I've list and I want to get the other element when I found the value I'm looking for. For example:
list = [1,2,3,4,5,6]
for element in list:
if element > 3:
variable = element
break
print(element)
4
So I want to get the 5. Is there a function to do that ?
result = [a for a in list if a > 3]
This returns an array of all elements you're looking for.
I am not entirely sure what you mean. Do you mean that you want to skip one element and return the second element that is found? If that is the case, your code should be
list = [1,2,3,4,5,6]
found1 = false
for element in list:
if element > 3:
if(!found1):
found1 = true;
else:
variable = element
break
print(variable)
One way you could do this is to use an index, i. By using enumerate I get back both the element and its index in the list. I call the list my_list rather than list since list is a reserved word already. Also using [:-1] to go one short of end of list, since if the last element matches there is no next element.
Other than that I tried to keep everything the same as you did it.
variable = None
my_list = [1,2,3,4,5,6]
for i, element in enumerate(my_list[:-1]):
if element > 3:
variable = my_list[i + 1]
break
print(variable)
The simplest solution I can think of:
list = [1,2,3,4,5,6]
found = False
for element in list:
if found:
break
if element > 3:
found = True
print(element)
This always shows the element that comes after the element that first satisfies your condition.
If you happen to know the exact value you're looking for, then simply use it as a parameter of index(), like this, to show the element coming after that:
list[list.index(4)+1]
But you can't use an if condition this way, you can only look up a specific value.
