True if data in a row

Question

I want this function to print True if 200 occurs 3 times in a row. False otherwise.

def aList(n):
    
    if n.count(200) >= 3:
        return True
    else:
        return False

print(aList([100, 200, 200, 300, 200]))

Currently it is returning True because 200 exists 3 or more times, however it should return False because it is not occurring in a row.

Edit: Even if I iterate over the list, how can I pull True if the occurrence happens x amount of times.

for i in range(len(n)):
        if i == 200:
            return True
    else:
        return False

Asked By: cnsc.bot

||

Source

Answer 1

So what you are trying to accomplish requires a consideration for positional elements… so you would need a counter to keep track of the counts and revert it back to zero when it does not match.. Little bit of pseudocode here, but something like this..

def nums_in_row(aList, target_num):
    count = 0
    for nums in aList:
        if count ==3:
            return True
        elif nums == target_num:
            count+=1
        else:
            count = 0
    return False

target_num would be 200 in your example and aList would be the list itself. You could generalize your function further, but based on your question, it seems to suffice.

Answered By: Aaron Gonzalez

Answer 2

You need a variable to keep track of how many times you have seen the number in a row. Add one when you match the value and reset the count to 0 if you don’t. Return true if you get to the threshold. If you finish the loop without hitting the threshold, return false.

def aList(n):
    count = 0
    for i in n:
        if i == 200:
            count += 1
        else:
            count = 0

        if count == 3:
            return True
    return False

print(aList([100, 200, 200, 300, 200]))  # -> False
print(aList([100, 200, 200, 200]))  # -> True

Answered By: Patrick

True if data in a row

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