Meaning "if each item is within range of other item with the same index".
price = [1, 2]
budget = [5, 7]
This works:
if price[0] in range(budget[0]) and price[1] in range(budget[1]):
affordable = True
I figure there's some way to just reference the whole array though. Like so: if price in budget:
You could use:
if all(x in range(y) for x,y in zip(price,budget)):
affordable = True
This will create tuples of price[i],budget[i] and then for each of these tuples we check that price[i] is in range(budget[i]). Nevertheless, you can optimize this further to:
if all(0 <= x < y for x,y in zip(price,budget)):
affordable = TrueNote that this makes the assumption that prices are all integers. If you however use x in range(y) it will fail if x is not an integer. So 0.7 in range(10) would fail whereas our second approach will succeed (but it depends of course on what you want).
0<=x<=y (or similar) instead of x in range(y) will deal with floats and not compile (and iterate; PY2) a list
if any(x not in range(y) for x, y in zip(price, budget)):range builds (compiles) a list of integers first, and then must iterate it again for the contains check, which seems like a lot work to check if some numerical values lies between two other numerical values. compile = build, put together, not in the computational sense of 'translate' ;)Assuming that both prices and budgets must be non-negative, using in range seems to be over-complicating things. Instead, you could just use the < operator.
Regardless of whether you use < or in range, it seems like the easiest approach would be to zip both lists and apply the condition on the pairs:
if (all([x[0] >= x[1] for x in zip(budget, price)])):
affordable = True
all; e.g.if all(price[i] in range(budget[i]) for i in range(...))orall(p in range(b) for p,b in zip(price, budget))if that's what you really want.