Finding the minimum of a nested list that has a constant in element[1]-CodePudding

I have a nested list:

list = [[4, 2], [3, 2], [1, 1]]

And I want to find the minimum number in index[0] that has a constant (value) of 2 in index[1].

For this list elements 1 and 2 have a value of 2 at index[1] within the nested list therefore the 4 and 3 fit the criteria and the min number is 3 so the output should be 3

    for val in freqList:
        print(val[0])

Gives me the values in index[0] of the list but I'm not sure how to only print the values of index[0] that have a value of 2 in index[1] and then how to select the minimum.

Any ideas on how to do this?

CodePudding user response：

If you want to get minimum of the first elements in the inner lists:

min([val[0] for val in freqList])

Also if you want to check inner lists for conditions:

min([val[0] for val in freqList if CONDITION)

which if your condition is val[1] == 2 (Also your question's answer)

min([val[0] for val in freqList if val[1] == 2])

or even you can check for multiple condition:

min([val[0] for val in freqList if ((val[1] == 2) and (len(val) == 2))])

CodePudding user response：

You can define function like this:

def find_minimum(where, const_id, const_val):
    values = []

    for i in where:
        if i[const_id] == const_val:
            values.append(i[0])

    return min(values)

Usage:

list = [[4, 2], [3, 2], [1, 1]]
minimum = find_minimum(list, 1, 2)
print(minimum)

CodePudding user response：

I am a beginner in Python so my solution can be too complicated.

I can suggest the following approach

lst2 = []
for sublist in lst:
    if len( sublist ) >= 2 and sublist[1] == 2:
        lst2.append( sublist[0] )
min_value = min( lst2 )

where lst is you original list of lists.

Using the built list lst2 you can output all values at index 0 of sub-lists that have at index 1 a value equal to 2.

For example

for value in lst2:
    print( value, end = ' ' )
print()
print( min( lst2 ) )