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  1. Prepare
  2. Python
  3. Collections
  4. Word Order
  5. Discussions

Word Order

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1588 Discussions

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  • syahrulmahar92
     + 0 comments

    from collections import Counter n = int(input()) l = [] for i in range(n): word = input() l.append(word) val = Counter(l) print(len(val.keys())) for x,y in val.items(): print(y, end=' ')

  • adrien_cordonni1
     + 0 comments

    Before Python 3.7, the order is not guaranteed in dictionaries:

    from collections import Counter, OrderedDict

class OrderedCounter(Counter, OrderedDict): 
    pass

words = OrderedCounter(input() for _ in range(int(input())))
print(len(words))
print(*words.values())

    Since Python 3.7, the order is guaranteed to be kept in dictionaries which makes the code even simpler:

    from collections import Counter

words = Counter(input() for _ in range(int(input())))
print(len(words))
print(*words.values())

  • msravindar2292
     + 0 comments

    Use OrderedDict to preserve the order of input words. Loop over input lines, setting each word's count to zero initially. Update each word's count as it appears. Finally, print the count of unique words and their occurrences.

  • sefamutlucu
     + 0 comments
    n = int(input())
words = {}

for i in range(n):    
    key = input()
    if key in words:
        words[key] += 1
    else:
        words[key] = 1

print(len(words))
print(*words.values())

  • imam_alfaridzi
     + 0 comments

    Here's my codes.

    using native:

    res = {}
for i in range(int(input())):
    u = input()
    if u in res:
        res[u] += 1
        continue
    res[u] = 1
print(len(res.keys()))
print(*res.values())

    with Counter func:

    from collections import Counter
res = Counter(input() for _ in range(int(input())))
print(len(res.keys()))
print(*res.values())