pandas 将 float64 列转换为日期时间熊猫
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Convert float64 column to datetime pandas
提问by JohnAndrews
I have the following pandas DataFramecolumn dfA['TradeDate']:
我有以下pandas DataFrame专栏dfA['TradeDate']：
0 20100329.0
1 20100328.0
2 20100329.0
...
and I wish to transform it to a datetime.
我希望将其转换为日期时间。
Based on another tread on SO, I convert it first to a string and then apply the strptimefunction.
基于 SO 上的另一个步骤，我首先将其转换为字符串，然后应用该strptime函数。
dfA['TradeDate'] = datetime.datetime.strptime( dfA['TradeDate'].astype('int').to_string() ,'%Y%m%d')
However this returns the error that my format is incorrect (ValueError).
但是，这会返回我的格式不正确的错误 ( ValueError)。
An issue that I spotted is that the column is not properly to string, but to an object.
我发现的一个问题是该列不能正确串接，而是串接一个对象。
When I try:
当我尝试：
dfA['TradeDate'] = datetime.datetime.strptime( dfA['TradeDate'].astype(int).astype(str),'%Y%m%d')
It returns: must be a Str and not Series.
它返回：必须是 Str 而不是系列。
回答by jezrael
You can use:
您可以使用：
df['TradeDate'] = pd.to_datetime(df['TradeDate'], format='%Y%m%d.0')
print (df)
TradeDate
0 2010-03-29
1 2010-03-28
2 2010-03-29
But if some bad values, add errors='coerce'for replace them to NaT
但是如果一些不好的值，添加errors='coerce'替换它们NaT
print (df)
TradeDate
0 20100329.0
1 20100328.0
2 20100329.0
3 20153030.0
4 yyy
df['TradeDate'] = pd.to_datetime(df['TradeDate'], format='%Y%m%d.0', errors='coerce')
print (df)
TradeDate
0 2010-03-29
1 2010-03-28
2 2010-03-29
3 NaT
4 NaT
回答by languitar
You can use to_datetimewith a custom format on a string representation of the values:
您可以to_datetime在值的字符串表示形式上使用自定义格式：
import pandas as pd
pd.to_datetime(pd.Series([20100329.0, 20100328.0, 20100329.0]).astype(str), format='%Y%m%d.0')
回答by aman sohane
strptimefunction works on a single value, not on series. You need to apply that function to each element of the column
strptime函数适用于单个值，而不适用于系列。您需要将该函数应用于列的每个元素
try pandas.to_datetimemethod
尝试pandas.to_datetime方法
eg
例如
dfA = pandas.DataFrame({"TradeDate" : [20100329.0,20100328.0]})
pandas.to_datetime(dfA['TradeDate'], format = "%Y%m%d")
or
或者
dfA['TradeDate'].astype(int).astype(str)\
.apply(lambda x:datetime.datetime.strptime(x,'%Y%m%d'))
回答by R Kiselev
In your first attempt you tried to convert it to string and then pass to strptime, which resulted in ValueError. This happens because dfA['TradeDate'].astype('int').to_string()creates a single string containing alldates as well as their row numbers. You can change this to
在您的第一次尝试中，您尝试将其转换为字符串，然后传递给strptime，这导致ValueError. 发生这种情况是因为dfA['TradeDate'].astype('int').to_string()创建了一个包含所有日期及其行号的字符串。您可以将其更改为
dates = dfA['TradeDate'].astype('int').to_string(index=False).split()
dates
[u'20100329.0', u'20100328.0', u'20100329.0']
to get a list of dates. Then use python list comprehensionto convert each element to datetime:
获取日期列表。然后使用python 列表推导将每个元素转换为datetime：
dfA['TradeDate'] = [datetime.strptime(x, '%Y%m%d.0') for x in dates]
