Pandas：时间戳索引四舍五入到最接近的第 5 分钟
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Pandas: Timestamp index rounding to the nearest 5th minute
提问by Plug4
I have a dfwith the usual timestamps as an index:
我有一个df通常的时间戳作为索引：
2011-04-01 09:30:00
2011-04-01 09:30:10
...
2011-04-01 09:36:20
...
2011-04-01 09:37:30
How can I create a column to this dataframe with the same timestamp but rounded to the nearest 5th minute interval? Like this:
如何使用相同的时间戳为该数据帧创建一列但四舍五入到最接近的第 5 分钟间隔？像这样：
index new_col
2011-04-01 09:30:00 2011-04-01 09:35:00
2011-04-01 09:30:10 2011-04-01 09:35:00
2011-04-01 09:36:20 2011-04-01 09:40:00
2011-04-01 09:37:30 2011-04-01 09:40:00
回答by cronos
The round_to_5min(t)solution using timedeltaarithmeticis correct but complicated and very slow. Instead make use of the nice Timstampin pandas:
round_to_5min(t)使用timedelta算术的解决方案是正确的，但复杂且非常缓慢。而是使用Timstamppandas中的 nice ：
import numpy as np
import pandas as pd
ns5min=5*60*1000000000 # 5 minutes in nanoseconds
pd.to_datetime(((df.index.astype(np.int64) // ns5min + 1 ) * ns5min))
Let's compare the speed:
我们来比较一下速度：
rng = pd.date_range('1/1/2014', '1/2/2014', freq='S')
print len(rng)
# 86401
# ipython %timeit
%timeit pd.to_datetime(((rng.astype(np.int64) // ns5min + 1 ) * ns5min))
# 1000 loops, best of 3: 1.01 ms per loop
%timeit rng.map(round_to_5min)
# 1 loops, best of 3: 1.03 s per loop
Just about 1000 times faster!
大约快 1000 倍！
回答by dustyrockpyle
You can try something like this:
你可以尝试这样的事情：
def round_to_5min(t):
delta = datetime.timedelta(minutes=t.minute%5,
seconds=t.second,
microseconds=t.microsecond)
t -= delta
if delta > datetime.timedelta(0):
t += datetime.timedelta(minutes=5)
return t
df['new_col'] = df.index.map(round_to_5min)
回答by Guido
回答by ShaharA
I had the same problem but with datetime64p[ns] timestamps.
我有同样的问题，但 datetime64p[ns] 时间戳。
I used:
我用了：
def round_to_5min(t):
""" This function rounds a timedelta timestamp to the nearest 5-min mark"""
t = datetime.datetime(t.year, t.month, t.day, t.hour, t.minute - t.minute%5, 0)
return t
followed by the the 'map' function
其次是“地图”功能
