pandas 将时间舍入到最接近的秒数 - Python
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Rounding time off to the nearest second - Python
提问by Jetman
I have a large dataset with more than 500 000 date & time stamps that look like this:
我有一个包含超过 500 000 个日期和时间戳的大型数据集，如下所示：
date time
2017-06-25 00:31:53.993
2017-06-25 00:32:31.224
2017-06-25 00:33:11.223
2017-06-25 00:33:53.876
2017-06-25 00:34:31.219
2017-06-25 00:35:12.634
How do I round these timestamps off to the nearest second?
如何将这些时间戳舍入到最接近的秒数？
My code looks like this:
我的代码如下所示：
readcsv = pd.read_csv(filename)
log_date = readcsv.date
log_time = readcsv.time
readcsv['date'] = pd.to_datetime(readcsv['date']).dt.date
readcsv['time'] = pd.to_datetime(readcsv['time']).dt.time
timestamp = [datetime.datetime.combine(log_date[i],log_time[i]) for i in range(len(log_date))]
So now I have combined the dates and times into a list of datetime.datetimeobjects that looks like this:
所以现在我将日期和时间组合成一个datetime.datetime对象列表，如下所示：
datetime.datetime(2017,6,25,00,31,53,993000)
datetime.datetime(2017,6,25,00,32,31,224000)
datetime.datetime(2017,6,25,00,33,11,223000)
datetime.datetime(2017,6,25,00,33,53,876000)
datetime.datetime(2017,6,25,00,34,31,219000)
datetime.datetime(2017,6,25,00,35,12,634000)
Where do I go from here?
The df.timestamp.dt.round('1s')function doesn't seem to be working?
Also when using .split()I was having issues when the seconds and minutes exceeded 59
我从这里去哪里？该df.timestamp.dt.round('1s')功能似乎不起作用？此外，.split()当秒和分钟超过 59 时，我遇到了问题
Many thanks
非常感谢
采纳答案by Srisaila
Using for loopand str.split():
使用for loop和str.split()：
dts = ['2017-06-25 00:31:53.993',
'2017-06-25 00:32:31.224',
'2017-06-25 00:33:11.223',
'2017-06-25 00:33:53.876',
'2017-06-25 00:34:31.219',
'2017-06-25 00:35:12.634']
for item in dts:
date = item.split()[0]
h, m, s = [item.split()[1].split(':')[0],
item.split()[1].split(':')[1],
str(round(float(item.split()[1].split(':')[-1])))]
print(date + ' ' + h + ':' + m + ':' + s)
2017-06-25 00:31:54
2017-06-25 00:32:31
2017-06-25 00:33:11
2017-06-25 00:33:54
2017-06-25 00:34:31
2017-06-25 00:35:13
>>>
You could turn that into a function:
你可以把它变成一个函数：
def round_seconds(dts):
result = []
for item in dts:
date = item.split()[0]
h, m, s = [item.split()[1].split(':')[0],
item.split()[1].split(':')[1],
str(round(float(item.split()[1].split(':')[-1])))]
result.append(date + ' ' + h + ':' + m + ':' + s)
return result
Testing the function:
测试功能：
dts = ['2017-06-25 00:31:53.993',
'2017-06-25 00:32:31.224',
'2017-06-25 00:33:11.223',
'2017-06-25 00:33:53.876',
'2017-06-25 00:34:31.219',
'2017-06-25 00:35:12.634']
from pprint import pprint
pprint(round_seconds(dts))
['2017-06-25 00:31:54',
'2017-06-25 00:32:31',
'2017-06-25 00:33:11',
'2017-06-25 00:33:54',
'2017-06-25 00:34:31',
'2017-06-25 00:35:13']
>>>
Since you seem to be using Python 2.7, to drop any trailing zeros, you may need to change:
由于您似乎在使用 Python 2.7，要删除任何尾随零，您可能需要更改：
str(round(float(item.split()[1].split(':')[-1])))
str(round(float(item.split()[1].split(':')[-1])))
to
到
str(round(float(item.split()[1].split(':')[-1]))).rstrip('0').rstrip('.')
str(round(float(item.split()[1].split(':')[-1]))).rstrip('0').rstrip('.')
I've just tried the function with Python 2.7 at repl.itand it ran as expected.
我刚刚在repl.it 上使用 Python 2.7 尝试了该函数，它按预期运行。
回答by electrovir
Without any extra packages, a datetime object can be rounded to the nearest second with the following simple function:
没有任何额外的包，日期时间对象可以使用以下简单函数四舍五入到最接近的秒：
import datetime
def roundSeconds(dateTimeObject):
newDateTime = dateTimeObject
if newDateTime.microsecond >= 500000:
newDateTime = newDateTime + datetime.timedelta(seconds=1)
return newDateTime.replace(microsecond=0)
回答by cs95
If you're using pandas, you can just roundthe data to the nearest second using dt.round-
如果您使用的是Pandas，则可以使用以下命令round将数据精确到最近的秒数dt.round-
df
timestamp
0 2017-06-25 00:31:53.993
1 2017-06-25 00:32:31.224
2 2017-06-25 00:33:11.223
3 2017-06-25 00:33:53.876
4 2017-06-25 00:34:31.219
5 2017-06-25 00:35:12.634
df.timestamp.dt.round('1s')
0 2017-06-25 00:31:54
1 2017-06-25 00:32:31
2 2017-06-25 00:33:11
3 2017-06-25 00:33:54
4 2017-06-25 00:34:31
5 2017-06-25 00:35:13
Name: timestamp, dtype: datetime64[ns]
If timestampisn't a datetimecolumn, convert it first, using pd.to_datetime-
如果timestamp不是datetime列，请先转换它，使用pd.to_datetime-
df.timestamp = pd.to_datetime(df.timestamp)
Then, dt.roundshould work.
那么，dt.round应该工作。
回答by mike rodent
The question doesn't say howyou want to round. Rounding down would often be appropriate for a time function. This is not statistics.
这个问题没有说明你想如何舍入。向下舍入通常适用于时间函数。这不是统计。
rounded_down_datetime = raw_datetime.replace(microsecond=0)
回答by Maciej Gumu?ka
If anyone wants to round a single datetime item off to the nearest second, this one works just fine:
如果有人想将单个日期时间项舍入到最近的秒数，那么这个就可以了：
pandas.to_datetime(your_datetime_item).round('1s')
回答by Rudradev Pal
If you are storing dataset into a file you can do like this:
如果您要将数据集存储到文件中，您可以这样做：
with open('../dataset.txt') as fp:
line = fp.readline()
cnt = 1
while line:
line = fp.readline()
print "\n" + line.strip()
sec = line[line.rfind(':') + 1:len(line)]
rounded_num = int(round(float(sec)))
print line[0:line.rfind(':') + 1] + str(rounded_num)
print abs(float(sec) - rounded_num)
cnt += 1
If you are storing dataset in a list:
如果您将数据集存储在列表中：
dts = ['2017-06-25 00:31:53.993',
'2017-06-25 00:32:31.224',
'2017-06-25 00:33:11.223',
'2017-06-25 00:33:53.876',
'2017-06-25 00:34:31.219',
'2017-06-25 00:35:12.634']
for i in dts:
line = i
print "\n" + line.strip()
sec = line[line.rfind(':') + 1:len(line)]
rounded_num = int(round(float(sec)))
print line[0:line.rfind(':') + 1] + str(rounded_num)
print abs(float(sec) - rounded_num)
回答by Gerardsson
An elegant solution that only requires the standard datetime module.
一个优雅的解决方案，只需要标准的 datetime 模块。
import datetime
currentimemili = datetime.datetime.now()
currenttimesecs = currentimemili - \
datetime.timedelta(microseconds=currentimemili.microsecond)
print(currenttimesecs)
回答by gerardw
Alternate version of @electrovir 's solution:
@electrovir 解决方案的替代版本：
import datetime
def roundSeconds(dateTimeObject):
newDateTime = dateTimeObject + datetime.timedelta(seconds=.5)
return newDateTime.replace(microsecond=0)
