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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)
for example, as it stands if,
myArray = [4, 8, 0, 1, 5]
then Position: 2, Value: 0
I want it to present position: 3, value: 1
948
M = min( A ) returns the minimum elements of an array. If A is a vector, then min(A) returns the minimum of A . If A is a matrix, then min(A) is a row vector containing the minimum value of each column of A .
The min() function returns the item with the lowest value, or the item with the lowest value in an iterable.
Python's inbuilt function allows us to find it in one line, we can find the minimum in the list using the min() function and then use the index() function to find out the index of that minimum element.
Try this index_of_small = lst. index(min(filter(lambda x : x > 0, lst))) . Filter before finding min. Save this answer.
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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