I am having a bit of troubles getting this part of my code to work, I need to find a way to count the amount of times than an H occurs in a row in a list, (H, H, H) would count as two occurrences. I feel like I'm getting somewhere but just can't get the j to count up.
list['H', 'H', 'T', 'H']
j = 0
n = 0
for letter in list:
if list[n] == ['H'] and list[n+1] == ['H']:
j = j + 1
print('it was this amount of times ' + str(j))
You could use a regex for this:
len(re.findall("H(?=H)", "HHH"))
The trick is using lookahead to match the second H, but still allow it to be matched again.
Before you think "regex is overkill for this and hard to read," two counter variables, indexing, and math has a lot more places for something to go wrong.
Try this
list= ['H', 'H', 'T', 'H']
j = 0
n = 0
for letter in list:
if letter == 'H' and list[n+1] == 'H':
j = j + 1
print('it was this amount of times ' + str(j))
Will give
it was this amount of times 3
n+1 will always check the second element in this example.Did not get how ('H', 'H', 'H') counted to 2, is it that you want to find how many consecutive H occurs after first H?
Try out this, just updated few parts of your code:
data = ['H', 'H', 'T', 'H']
j = 0
for index in range(len(data)-1):
if data[index] == 'H' and data[index+1] == 'H':
j = j + 1
print('it was this amount of times ' + str(j))
Output:
it was this amount of times 1
You could do something clever with zip and sum:
>>> data = ['H', 'H', 'T', 'H', 'H', 'H', 'T']
>>> sum(a == b == 'H' for a, b in zip(data[1:], data[:-1]))
3
Same thing with an index:
sum(data[i-1] == data[i] == 'H' for i in range(1, len(data)))
[h, h, h]not count as 3 occurrences?['H', 'H', 'T', 'H', 'H', 'H']? 2 or 3?