在有空值的日期上使用 lambda 和 strftime (Pandas)
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Using lambda and strftime on dates when there are null values (Pandas)
提问by FortuneFaded
I'm trying to change the format of a datetime column in my Dataframe using lambda and strftime like below
我正在尝试使用如下所示的 lambda 和 strftime 更改我的 Dataframe 中日期时间列的格式
df['Date Column'] = df['Date Column'].map(lambda x: x.strftime('%m/%d/%Y'))
However, since I have null values in some of these fields, this is giving me an error. I cannot drop these null rows because I still need them for the data in the other columns. Is there a way around this error without dropping the nulls.
但是，由于我在其中一些字段中有空值，这给了我一个错误。我无法删除这些空行，因为其他列中的数据仍然需要它们。有没有办法在不删除空值的情况下解决这个错误。
Perhaps something like
也许像
df['Date Column'].map(lambda x: x.strftime('%m/%d/%Y') if x != null else "")
?
?
The method I've used is to drop the nulls, format the column, then merge it back onto the original dataset, but this seems like a very inefficient method.
我使用的方法是删除空值，格式化列，然后将其合并回原始数据集，但这似乎是一种非常低效的方法。
采纳答案by Stop harming Monica
You should be not checking for nan/nat (un)equality, but .notnull()should work and it does for me:
你不应该检查 nan/nat (un)equality，但.notnull()应该工作，它对我有用：
s = pd.date_range('2000-01-01', periods=5).to_series().reset_index(drop=True)
s[2] = None
s
0 2000-01-01
1 2000-01-02
2 NaT
3 2000-01-04
4 2000-01-05
dtype: datetime64[ns]
s.map(lambda x: x.strftime('%m/%d/%Y') if pd.notnull(x) else '')
0 01/01/2000
1 01/02/2000
2
3 01/04/2000
4 01/05/2000
dtype: object
This returns the same that the answers by @Alexander and @Batman but is more explicit. It may also be slightly slower for large series.
这返回与@Alexander 和@Batman 的答案相同，但更明确。对于大系列，它也可能稍微慢一些。
Alternatively you can use the .dtaccesor. The null values will be formatted as NaT.
或者，您可以使用.dtaccesor。空值将被格式化为NaT.
s.dt.strftime('%m/%d/%Y')
0 01/01/2000
1 01/02/2000
2 NaT
3 01/04/2000
4 01/05/2000
dtype: object
回答by Batman
Personally I'd just define a small function, and then use that.
我个人只是定义一个小函数，然后使用它。
def to_string(date):
if date:
string = date.strftime('%Y%m%d')
else:
string = ""
return string
Then
然后
df['Date Column'].map(to_string)
Otherwise
除此以外
df['Date Column'].map(lambda x: x.strftime('%Y%m%d') if x else "")
回答by Alexander
You can use a conditional assignment (ternary).
您可以使用条件赋值（三元）。
df['Date Column'] = df['Date Column'].map(lambda x: x.strftime('%m/%d/%Y') if x else '')
