如何使用 pandas.date_range() 在指定的开始日期和结束日期之间获取具有 n 个指定周期（相等）的时间序列
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How can I use pandas.date_range() to obtain a time series with n specified periods (equal) between a specified start and end date
提问by arosner09
I'd like to get a list or series of n dates between a start and end date (inclusive of those bounds), but
我想在开始日期和结束日期（包括这些界限）之间获得一个列表或一系列 n 个日期，但是
dateIndex=pd.date_range(start=dt.datetime.today().date(), end=pd.to_datetime(expiry).date(), periods=n)
results with ValueError: Must specify two of start, end, or periods. I cannot use freq=Freq argument because my date range won't be uniform - it may be anywhere from a month to 2 years span, thus I'd like an equally spaced time series with n points.
带有 ValueError 的结果：必须指定开始、结束或句点中的两个。我不能使用 freq=Freq 参数，因为我的日期范围不会是统一的 - 它可能是一个月到 2 年的任何时间跨度，因此我想要一个具有 n 点的等距时间序列。
Thanks!
谢谢！
回答by Andy Hayden
I don't think you can do this just with date_range, but why not use numpy's linspace:
我认为你不能只用 来做到这一点date_range，但为什么不使用 numpy 的linspace：
In [11]: start = pd.Timestamp('2012-01-01')
In [12]: end = pd.Timestamp('2012-02-01')
In [13]: np.linspace(start.value, end.value, 10) # 10 dates inclusive
Out[13]:
array([ 1.32537600e+18, 1.32567360e+18, 1.32597120e+18,
1.32626880e+18, 1.32656640e+18, 1.32686400e+18,
1.32716160e+18, 1.32745920e+18, 1.32775680e+18,
1.32805440e+18])
In [14]: pd.to_datetime(np.linspace(start.value, end.value, 10))
Out[14]:
<class 'pandas.tseries.index.DatetimeIndex'>
[2012-01-01 00:00:00, ..., 2012-02-01 00:00:00]
Length: 10, Freq: None, Timezone: None
You couldpass this as a freq, but this may/will be inaccurate for times which don't evenly divide:
您可以将其作为频率传递，但是对于不均匀划分的时间，这可能/将不准确：
In [21]: (end - start)/ 9
Out[21]: datetime.timedelta(3, 38400)
In [22]: ((end - start)/ 9).total_seconds()
Out[22]: 297600.0
# Note: perhaps there's a better way to pass this as a freq?
In [23]: pd.date_range(start=start, end=end, freq='%iS' % ((end - start)/ 9).total_seconds())
Out[23]:
<class 'pandas.tseries.index.DatetimeIndex'>
[2012-01-01 00:00:00, ..., 2012-02-01 00:00:00]
Length: 10, Freq: 297600S, Timezone: None
回答by Joooeey
As of Pandas 0.23 (or earlier), you can just use pandas.date_rangelike you originally tried. It does not raise an error and behaves as you would expect. Example:
从 Pandas 0.23（或更早版本）开始，您可以pandas.date_range像最初尝试的那样使用。它不会引发错误并按您的预期运行。例子：
pd.date_range('2016-01-01', '2017-01-01', periods=13, tz='utc')
Out[44]:
DatetimeIndex(['2016-01-01 00:00:00+00:00', '2016-01-31 12:00:00+00:00',
'2016-03-02 00:00:00+00:00', '2016-04-01 12:00:00+00:00',
'2016-05-02 00:00:00+00:00', '2016-06-01 12:00:00+00:00',
'2016-07-02 00:00:00+00:00', '2016-08-01 12:00:00+00:00',
'2016-09-01 00:00:00+00:00', '2016-10-01 12:00:00+00:00',
'2016-11-01 00:00:00+00:00', '2016-12-01 12:00:00+00:00',
'2017-01-01 00:00:00+00:00'],
dtype='datetime64[ns, UTC]', freq=None)
There were 366 days in 2016 (a leap year), so time stamps are 30.5 days apart.
2016 年有 366 天（闰年），因此时间戳相隔 30.5 天。
