在Python中查找给定数组中最小值的索引
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Find the index of minimum values in given array in Python
提问by user2766019
I need to find the index of more than one minimum values that occur in an array. I am pretty known with np.argminbut it gives me the index of very first minimum value in a array. For example.
我需要找到数组中出现的多个最小值的索引。我很熟悉，np.argmin但它给了我数组中第一个最小值的索引。例如。
a = np.array([1,2,3,4,5,1,6,1])
print np.argmin(a)
This gives me 0, instead I am expecting, 0,5,7.
这给了我 0，而不是我期待，0,5,7。
Thanks!
谢谢！
采纳答案by Tom Swifty
This should do the trick:
这应该可以解决问题：
a = np.array([1,2,3,4,5,1,6,1])
print np.where(a == a.min())
argmin doesn't return a list like you expect it to in this case.
在这种情况下，argmin 不会像您期望的那样返回列表。
回答by tonjo
Maybe
也许
mymin = np.min(a)
min_positions = [i for i, x in enumerate(a) if x == mymin]
It will give [0,5,7].
它将给出 [0,5,7]。
回答by jrk0414
I think this would be the easiest way, although it doesn't use any fancy numpy function
我认为这将是最简单的方法，尽管它没有使用任何花哨的 numpy 函数
a = np.array([1,2,3,4,5,1,6,1])
min_val = a.min()
print "min_val = {0}".format(min_val)
# Find all of them
min_idxs = [idx for idx, val in enumerate(a) if val == min_val]
print "min_idxs = {0}".format(min_idxs)
