Python Lambda函数从直接代码返回不同的输出

对于日期的相同输出dd ()，因为使用Timestamp.date而不是Series.dt.date:

#if omit () get built methods
print (df.apply(lambda x : x['date1'].date, axis = 1))
 0    <built-in method date of Timestamp object at 0...
 1    <built-in method date of Timestamp object at 0...
 2    <built-in method date of Timestamp object at 0...
 3    <built-in method date of Timestamp object at 0...
 4    <built-in method date of Timestamp object at 0...
 dtype: object

#get dates
print (df.apply(lambda x : x['date1'].date(), axis = 1))
0    2023-05-11
1    2023-02-24
2    2023-07-09
3    2023-01-19
4    2023-02-10
dtype: object

df.apply(lambda x : (abs(x['value1'] - x['value2']) > .5) &
                    (x['date1'].date() != x['date2'].date()), axis = 1)

更好的方法是您的第二种矢量化方法:

(abs(df['value1'] - df['value2']) > .5) & (df['date1'].dt.date != df['date2'].dt.date)

如果需要返回yes/no:

np.where((abs(df['value1'] - df['value2']) > .5) & (df['date1'].dt.date != df['date2'].dt.date), 'yes', 'no')

df.apply(lambda x : 'yes' if (abs(x['value1'] - x['value2']) > .5) & (x['date1'].date() != x['date2'].date()) else 'no', axis = 1)

编辑:如果使用Boolean，&和and工作相同，感谢@Mustafa Aydın的 comments :

在布尔AND运算中使用and比使用&更清楚，因为&主要用于按位运算--在纯Python中，重载用于矢量化--以及在NumPy(和Pandas)中

print (True and True)
True
print (False and True)
False
print (True and False)
False
print (False and False)False

print (True & True)
True
print (False & True)
False
print (True & False)
False
print (False & False)
False