将日期时间格式转换为Unix时间戳Pandas

我试图将正常的日期时间转换为熊猫中的unix时间戳。同时寻找一些样本我只能找到一个例子here，但我不能在我的上下文中使用。该数据集没有标题，最后的2 columns需要转换UNIX time stamp并与前3列一起生成新的输出。将日期时间格式转换为Unix时间戳Pandas

1466f7b93975983f6e292a8a4faaa4b2,1619b4d0d283c0dddb17d24a359a3b49,36db348cde68592a31d502366fc52932,2010-03-08 17:09:00.472544,2010-03-12 16:09:58.122987 
367c13356a5d22158f0ae56977134e2c,eedb7d0714796b64767a8710ea3844a7,925476200929fd346ea312cbe9a046fe,2010-03-08 17:08:29.174236,2010-03-12 16:09:58.122987 
edf6b1e4f67b0e8a5080d299c9f9aeb2,7cb7681b90388a7522d0f06578591567,ffde0649a72ded8e33522c503a4d5cbe,2010-03-08 17:08:22.030524,2010-03-12 16:09:58.122987 
6bb2ad8bc78897e99072d4d76cf0f19c,b644947ac4db03bdb518cfa71765f8c8,eb25089d396c06255cbb5f1bad801cc4,2010-03-08 17:07:55.819137,2010-03-12 16:09:58.122987

输入文件拥有数百万行，只有少数我已经发布在这里。任何建议将是有价值的。在此先感谢

来源

2016-05-12 Sitz Blogz

您可以先read_csv然后将最后两列转换为除以10**9。对于写入文件时使用to_csv：

import pandas as pd 
import numpy as np 
import io 

temp=u"""1466f7b93975983f6e292a8a4faaa4b2,1619b4d0d283c0dddb17d24a359a3b49,36db348cde68592a31d502366fc52932,2010-03-08 17:09:00.472544,2010-03-12 16:09:58.122987 
367c13356a5d22158f0ae56977134e2c,eedb7d0714796b64767a8710ea3844a7,925476200929fd346ea312cbe9a046fe,2010-03-08 17:08:29.174236,2010-03-12 16:09:58.122987 
edf6b1e4f67b0e8a5080d299c9f9aeb2,7cb7681b90388a7522d0f06578591567,ffde0649a72ded8e33522c503a4d5cbe,2010-03-08 17:08:22.030524,2010-03-12 16:09:58.122987 
6bb2ad8bc78897e99072d4d76cf0f19c,b644947ac4db03bdb518cfa71765f8c8,eb25089d396c06255cbb5f1bad801cc4,2010-03-08 17:07:55.819137,2010-03-12 16:09:58.122987"""

#after testing replace io.StringIO(temp) to filename 
df = pd.read_csv(io.StringIO(temp), 
       header=None, #no header in csv 
       names=['a','b','c','d', 'e'], #set custom column names 
       parse_dates=['d','e']) #parse columns d, e to datetime 
print df 
            a         b \ 
0 1466f7b93975983f6e292a8a4faaa4b2 1619b4d0d283c0dddb17d24a359a3b49 
1 367c13356a5d22158f0ae56977134e2c eedb7d0714796b64767a8710ea3844a7 
2 edf6b1e4f67b0e8a5080d299c9f9aeb2 7cb7681b90388a7522d0f06578591567 
3 6bb2ad8bc78897e99072d4d76cf0f19c b644947ac4db03bdb518cfa71765f8c8 

            c       d \ 
0 36db348cde68592a31d502366fc52932 2010-03-08 17:09:00.472544 
1 925476200929fd346ea312cbe9a046fe 2010-03-08 17:08:29.174236 
2 ffde0649a72ded8e33522c503a4d5cbe 2010-03-08 17:08:22.030524 
3 eb25089d396c06255cbb5f1bad801cc4 2010-03-08 17:07:55.819137 

          e 
0 2010-03-12 16:09:58.122987 
1 2010-03-12 16:09:58.122987 
2 2010-03-12 16:09:58.122987 
3 2010-03-12 16:09:58.122987 


df['d'] = df['d'].astype(np.int64) // 10**9 
df['e'] = df['e'].astype(np.int64) // 10**9

print df 
            a         b \ 
0 1466f7b93975983f6e292a8a4faaa4b2 1619b4d0d283c0dddb17d24a359a3b49 
1 367c13356a5d22158f0ae56977134e2c eedb7d0714796b64767a8710ea3844a7 
2 edf6b1e4f67b0e8a5080d299c9f9aeb2 7cb7681b90388a7522d0f06578591567 
3 6bb2ad8bc78897e99072d4d76cf0f19c b644947ac4db03bdb518cfa71765f8c8 

            c   d   e 
0 36db348cde68592a31d502366fc52932 1268068140 1268410198 
1 925476200929fd346ea312cbe9a046fe 1268068109 1268410198 
2 ffde0649a72ded8e33522c503a4d5cbe 1268068102 1268410198 
3 eb25089d396c06255cbb5f1bad801cc4 1268068075 1268410198 

df.to_csv('filename', header=None, index=False)

来源

2016-05-12 16:22:21 jezrael

太感谢你了..让我与文件读取运行它，并回信。 –

再次感谢您的编辑。我是熊猫初学者..通过阅读CSV文件的方式不是问题，但写入输出文件是;） –

没问题，我编辑解决方案。 – jezrael

Unix的日期时间正好是自1月1日的秒数，从1970年正确

所以要保证转换日期：

def dt2ut(dt): 
    epoch = pd.to_datetime('1970-01-01') 
    return (dt - epoch).total_seconds()

然后

import pandas as pd 
import numpy as np 
import io 

temp=u"""1466f7b93975983f6e292a8a4faaa4b2,1619b4d0d283c0dddb17d24a359a3b49,36db348cde68592a31d502366fc52932,2010-03-08 17:09:00.472544,2010-03-12 16:09:58.122987 
367c13356a5d22158f0ae56977134e2c,eedb7d0714796b64767a8710ea3844a7,925476200929fd346ea312cbe9a046fe,2010-03-08 17:08:29.174236,2010-03-12 16:09:58.122987 
edf6b1e4f67b0e8a5080d299c9f9aeb2,7cb7681b90388a7522d0f06578591567,ffde0649a72ded8e33522c503a4d5cbe,2010-03-08 17:08:22.030524,2010-03-12 16:09:58.122987 
6bb2ad8bc78897e99072d4d76cf0f19c,b644947ac4db03bdb518cfa71765f8c8,eb25089d396c06255cbb5f1bad801cc4,2010-03-08 17:07:55.819137,2010-03-12 16:09:58.122987""" 
#after testing replace io.StringIO(temp) to filename 
df = pd.read_csv(io.StringIO(temp), header=None, names=['a','b','c','d', 'e']) 

df['d'] = df['d'].apply(dt2ut).astype(np.int64) 
df['e'] = df['e'].apply(dt2ut).astype(np.int64)

来源

2016-05-12 16:27:39 piRSquared

我尝试比较解决方案，但我们的输出是不同的......你能检查你的解决方案吗？ – jezrael

我的appologies，转换功能上的错字。属性'秒'应该是方法'total_seconds（）' – piRSquared

非常感谢你的解决方案。 :) –

将日期时间格式转换为Unix时间戳Pandas

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