In a Django template, how do I return the first element of a list that matches a condition? The list could be empty. I want to know if there is a filter that finds the first element in an iterable to match a particular condition, then stops checking, similar to Python's next (https://stackoverflow.com/a/2364277/1337422).
>>> lis = ['a', 1, 40, 'three', 5]
>>> next(x for x in lis if x == 1)
1
There isn't a built in Django template filter that does this.
If it is a particular condition you want to match, then it would be easy to write a custom template filter. For example:
from django import template
register = template.Library()
@register
def first_one_in_list(l):
return next((x for x in l if x==1), default=None) # specify default to avoid stop iteration
If you want to specify an arbitrary condition in the template, I think you'd need to do something hacky. The Django template language is designed to prevent complex logic in templates.
