在python中将numpy.datetime64转换为字符串对象
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/19502506/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Convert numpy.datetime64 to string object in python
提问by cs_newbie
I am having trouble converting a python datetime64object into a string. For example:
我在将 pythondatetime64对象转换为字符串时遇到问题。例如：
t = numpy.datetime64('2012-06-30T20:00:00.000000000-0400')
Into:
进入：
'2012.07.01' as a string. (note time difference)
I have already tried to convert the datetime64object to a datetimelong then to a string, but I seem to get this error:
我已经尝试将datetime64对象转换为datetimelong 然后转换为字符串，但我似乎收到此错误：
dt = t.astype(datetime.datetime) #1341100800000000000L
time.ctime(dt)
ValueError: unconvertible time
回答by cs_newbie
Solution was:
解决方案是：
import pandas as pd
ts = pd.to_datetime(str(date))
d = ts.strftime('%Y.%m.%d')
回答by Wirawan Purwanto
There is a route without using pandas; but see caveat below.
有一条不使用pandas的路由；但请参阅下面的警告。
Well, the tvariable has a resolution of nanoseconds, which can be shown by inspection in python:
好吧，该t变量的分辨率为纳秒，可以通过在 python 中检查来显示：
>>> numpy.dtype(t)
dtype('<M8[ns]')
This means that the integral value of this value is 10^9 times the UNIX timestamp. The value printed in your question gives that hint. Your best bet is to divide the integral value of tby 1 billion then you can use time.strftime:
这意味着该值的整数值是 UNIX 时间戳的 10^9 倍。您的问题中打印的值给出了提示。最好的办法是将 的积分值t除以 10 亿，然后您可以使用time.strftime：
>>> import time
>>> time.strftime("%Y.%m.%d", time.gmtime(t.astype(int)/1000000000))
2012.07.01
In using this, be conscious of two assumptions:
在使用它时，请注意两个假设：
1) the datetime64 resolution is nanosecond
1) datetime64 分辨率为纳秒
2) the time stored in datetime64 is in UTC
2) datetime64 中存储的时间是UTC
Side note 1: Interestingly, the numpy developers decided [1] that datetime64object that has a resolution greater than microsecond will be cast to a longtype, which explains why t.astype(datetime.datetime)yields 1341100800000000000L. The reason is that datetime.datetimeobject can't accurately represent a nanosecond or finer timescale, because the resolution supported by datetime.datetimeis only microsecond.
旁注 1：有趣的是，numpy 开发人员决定 [1]datetime64分辨率大于微秒的对象将被强制转换为long类型，这解释了为什么t.astype(datetime.datetime)yield 1341100800000000000L。原因是datetime.datetime对象不能准确表示纳秒或更精细的时间尺度，因为其支持的分辨率datetime.datetime只有微秒。
Side note 2: Beware the different conventions between numpy 1.10 and earlier vs 1.11 and later:
旁注 2：注意 numpy 1.10 及更早版本与 1.11 及更高版本之间的不同约定：
in numpy <= 1.10, datetime64 is stored internally as UTC, and printed as local time. Parsing is assuming local time if no TZ is specified, otherwise the timezone offset is accounted for.
in numpy >= 1.11, datetime64 is stored internally as timezone-agnostic value (seconds since 1970-01-01 00:00 in unspecified timezone), and printed as such. Time parsing does not assume the timezone, although
+NNNNstyle timezone shift is still permitted and that the value is converted to UTC.
在 numpy <= 1.10 中，datetime64 在内部存储为 UTC，并打印为本地时间。如果未指定 TZ，则解析假定本地时间，否则会考虑时区偏移。
在 numpy >= 1.11 中，datetime64 在内部存储为与时区无关的值（未指定时区中自 1970-01-01 00:00 以来的秒数），并按原样打印。时间解析不假定时区，尽管
+NNNN仍然允许样式时区转换并且该值转换为 UTC。
[1]: https://github.com/numpy/numpy/blob/master/numpy/core/src/multiarray/datetime.csee routine convert_datetime_to_pyobject.
[1]：https: //github.com/numpy/numpy/blob/master/numpy/core/src/multiarray/datetime.c见例程convert_datetime_to_pyobject。
回答by citynorman
If you don't want to do that conversion gobbledygook and are ok with just one date format, this was the best solution for me
如果您不想进行这种转换 gobbledygook 并且只使用一种日期格式就可以了，那么这对我来说是最好的解决方案
str(t)[:10]
Out[11]: '2012-07-01'
As noted this works for pandas too
如前所述，这也适用于熊猫
df['d'].astype(str).str[:10]
df['d'].dt.strftime('%Y-%m-%d') # equivalent
回答by four43
I wanted an ISO 8601 formatted string without needing any extra dependencies. My numpy_array has a single element as a datetime64. With help from @Wirawan-Purwanto, I added just a bit:
我想要一个 ISO 8601 格式的字符串，而不需要任何额外的依赖项。我的 numpy_array 有一个作为 datetime64 的元素。在@Wirawan-Purwanto 的帮助下，我补充了一点：
from datetime import datetime
ts = numpy_array.values.astype(datetime)/1000000000
return datetime.utcfromtimestamp(ts).isoformat() # "2018-05-24T19:54:48"
回答by Julie
You can use Numpy's datetime_as_stringfunction. The unit='D'argument specifies the precision, in this case days.
您可以使用 Numpy 的datetime_as_string功能。该unit='D'参数指定的精度，在这种情况下天。
>>> t = numpy.datetime64('2012-06-30T20:00:00.000000000-0400')
>>> numpy.datetime_as_string(t, unit='D')
'2012-07-01'
回答by tommy.carstensen
Building on this answerI would do the following:
基于此答案，我将执行以下操作：
import numpy
import datetime
t = numpy.datetime64('2012-06-30T20:00:00.000000000')
datetime.datetime.fromtimestamp(t.item() / 10**9).strftime('%Y.%m.%d')
The division by a billion is to convert from nanoseconds to seconds.
除以十亿是将纳秒转换为秒。
