5
5   female    squirrel  cat       0
6   squirrel  dog       cat       0

我如何做到这一点？

Answer 1

`numpy.select`

2020年答案

这是np.select的完美案例，其中我们可以根据多个条件创建一列，当条件更多时，这是一种可读的方法：

conditions = [
    df['gender'].eq('male') & df['pet1'].eq(df['pet2']),
    df['gender'].eq('female') & df['pet1'].isin(['cat', 'dog'])
]

choices = [5,5]

df['points'] = np.select(conditions, choices, default=0)

print(df)
     gender      pet1      pet2  points
0      male       dog       dog       5
1      male       cat       cat       5
2      male       dog       cat       0
3    female       cat  squirrel       5
4    female       dog       dog       5
5    female  squirrel       cat       0
6  squirrel       dog       cat       0

Answer 2

您可以使用np.where执行此操作，条件使用按位&和|用于and和or，并在多个条件周围使用括号优先。因此，如果条件为真，则返回5，否则返回0：

In [29]:
df['points'] = np.where( ( (df['gender'] == 'male') & (df['pet1'] == df['pet2'] ) ) | ( (df['gender'] == 'female') & (df['pet1'].isin(['cat','dog'] ) ) ), 5, 0)
df

Out[29]:
     gender      pet1      pet2  points
0      male       dog       dog       5
1      male       cat       cat       5
2      male       dog       cat       0
3    female       cat  squirrel       5
4    female       dog       dog       5
5    female  squirrel       cat       0
6  squirrel       dog       cat       0

Answer 3

使用apply。

def f(x):
  if x['gender'] == 'male' and x['pet1'] == x['pet2']: return 5
  elif x['gender'] == 'female' and (x['pet1'] == 'cat' or x['pet1'] == 'dog'): return 5
  else: return 0

data['points'] = data.apply(f, axis=1)

Answer 4

@RuggeroTurra描述的apply方法对于500k行需要更长的时间。我最终使用了像

这样的东西

df['result'] = ((df.a == 0) & (df.b != 1)).astype(int) * 2 + \
               ((df.a != 0) & (df.b != 1)).astype(int) * 3 + \
               ((df.a == 0) & (df.b == 1)).astype(int) * 4 + \
               ((df.a != 0) & (df.b == 1)).astype(int) * 5

其中apply方法需要25秒，上面的方法大约需要18ms。

Answer 5

您还可以使用apply功能。例如：

def myfunc(gender, pet1, pet2):
    if gender=='male' and pet1==pet2:
        myvalue=5
    elif gender=='female' and (pet1=='cat' or pet1=='dog'):
        myvalue=5
    else:
        myvalue=0
    return myvalue

然后通过设置axis=1

使用Apply功能

df['points'] = df.apply(lambda x: myfunc(x['gender'], x['pet1'], x['pet2']), axis=1)

我们得到：

     gender      pet1      pet2  points
0      male       dog       dog       5
1      male       cat       cat       5
2      male       dog       cat       0
3    female       cat  squirrel       5
4    female       dog       dog       5
5    female  squirrel       cat       0
6  squirrel       dog       cat       0

熊猫：如何根据现有列的多个条件分配值？

5 个答案:

`numpy.select`