Python inner loop and outer loop with counter to iterate over list

Question

I have specific issue where, im trying to find solution on inner loop(execute 3 time) and continue outer to process rest of the list in for loop:

strings = ['A','A','A','A','A','A','A','A','A','A','B','B','B','B','B','B','B','B','B','B',\
           'A','A','A','A','A','A','A','A','A','A','B','B','B','B','B','B','B','B','B','B']

i=0
for string in strings:
    global i
    if string == 'A':
        while i < 3:
            print(string, i)
            i+=1
            if i==3: continue
    elif string== 'B':
         while i < 3:
            print(string,i)
            i+=1
            if i==3: continue
#         print(string)

Current result: A 0 A 1 A 2

Expected to have continued over list once the inner loop complete and process from next:

A 0
A 1
A 2

B 0
B 1
B 2

A 0
A 1
A 2

B 0
B 1
B 2

Can you explain a little bit about what strings is supposed to be? There are several alternate ways to get your expected output — will-hedges
– will-hedges, Commented Jun 30, 2022 at 11:37
Replace global i with i = 0; you need to reset it per inner loop. Or just replace manual i management using while with for i in range(3):. Either way, get rid of if i == 3: continue (which accomplishes nothing). — ShadowRanger
– ShadowRanger, Commented Jun 30, 2022 at 11:39
Hi ShadownRanger,Can you explain with sample code. :strings = ['A','A','A','A','A','A','A','A','A','A','B','B','B','B','B','B','B','B','B','B',\ 'A','A','A','A','A','A','A','A','A','A','B','B','B','B','B','B','B','B','B','B'] for string in strings: i=0 if string == 'A': while i < 3: print(string, i) i+=1 i=0 elif string== 'B': while i < 3: print(string,i) i+=1 i=0 — mutt.dropbox
– mutt.dropbox, Commented Jun 30, 2022 at 12:15

mozway · Accepted Answer · 2022-06-30 11:45:05Z

2

If I understand correctly the logic, you could use itertools.groupby to help you form the groups:

variant #1

from itertools import groupby

MAX = 3

for k,g in groupby(strings):
    for i in range(min(len(list(g)), MAX)):
        print(f'{k} {i}')
    print()

variant #2

from itertools import groupby

MAX = 3

for k,g in groupby(strings):
    for i,_ in enumerate(g):
        if i >= MAX:
            break
        print(f'{k} {i}')
    print()

output:

A 0
A 1
A 2

B 0
B 1
B 2

A 0
A 1
A 2

B 0
B 1
B 2

variant #3: without import

prev = None
count = 0
MAX = 3
for s in strings:
    if s == prev:
        if count < MAX:
            print(f'{s} {count}')
            count += 1
    elif prev:
        count = 0
        print()
    prev = s

edited Jun 30, 2022 at 11:45

answered Jun 30, 2022 at 11:37

mozway

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4 Comments

mutt.dropbox Over a year ago

Hi Mozway, is there way to count based on, if string == 'A': count and output 3 following string and exit to continue iteration on other list?

mozway Over a year ago

not sure what you mean by "other list", maybe this should be a new question where you provide all the reproducible details?

mutt.dropbox Over a year ago

Result is correct as expected, but as i need to check also if there is A and B only in this list. If there is anything else i need to skip the counter. As A & B is relevant not others in the given list. Sorry this is only details i can share. count only if A and B

mozway Over a year ago

groupby return the key (here k), you can check that k in {'A', 'B'} in your loop and continue otherwise

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