在数组>中查找最小值0 [英] Find min value in array > 0

问题描述

我正在寻找数组中最低的正值及其在列表中的位置.如果列表中的值重复，则仅关注FIRST实例.这就是我想要的，但是要包含0.

I am looking to find the lowest positive value in an array and its position in the list. If a value within the list is duplicated, only the FIRST instance is of interest. This is what I have which does what I want but includes 0.

print "Position:", myArray.index(min(myArray))
print "Value:", min(myArray)

例如，按现状显示

myArray = [4, 8, 0, 1, 5]

然后位置:2，值:0

then Position: 2, Value: 0

我希望它显示位置:3，值:1

I want it to present position: 3, value: 1

推荐答案

您可以使用生成器表达式和min.这会将m设置为a中的最小值，该值大于0.然后使用list.index查找该值首次出现的索引.

You can use a generator expression with min. This will set m as the minimum value in a that is greater than 0. It then uses list.index to find the index of the first time this value appears.

a = [4, 8, 0, 1, 5]

m = min(i for i in a if i > 0)

print("Position:", a.index(m))
print("Value:", m)
# Position: 3
# Value: 1

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