Python 我们有嵌套列表创建字典关键字作为单词列表，以及它们作为值出现的次数

实际代码为：

big_list = [[['one', 'two'], ['seven', 'eight']], [['nine', 'four'], ['three', 'one']], [['two', 'eight'], ['seven', 'four']], [['five', 'one'], ['four', 'two']], [['six', 'eight'], ['two', 'seven']], [['three', 'five'], ['one', 'six']], [['nine', 'eight'], ['five', 'four']], [['six', 'three'], ['four', 'seven']]]

word_counts = {}  
for n in big_list:
    for i in big_list:
        if n == i:
            word_counts[i] = word_counts[i] + 1


print(word_counts)                       

错误为：

不可损坏类型：第8行的“列表”

预期结果：

像那样

所以请帮我找到正确的解决方案

可以使用以展平嵌套列表并计算每个元素出现的次数：

from collections import Counter
from itertools import chain

Counter(chain(*chain(*big_list)))

输出

---

对于没有导入的解决方案，您可以执行以下操作：

d = {}
for i in big_list:
    for j in i:
        for k in j:
            if not d.get(k):
                d[k] = 1
            else:
                d[k] += 1

print(d)
# {'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'nine': 2, 
#  'four': 5, 'three': 3, 'five': 3, 'six': 3}

---

您可以尝试以下代码：

big_list = [[['one', 'two'], ['seven', 'eight']], [['nine', 'four'], ['three', 'one']], [['two', 'eight'], ['seven', 'four']], [['five', 'one'], ['four', 'two']], [['six', 'eight'], ['two', 'seven']], [['three', 'five'], ['one', 'six']], [['nine', 'eight'], ['five', 'four']], [['six', 'three'], ['four', 'seven']]]

word_counts= {}

for ch in big_list:

    for word in ch:

              for req in word:

                     if req not in word_counts:

                                word_counts[req] = 0

                     word_counts[req]+=1

print(word_counts)

---

我们可以不使用module/libaray吗？让我发布另一个可选的if嵌套字典，而不是列表，那么什么时候如何计算值相同的键，但不同的子字典我不确定你的意思@MahendraSeervi。。。不幸的是，这似乎是一个不同的问题，所以我建议你问一个新问题，你肯定会很快得到答案：-）Hii@yatu如何使用循环和打印键和值迭代嵌套字典请阅读以改进答案的格式，并查看以改进答案。

big_list = [[['one', 'two'], ['seven', 'eight']], [['nine', 'four'], ['three', 'one']], [['two', 'eight'], ['seven', 'four']], [['five', 'one'], ['four', 'two']], [['six', 'eight'], ['two', 'seven']], [['three', 'five'], ['one', 'six']], [['nine', 'eight'], ['five', 'four']], [['six', 'three'], ['four', 'seven']]]

word_counts= {}

for ch in big_list:

    for word in ch:

              for req in word:

                     if req not in word_counts:

                                word_counts[req] = 0

                     word_counts[req]+=1

print(word_counts)