CODEWITHSUNDEEP
pythonlistduplicates
Gibbs
Gibbs

Reputation: 17

List Returns firts Item of Duplicates Work Around Python

Good[ Morning : Evening ]

Im trying to make a small programme that :

-reads a text file containing numbers -assigned each line into an element of int_list

-takes a given list (max_quantity) and divides all the elements by 2 resulting in (max_q_in_half)

-then make all the elements that are below 50% of their maximum value into a new list (new_list)

-then return the smallest number in the (new_list)

-then uses that number to get it's index in the original (int_list)

-then uses that index to match the number to it's element in (metal_list)

HERE IS THE PROBLME

As you see in the output of this code the numbers below 50% are [41, 1]
the lowest value is 1 which has the index 6

but when i search int_list for this value it returns the firt occurence of 1 which is at index 3 which coresponds to BRASS but the correct answer should be Titanium

im triying to work around this issue by assigning a weight to each line in the text file so when i search the index of a value that value would be unique take into account that the lines vary in number in the original text file

could somebody help me make a programe that assignes wheights of X.01 to line one and X.02 to line two ...

-41.01 #Line_1

-20.02 #Line_2

-6.03 #Line_3

-1.04 #Line_4

-14.05 #Line_5

-6.06 #Line_6

-1.07 #Line_7 ...

int_list = [41, 20, 6, 1, 14, 6, 1]

metal_list = ['aluminium', 'iron', 'zinc', 'brass', 'stainless steel','copper', 'titanium']

max_quantity = [ 100, 30, 6, 2, 18, 7, 15]

max_q_in_half = [x // 2 for x in max_quantity]


new_list = [i for i, j in zip(int_list, max_q_in_half) if i < j]

x = min(new_list)

v = int_list.index(x)
w = metal_list.__getitem__(v)


print("Current List ", int_list)

print("what is below 50%",new_list)
print('lowest value ',x)


print("position in int_list  ",v)

print("position Metal_list ",w)

Upvotes: 1

Views: 43

Answers (1)

Сергей Кох
Сергей Кох

Reputation: 1723

You can divide the elements of the list by 2 when checking. It is better to number it when creating a list, but with the second member, so that the number does not interfere when checking for a minimum.

int_list = [41, 20, 6, 1, 14, 6, 1]
metal_list = ['aluminium', 'iron', 'zinc', 'brass', 'stainless steel', 'copper', 'titanium']
max_quantity = [100, 30, 6, 2, 18, 7, 15]

x = min([(i, n) for n, (i, j) in enumerate(zip(int_list, max_quantity)) if i < j//2])
print(x)
w = metal_list[x[1]]
print("position Metal_list ", w)

(1, 6)
position Metal_list  titanium

Upvotes: 1

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