Python 查找最接近给定日期的日期
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Find the closest date to a given date
提问by user3600497
I have an array of datetime objects, and I would like to find which element in the array is the closest to a given date (e.g datetime.datetime(2014,12,16))
我有一个 datetime 对象数组，我想找到数组中最接近给定日期的元素（例如datetime.datetime(2014,12,16)）
Thispost shows how to find the nearest date which is not before the given date. How can I alter this code so that it can return dates that are before a given date?
这篇文章展示了如何找到不在给定日期之前的最近日期。如何更改此代码以便它可以返回给定日期之前的日期？
For example, if the array housed elements datetime.datetime(2014,12,10)and datetime.datetime(2014,12,28), the former item should be returned because it is closest to datetime.datetime(2014,12,16)in absolute value.
例如，如果数组包含元素datetime.datetime(2014,12,10)and datetime.datetime(2014,12,28)，则应返回前一项，因为它datetime.datetime(2014,12,16)在绝对值上最接近。
采纳答案by Tamas Hegedus
This function will return the datetimein itemswhich is the closest to the date pivot.
该函数将返回datetime在items其最接近的日期pivot。
def nearest(items, pivot):
return min(items, key=lambda x: abs(x - pivot))
The good part this function works on types other than datetimetoo out of the box, if the type supports comparison, subtraction and abs, e.g.: numbers and vector types.
datetime如果类型支持比较、减法和abs，例如：数字和向量类型，则此函数适用于除开箱即用以外的类型。
回答by 3442
def nearestDate(base, dates):
nearness = { abs(base.timestamp() - date.timestamp()) : date for date in dates }
return nearness[min(nearness.keys())]
回答by Kevin Zhu
As answered on this linklink, 'truncate' function is there for you.
正如此链接链接所回答的那样，“截断”功能就在您身边。
df.truncate(before='2012-01-07')
df.truncate(before='2012-01-07')
Or you can use get_locwith 'nearest' option.
或者您可以将get_loc与 'nearest' 选项一起使用。
df.iloc[df.index.get_loc(datetime.datetime(2016,02,02),method='nearest')]
回答by MarMat
To find a closest date andreturn the timedelta (difference between two dates) I did the following:
要找到最近的日期并返回时间增量（两个日期之间的差异），我执行了以下操作：
def nearest_date(items,pivot):
nearest=min(items, key=lambda x: abs(x - pivot))
timedelta = abs(nearest - pivot)
return nearest, timedelta
This may be useful when you have a minimum threshold for nearness for your app like I did.
当您像我一样对应用程序有一个最小接近度阈值时，这可能很有用。
回答by Lorenzo Meneghetti
My solution to find the closest index instead of the value
我的解决方案是找到最接近的索引而不是值
def nearest_ind(items, pivot):
time_diff = np.abs([date - pivot for date in items])
return time_diff.argmin(0)
回答by Chiel
This code returns the nearest date beforethe given date:
此代码返回给定日期之前最近的日期：
def nearest(items, pivot):
return min([i for i in items if i < pivot], key=lambda x: abs(x - pivot))
