计算列表中单词的频率并按频率进行排序

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The ideal way is to use a dictionary that maps a word to it's count. But if you can't use that, you might want to use 2 lists - 1 storing the words, and the other one storing counts of words. Note that order of words and counts matters here. Implementing this would be hard and not very efficient.

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Using Counter would be the best way, but if you don't want to do that, you can implement it yourself this way.

# The list you already have
word_list = ['words', ..., 'other', 'words']
# Get a set of unique words from the list
word_set = set(word_list)
# create your frequency dictionary
freq = {}
# iterate through them, once per unique word.
for word in word_set:
freq[word] = word_list.count(word) / float(len(word_list))

freq will end up with the frequency of each word in the list you already have.

You need float in there to convert one of the integers to a float, so the resulting value will be a float.

Edit:

If you can't use a dict or set, here is another less efficient way:

# The list you already have
word_list = ['words', ..., 'other', 'words']
unique_words = []
for word in word_list:
if word not in unique_words:
unique_words += [word]
word_frequencies = []
for word in unique_words:
word_frequencies += [float(word_list.count(word)) / len(word_list)]
for i in range(len(unique_words)):
print(unique_words[i] + ": " + word_frequencies[i])

The indicies of unique_words and word_frequencies will match.

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words = file("test.txt", "r").read().split() #read the words into a list.
uniqWords = sorted(set(words)) #remove duplicate words and sort
for word in uniqWords:
print words.count(word), word

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One way would be to make a list of lists, with each sub-list in the new list containing a word and a count:

list1 = []    #this is your original list of words
list2 = []    #this is a new list


for word in list1:
if word in list2:
list2.index(word)[1] += 1
else:
list2.append([word,0])

Or, more efficiently:

for word in list1:
try:
list2.index(word)[1] += 1
except:
list2.append([word,0])

This would be less efficient than using a dictionary, but it uses more basic concepts.

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You can use

from collections import Counter

It supports Python 2.7，read more information here

1.

>>>c = Counter('abracadabra')
>>>c.most_common(3)
[('a', 5), ('r', 2), ('b', 2)]

use dict

>>>d={1:'one', 2:'one', 3:'two'}
>>>c = Counter(d.values())
[('one', 2), ('two', 1)]

But, You have to read the file first, and converted to dict.

2. it's the python docs example,use re and Counter

# Find the ten most common words in Hamlet
>>> import re
>>> words = re.findall(r'\w+', open('hamlet.txt').read().lower())
>>> Counter(words).most_common(10)
[('the', 1143), ('and', 966), ('to', 762), ('of', 669), ('i', 631),
('you', 554),  ('a', 546), ('my', 514), ('hamlet', 471), ('in', 451)]

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use this

from collections import Counter
list1=['apple','egg','apple','banana','egg','apple']
counts = Counter(list1)
print(counts)
# Counter({'apple': 3, 'egg': 2, 'banana': 1})

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Yet another solution with another algorithm without using collections:

def countWords(A):
dic={}
for x in A:
if not x in  dic:        #Python 2.7: if not dic.has_key(x):
dic[x] = A.count(x)
return dic


dic = countWords(['apple','egg','apple','banana','egg','apple'])
sorted_items=sorted(dic.items())   # if you want it sorted

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You can use reduce() - A functional way.

words = "apple banana apple strawberry banana lemon"
reduce( lambda d, c: d.update([(c, d.get(c,0)+1)]) or d, words.split(), {})

returns:

{'strawberry': 1, 'lemon': 1, 'apple': 2, 'banana': 2}

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Try this:

words = []
freqs = []


for line in sorted(original list): #takes all the lines in a text and sorts them
line = line.rstrip() #strips them of their spaces
if line not in words: #checks to see if line is in words
words.append(line) #if not it adds it to the end words
freqs.append(1) #and adds 1 to the end of freqs
else:
index = words.index(line) #if it is it will find where in words
freqs[index] += 1 #and use the to change add 1 to the matching index in freqs

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the best thing to do is :

def wordListToFreqDict(wordlist):
wordfreq = [wordlist.count(p) for p in wordlist]
return dict(zip(wordlist, wordfreq))

then try to : wordListToFreqDict(originallist)

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Here is code support your question is_char() check for validate string count those strings alone, Hashmap is dictionary in python

def is_word(word):
cnt =0
for c in word:


if 'a' <= c <='z' or 'A' <= c <= 'Z' or '0' <= c <= '9' or c == '$':
cnt +=1
if cnt==len(word):
return True
return False


def words_freq(s):
d={}
for i in s.split():
if is_word(i):
if i in d:
d[i] +=1
else:
d[i] = 1
return d


print(words_freq('the the sky$ is blue not green'))

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Pandas answer:

import pandas as pd
original_list = ["the", "car", "is", "red", "red", "red", "yes", "it", "is", "is", "is"]
pd.Series(original_list).value_counts()

If you wanted it in ascending order instead, it is as simple as:

pd.Series(original_list).value_counts().sort_values(ascending=True)

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for word in original_list:
words_dict[word] = words_dict.get(word,0) + 1


sorted_dt = {key: value for key, value in sorted(words_dict.items(), key=lambda item: item[1], reverse=True)}


keys = list(sorted_dt.keys())
values = list(sorted_dt.values())
print(keys)
print(values)

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Simple way

d = {}
l = ['Hi','Hello','Hey','Hello']
for a in l:
d[a] = l.count(a)
print(d)
Output : {'Hi': 1, 'Hello': 2, 'Hey': 1}

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word and frequency if you need

def counter_(input_list_):
lu = []
for v in input_list_:
ele = (v, lc.count(v)/len(lc)) #if you don't % remove <</len(lc)>>
if ele not in lu:
lu.append(ele)
return lu


counter_(['a', 'n', 'f', 'a'])

output:

[('a', 0.5), ('n', 0.25), ('f', 0.25)]