I have this grid array which contains numbers from 0 to 7. A slot can be considered free on the grid only if it is a 0 and there's no different number before it:
[0, 0, 0, 0, 1, 0, 0, 0] # every 0 before the 1 is a free slot
I want to make it so that the program can find the first slot that contains a number from 1 to 7, so that it knows that all the previous slots are free. I can make it work using a loop, like:
array = [0, 0, 0, 0, 1, 0, 0, 0]
freeslot = len(array)-1 # initializing in case it does not find a match
for n in range(1, 8):
if (n in array) and (array.index(n)-1 < freeslot):
freeslot = array.index(n)-1
The thing is, whenever I'm running this to check the grid, I check for 15 different arrays, so I feel that a loop makes it too slow. Is there a way to make it not using loops, or at least make it more efficient?
CodePudding user response:
This should work. freeslot = list(map(lambda x: True if 7 >= x > 0 else False, array)).index(True)
. Basically what it does is replace everything in a array list copy that is less than 7 but over 1 and replaces it with True. Then it indexes the first case of True in the new copy.
CodePudding user response:
A possible solution is to use a numpy array and have a look at this solution: Numpy first occurrence of value greater than existing value
CodePudding user response:
ans = len(arr) - 1
for i, v in enumerate(arr):
if v > 0:
ans = i - 1
break
ans
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