Python 熊猫从日期时间转换为整数时间戳
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pandas convert from datetime to integer timestamp
提问by Francesco Boi
Considering a pandas dataframe in python having a column named timeof type integer, I can convert it to a datetimeformat with the following instruction.
考虑到 python 中的 Pandas 数据框有一个名为timeinteger 类型的列，我可以datetime使用以下指令将其转换为格式。
df['time'] = pandas.to_datetime(df['time'], unit='s')
so now the column has entries like: 2019-01-15 13:25:43.
所以现在列有类似的条目：2019-01-15 13:25:43。
What is the command to revert the string to an integer timestamp value (representing the number of seconds elapsed from 1970-01-01 00:00:00)?
将字符串恢复为整数时间戳值（表示从 开始经过的秒数1970-01-01 00:00:00）的命令是什么？
I checked pandas.Timestampbut could not find a conversion utility and I was not able to use pandas.to_timedeltafor this.
我检查pandas.Timestamp但找不到转换实用程序，我无法使用pandas.to_timedelta它。
Is there any utility for this conversion?
这种转换有什么用吗？
回答by Always Sunny
You can typecast to int using astype(int)and divide it by 10**9to get the number of seconds to the unix epoch start.
您可以使用类型转换为 intastype(int)并将其除以10**9获得 unix epoch 开始的秒数。
import pandas as pd
df = pd.DataFrame({'time': [pd.to_datetime('2019-01-15 13:25:43')]})
df_unix_sec = pd.to_datetime(df['time']).astype(int)/ 10**9
print(df_unix_sec)
回答by ALollz
Use .dt.total_seconds()on a timedelta64:
使用.dt.total_seconds()上timedelta64：
import pandas as pd
df = pd.DataFrame({'time': [pd.to_datetime('2019-01-15 13:25:43')]})
# pd.to_timedelta(df.time).dt.total_seconds() # Is deprecated
(df.time - pd.to_datetime('1970-01-01')).dt.total_seconds()
Output
输出
0 1.547559e+09
Name: time, dtype: float64
回答by Ignacio Peletier
The easiest way is to use .value
最简单的方法是使用 .value
pd.to_datetime('1970-01-01').value
