Question

我有5个单词清单。我需要查找出现在2个以上列表中的所有单词。任何单词在列表中都可以出现多次。

我用过collections.Counter，但是它只返回单个列表中所有单词的频率。

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例如，这些列表的输出应为['log'：4，'branch'：3]，因为'log'存在于4个列表中，而'branch'存在于3个列表中。

Answer 1

没有Counter：

a = ['wood', 'tree', 'bark', 'log']
b = ['branch', 'mill', 'boat', 'boat', 'house']
c = ['log', 'tree', 'water', 'boat']
d = ['water', 'log', 'branch', 'water']
e = ['branch', 'rock', 'log']

all_lists = [a, b, c, d, e]
all_words = set().union(w for l in all_lists for w in l)

out = {}
for word in all_words:
    s = sum(word in l for l in all_lists)
    if s > 2:
        out[word] = s

print(out)

打印：

{'branch': 3, 'log': 4}

编辑（以打印列表名称）：

a = ['wood', 'tree', 'bark', 'log']
b = ['branch', 'mill', 'boat', 'boat', 'house']
c = ['log', 'tree', 'water', 'boat']
d = ['water', 'log', 'branch', 'water']
e = ['branch', 'rock', 'log']

all_lists = {'a':a, 'b':b, 'c':c, 'd':d, 'e':e}
all_words = set().union(w for l in all_lists.values() for w in l)

out = {}
for word in all_words:
    s = sum(word in l for l in all_lists.values())
    if s > 2:
        out[word] = s

for k, v in out.items():
    print('Word : {}'.format(k))
    print('Count: {}'.format(v))
    print('Lists: {}'.format(', '.join(kk for kk, vv in all_lists.items() if k in vv )))
    print()

打印：

Word : log
Count: 4
Lists: a, c, d, e

Word : branch
Count: 3
Lists: b, d, e

Answer 2

您可以sum个计数器-从空的Counter()开始：

from collections import Counter

lists = [a, b, c, d, e]
total = sum((Counter(set(lst)) for lst in lists), Counter())
# Counter({'log': 4, 'branch': 3, 'tree': 2, 'boat': 2, 'water': 2,
#          'wood': 1, 'bark': 1, 'house': 1, 'mill': 1, 'rock': 1})
res = {word: occ for word, occ in total.items() if occ > 2}
# {'log': 4, 'branch': 3}

请注意，我首先将所有列表转换为set，以避免重复计数同一列表中多次出现的单词。

如果您需要知道单词的来源列表，可以尝试以下操作：

lists = {"a": a, "b": b, "c": c, "d": d, "e": e}

total = sum((Counter(set(lst)) for lst in lists.values()), Counter())
# Counter({'log': 4, 'branch': 3, 'tree': 2, 'boat': 2, 'water': 2,
#          'wood': 1, 'bark': 1, 'house': 1, 'mill': 1, 'rock': 1})
res = {word: occ for word, occ in total.items() if occ > 2}
# {'log': 4, 'branch': 3}

word_appears_in = {
    word: [key for key, value in lists.items() if word in value] for word in res
}
# {'log': ['a', 'c', 'd', 'e'], 'branch': ['b', 'd', 'e']}

在多个列表中查找常用单词

2 个答案: