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time loc 0 2014-12-08 18 ad 1 2014-12-09 12 as 2 2014-12-12 12 xs 处理为
time loc 0 1 ad 1 2 as 2 5 xs 其中 "2014-12-08 18" 代表 14 年 12 月 8 日的 18 点，想把时间这列更新为从12 月 8 日算起的第X天，应该如何操作？
9 replies 2018-04-09 17:42:39 +08:00
 | 1 aaronzjw Sep 1, 2016 import time 应该可以吧 |
 | 2 20150517 Sep 2, 2016 via iPhone apply |
 | 3 wickila Sep 2, 2016  1 import pandas as pd import datetime def date2day(x): sd = datetime.datetime.strptime('2014-12-08 0', '%Y-%M-%d %H') d = datetime.datetime.strptime(x['time'], '%Y-%M-%d %H') x['time'] = (d - sd).days + 1 return x df = pd.DataFrame([['2014-12-08 18', 'ad'], ['2014-12-09 12', 'as'], ['2014-12-12 12', 'xs']], columns=['time', 'loc']) df = df.apply(date2day, axis=1) print df |
 | 4 xixijun Sep 2, 2016 df['date_diff']=df['time'].diff().fillna(0)+pd.Timedelta('1 days') |
 | 9 weimao Apr 9, 2018 td = _df['time'] time = pd.to_datetime(td) start = pd.datetime(2014, 12, 8) day = time - start _df['time'] = day.dt.days |
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