Question

我有多个简单功能需要在数据框某些列的每一行上实现。数据框非常像一千万行。我的数据框是这样的：

Date      location   city        number  value
12/3/2018   NY       New York      2      500
12/1/2018   MN       Minneapolis   3      600
12/2/2018   NY       Rochester     1      800
12/3/2018   WA       Seattle       2      400

我具有以下功能：

def normalized_location(row):
    if row['city'] == " Minneapolis":
        return "FCM"
    elif row['city'] == "Seattle":
        return "FCS"
    else:
        return "Other"

然后我使用：

df['Normalized Location'] =df.apply (lambda row: normalized_location (row),axis=1)

这太慢了，我该如何提高效率？

Answer 1

我们可以将map与defaultdict配合使用来快速完成此功能。

from collections import defaultdict

d = defaultdict(lambda: 'Other')
d.update({"Minneapolis": "FCM", "Seattle": "FCS"})

df['normalized_location'] = df['city'].map(d)

print(df)
        Date location         city  number  value normalized_location
0  12/3/2018       NY     New York       2    500               Other
1  12/1/2018       MN  Minneapolis       3    600                 FCM
2  12/2/2018       NY    Rochester       1    800               Other
3  12/3/2018       WA      Seattle       2    400                 FCS

...出于性能原因，规避了fillna通话。这种方法很容易推广到多个替代品。

Answer 2

您可能要使用np.select：

conds = [df.city == 'Minneapolis', df.city == 'Seattle']
choices = ['FCM', 'FCS']

df['normalized_location'] = np.select(conds, choices, default='other')

>>> df
        Date location         city  number  value normalized_location
0  12/3/2018       NY     New York       2    500               other
1  12/1/2018       MN  Minneapolis       3    600                 FCM
2  12/2/2018       NY    Rochester       1    800               other
3  12/3/2018       WA      Seattle       2    400                 FCS

Answer 3

您可以使用嵌套的np.where()：

df['city'] = np.where(df['city']=='Minneapolis', 'FCM', np.where(df['city']=='Seattle', 'FCS', 'Other'))

Answer 4

尝试以下方法：

map_ = {'Minneapolis':'FCM', 'Seattle':'FCS'}
df.loc[:,'city'] = df.loc[:,'city'].map(map_).fillna('Other')

print(df)
    Date      location  city    number  value
0  12/3/2018       NY  Other       2    500
1  12/1/2018       MN    FCM       3    600
2  12/2/2018       NY  Other       1    800
3  12/3/2018       WA    FCS       2    400

将熊猫列中的值替换为缺少键的默认值

4 个答案: