如何在列表压缩中存储来自调用 function 的多个返回变量中的一个变量？

Question

I want outputs like output from following code function:

def addMulDiv(i, k):
    add=k+i
    mul=k*i
    div=k//2
    return add, mul, div

Lst = []
for i in range(1, 10):
    add, _, _ = addMulDiv(i, 500)
    Lst.append(add)
    
print(Lst)

output: [501, 502, 503, 504, 505, 506, 507, 508, 509]

But I am trying with following code in List compression but I failed.

Lst = [add for i in range(1, 10) add, _, _ = addMulDiv(i, 500)]

How can I do this easily?

Thanks.

Answer 1

Your function addMulDiv() returns a tuple, so you can just use

Lst = [addMulDiv(i, 500)[0] for i in range(1, 10)]

Answer 2

Your function is actually returning a tuple. Hence, you can fetch the first element using [0]

Lst1 = [addMulDiv(i,500)[0] for i in range(1,10)]

Answer 3

You don't fully understood how List compressions works. For instance,

[i ** 2 for i in range(10) if i % 2 == 0]

This means that you get only even numbers from range and raise it to the power of 2.

output : [4, 16, 36, 64]

Solution for your problem:

def addMulDiv(i, k):
    add = k + i
    mul = k * i
    div = k // 2
    return add, mul, div


Lst = [addMulDiv(i, 500)[0] for i in range(1, 10)]
print(Lst)

This means that you get only add result from you function return, because add result has 0 index in returned tuple