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@cowbert
Last active November 27, 2019 08:49
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def countAllList(lst):
# define the list of accumulators:
counts = [0]*10
# counts == [0,0,0...0] length of 10
# iterate over lst to process each string at a time
for s in lst:
# lets turn the string into a list of letters
s_list = list(s)
# now count the number of occurances for each digit
# instead of typing out each digit lets have python
# do it for us by looping!
for i in range(10):
# this loops i from 0 to 9
# str(i) turns the number into a string
counts[i] += s_list.count(str(i))
# since when i == 0, we can kill 2 birds with 1 stone and refer to
# counts[i] as the first element of counts.
# At the same time we count the occurance of str(i)
# in our listified string
# increment i and repeat (i==1, add the result of s_list.count('1') to counts[1] )
return counts
lres = countAllList(["111","2222","009"])
print(lres)
assert countAllList(["111","2222","009"]) == [2, 3, 4, 0, 0, 0, 0, 0, 0, 1]
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