创建字符串中单词的字典，按出现次数排序，仅显示包含 4 个或更多字母的单词

Question

方向：字典的键应该是单词，值应该是单词在段落中出现的次数，按降序对字典进行排序。仅显示包含 4 个或更多字母的单词。

text = """The goal is to turn data into information and information into insight . 
You can have data without information but you cannot have information without data ."""

我一直在解决这个问题，但似乎不能只显示包含 4 个或更多字母的单词。任何帮助将不胜感激，谢谢。

输出应该是

这是我目前所做的

# clean up string 'text'
for char in '-.,\n':
    text=text.replace(char,' ')
text = text.lower()
words_list = text.split(' ')

#define dictionary
words_dict = {}

#Count number of times each word comes up in list of words (in dictionary)
for word in words_list:
    if word not in words_dict:
        words_dict[word] = 0
    words_dict[word] += 1

#sort dictionary by number of occurences
sorted(words_dict.items(), key = lambda x: x[1], reverse = True)

Answer 1

对已发布代码的最小更改

只在原始代码中添加了一个长度过滤器

text = """The goal is to turn data into information and information into insight . 
You can have data without information but you cannot have information without data ."""

# clean up string 'text'
for char in '-.,\n':
    text=text.replace(char,' ')
text = text.lower()
words_list = text.split(' ')

#define dictionary
words_dict = {}

#Count number of times each word comes up in list of words (in dictionary)
for word in words_list:
    if word not in words_dict:
        words_dict[word] = 0
    words_dict[word] += 1

# Filter words
filtered = {word:count for word,count in words_dict.items() if len(word) >= 4}

#sort dictionary by number of occurences
result = sorted(filtered.items(), key = lambda x: x[1], reverse = True)

print('{:<15} {:<4}'.format('Word', 'Count'))
for word, count in result:
  print(f'{word:15} {count:4}')

输出

Word            Count
information        4
data               3
into               2
have               2
without            2
goal               1
turn               1
insight            1
cannot             1

Answer 2

您可以在此处使用collections.Counter 和filter。

from collections import Counter
text = """The goal is to turn data into information and information into insight . 
You can have data without information but you cannot have information without data ."""

count = Counter(filter(lambda x:len(x)>=4, text.split()))

sorted(count.items(),key = lambda x:x[1],reverse=True)
[('information', 4),
 ('data', 3),
 ('into', 2),
 ('have', 2),
 ('without', 2),
 ('goal', 1),
 ('turn', 1),
 ('insight', 1),
 ('cannot', 1)]

编辑： 不使用collections。您可以使用dict.setdefault 模仿Counter 的行为

text = text.split()
new = dict()

for t in filter(lambda x:len(x)>=4,text):
    new[t] = new.setdefault(t,0) + 1

sorted(new.items(),key=lambda x:x[1],reverse=True)