This is what I try:
example1:
a=['1','2','3','4']
for m in range(len(a)):
int(a[m])
print(a)
But then I realized that for example in:
example 2:
a=['1','2','3','4']
b= []
int(a[0])
b = b + [a[0]]
print(b)
#Output b = ['1']
Which means that int (a [0]) does not do anything that I could check. Could you give me advice on what I could do? I'm a little lost here.
I think you're under the impression of doing this:
int(a[0])
before this:
b = b + [a[0]]
It has the effect of changing the first element of the array a from string to int.
This is not the case, in fact, int(a[0]) returns you a 1 , which is the result of transforming the component 0 of the array a into%. But this operation does not change the array element in the array itself, but returns a copy of a[0] already transformed to integer type.
To achieve what you want, you must do this:
b = b + [int(a[0])]
In this way, you create a list that contains the number 1 by means of the expression [int(a[0])] and then unes the elements of the list b and the list [int(a[0])] with the expression b = b + [int(a[0])] .
with the function int () combierte string a int
a=['1','2','3','4']
b= [int(x) for x in a]
print(b)