寻找第二个最接近价值的指数 [英] Find the second closest index to value

问题描述

我正在使用

index = (np.abs(array - value)).argmin()

在数组中找到与值具有最小绝对差的索引.

to find the index in an array with the smallest absolute difference to a value.

但是，有没有一种类似的好方法可以找到与该值最接近的 second 索引?

However, is there a nice clean way such as this for finding the second closest index to the value?

推荐答案

我认为这可行

a = np.linspace(0,10,30)
array([  0.        ,   0.34482759,   0.68965517,   1.03448276,
         1.37931034,   1.72413793,   2.06896552,   2.4137931 ,
         2.75862069,   3.10344828,   3.44827586,   3.79310345,
         4.13793103,   4.48275862,   4.82758621,   5.17241379,
         5.51724138,   5.86206897,   6.20689655,   6.55172414,
         6.89655172,   7.24137931,   7.5862069 ,   7.93103448,
         8.27586207,   8.62068966,   8.96551724,   9.31034483,
         9.65517241,  10.        ])
n = np.pi
a[np.argsort(np.abs(a-n))[1]]
# Output 3.4482758620689657
# the closest value is 3.103...

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