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how can I found the most repeated word and how much repeated it [duplicate]

I am using Python 3.3
I need to create two lists, one for the unique words and the other for the frequencies of the word.
I have to sort the unique word list based on the frequencies list so that the word with the highest frequency is first in the list.
I have the design in text but am uncertain how to implement it in Python.
The methods I have found so far use either Counter or dictionaries which we have not learned. I have already created the list from the file containing all the words but do not know how to find the frequency of each word in the list. I know I will need a loop to do this but cannot figure it out.
Here's the basic design:
original list = ["the", "car",....]
newlst = []
frequency = []
for word in the original list
if word not in newlst:
newlst.append(word)
set frequency = 1
else
increase the frequency
sort newlst based on frequency list

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})

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)]

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

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)

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

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.

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}

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.

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.

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

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'))

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)

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}

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)]

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)

Anagrams in Python using lists

Imagine we have following list of strings:
Input: ["eat", "tea", "tan", "ate", "nat", "bat"]
The output of our program should group each set of anagram and return them all together as a list as following:
Output:
[
["ate","eat","tea"],
["nat","tan"],
["bat"]
]
My current solution finds the first set of anagrams but fails to detect the other two and instead, duplicates the first groups into the list:
class Solution(object):
def groupAnagrams(self, strs):
allResults=[]
results=[]
temp=''
for s in strs:
temp=s[1:]+s[:1]
for i in range(0,len(strs)):
if temp==strs[i]:
results.append(strs[i])
allResults.append(results)
return allResults
and the output is:
[["ate","eat","tea"],["ate","eat","tea"],["ate","eat","tea"],["ate","eat","tea"],["ate","eat","tea"],["ate","eat","tea"]]
How to fix this issue?
EDIT:
I have fixed the duplication in appending by appending the results into allResults outside of second loop:
class Solution(object):
def groupAnagrams(self, strs):
allResults=[]
results=[]
temp=''
for s in strs:
temp=s[1:]+s[:1]
for i in range(0,len(strs)):
if temp==strs[i]:
results.append(strs[i])
allResults.append(results)
print(results)
return allResults
Yet, it does not detect the other two sets of anagrams.

you can do it using defaultdict of python in-built collections library and sorted :
In [1]: l = ["eat", "tea", "tan", "ate", "nat", "bat"]
In [2]: from collections import defaultdict
In [3]: d = defaultdict(list)
In [4]: for x in l:
...: d[str(sorted(x))].append(x)
In [5]: d.values()
Out[5]: dict_values([['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']])
to fix your the solution you need add the variable to check is allready added, for exanmple(and the while walk through the strs i use enumerate for little performance in the search of the anagrams):
class Solution(object):
def groupAnagrams(self, strs):
allResults = []
added = set([])
temp=''
for i, s in enumerate(strs):
results = []
unique_s = "".join(sorted(s))
if unique_s in added:
continue
else:
added.add(unique_s)
for x in strs[i:]:
if unique_s=="".join(sorted(x)):
results.append(strs[i])
allResults.append(results)
print(added)
return allResults

Use itertools.groupby
>>> lst = ["eat", "tea", "tan", "ate", "nat", "bat"]
>>>
>>> from itertools import groupby
>>> f = lambda w: sorted(w)
>>> [list(v) for k,v in groupby(sorted(lst, key=f), f)]
[['bat'], ['eat', 'tea', 'ate'], ['tan', 'nat']]

Using only lists, as requested in the title of the question:
The second line s_words takes all the letters of each word in words, sorts them, and recreates a string composed of the sorted letters of the word; it creates a list of all the these sorted letters strings, in the same order as the original sequence of words --> this will be used to compare the possible anagrams (the letters of anagrams produce the same string when sorted)
The 3rd line indices hold True or False values, to indicate if the corresponding word has been extracted already, and avoid duplicates.
The following code is a double loop that for each s_word, determines which other s_word is identical, and uses its index to retrieve the corresponding word in the original list of words; it also updates the truth value of the indices.
words = ["eat", "tea", "tan", "ate", "nat", "bat"]
s_words = [''.join(sorted(list(word))) for word in words]
indices = [False for _ in range(len(words))]
anagrams = []
for idx, s_word in enumerate(s_words):
if indices[idx]:
continue
ana = [words[idx]]
for jdx, word in enumerate(words):
if idx != jdx and not indices[jdx] and s_word == s_words[jdx]:
ana.append(words[jdx])
indices[jdx] = True
anagrams.append(ana)
print(anagrams)
output:
[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]

The way you implemented your function, you are only looking at rotations of the strings (that is you shift a letter from the beginning to the end, e.g. a-t-e -> t-e-a -> e-a-t). What your algorithm cannot detect is single permutations were you only switch two letters (n-a-t -> t-a-n). In mathematical language you only consider the even permutations of the three letter strings and not the odd permutations.
A modification of your code could for example be:
def get_list_of_permutations(input_string):
list_out = []
if len(input_string) > 1:
first_char = input_string[0]
remaining_string = input_string[1:]
remaining_string_permutations = get_list_of_permutations(remaining_string)
for i in range(len(remaining_string)+1):
for permutation in remaining_string_permutations:
list_out.append(permutation[0:i]+first_char+permutation[i:])
else:
return [input_string]
return list_out
def groupAnagrams(strs):
allResults=[]
for s in strs:
results = []
list_of_permutations = get_list_of_permutations(s)
for i in range(0,len(strs)):
if strs[i] in list_of_permutations:
results.append(strs[i])
if results not in allResults:
allResults.append(results)
return allResults
The output is
Out[218]: [['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]
Edit: modified the code to work with all lengths of strings.

https://docs.python.org/3/library/itertools.html#itertools.permutations
from itertools import permutations
word_list = ["eat", "tea", "tan", "ate", "nat", "bat"]
anagram_group_list = []
for word in word_list:
if word == None:
pass
else:
anagram_group_list.append([])
for anagram in permutations(word):
anagram = ''.join(anagram)
try:
idx = word_list.index(anagram)
word_list[idx] = None
anagram_group_list[-1].append(anagram)
except ValueError:
pass # this anagram is not present in word_list
print(anagram_group_list)
# [['eat', 'ate', 'tea'], ['tan', 'nat'], ['bat']]
after refactoring code and stopping it from producing redundant result your code still doesn't give expected result as logic for producing anagram is not completely correct
def groupAnagrams(word_list):
allResults=[]
results=[]
for idx,s in enumerate(word_list):
if s == None:
pass
else:
results = [s] # word s is added to anagram list
# you were generating only 1 anagram like for tan --> ant but in word_list only nat was present
for i in range(1,len(s),1):
temp = s[i:]+s[:i] #anagram
# for s = 'tan' it generates only 'ant and 'nta'
# when it should generate all six tna ant nta _nat_ atn tan
if temp in word_list:
results.append(temp)
word_list[word_list.index(temp)] = None
allResults.append(results)
return allResults
print(groupAnagrams(["eat", "tea", "tan", "ate", "nat", "bat"]))
# [['eat', 'ate', 'tea'], ['tan'], ['nat'], ['bat']]

The detection of anagrams of words consisting of unique characters can be done by comparison between sets. [See comment for a general solution]
words = ["eat", "tea", "tan", "ate", "nat", "bat"]
anagrams = []
for w in words:
m = [w2 for w2 in words if set(w2) == set(w)]
if m not in anagrams:
anagrams += [m]
print(anagrams)
Output
[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]
EDIT
For words with duplicate characters a multi-set approach can be used. A multi-set can be modeled with collections.Counter.
from collections import Counter
words = ["eat", "tea", "tan", "ate", "nat", "bat", "cia", "aci"]
# group per index
d = {}
multi_sets = list(map(Counter, words))
for i, w in enumerate(words):
i_reference = multi_sets.index(Counter(w)) # always 1st match
d.setdefault(i_reference, []).append(words[i])
anagrams = list(d.values())
# inplace sort: group per size of family of anagrams
anagrams.sort(key=len, reverse=True)
print(anagrams)
Remark: ordering a multi-set is highly non-trivial and the usual methods __lt__, __gt__, ... are not implemented. As a consequence sorted cannot be used. Comparison is still possible with __eq__ or __ne__ which are both naturally supported by Counter.

How to count elements on each position in lists

I have a lot of lists like:
SI821lzc1n4
MCap1kr01lv
All of them have the same length. I need to count how many times each symbol appears on each position. Example:
abcd
a5c1
b51d
Here it'll be a5cd

One way is to use zip to associate characters in the same position. We can then send all of the characters from each position to a Counter, then use Counter.most_common to get the most common character
from collections import Counter
l = ['abcd', 'a5c1', 'b51d']
print(''.join([Counter(z).most_common(1)[0][0] for z in zip(*l)]))
# a5cd

from statistics import mode
[mode([x[i] for x in y]) for i in xrange(len(y[0]))]
where y is your list.
Python 3.4 and up

You could use combination of zip and Counter
a = ("abcd")
b = ("a5c1")
c = ("b51d")
from collections import Counter
zippedList = list(zip(a,b,c))
print("zipped: {}".format(zippedList))
final = ""
for x in zippedList:
countLetters = Counter(x)
print(countLetters)
final += countLetters.most_common(3)[0][0]
print("output: {}".format(final))
output:
zipped: [('a', 'a', 'b'), ('b', '5', '5'), ('c', 'c', '1'), ('d', '1', 'd')]
Counter({'a': 2, 'b': 1})
Counter({'5': 2, 'b': 1})
Counter({'c': 2, '1': 1})
Counter({'d': 2, '1': 1})
output: a5cd

This all depends on where your list is. Is your list coming from another file or is it an actual array? At the end of the day, the best way to do this simply is going to be to use a dictionary and a for loop.
new_dict = {}
for i in range(len(line)):
if i in new_dict:
new_dict[i].append(line[i])
else:
new_dict[i] = [line[i]]
Then after that I'm assuming that you'd like to output the four most common element appearances. For that I'd recommend importing statistics and using the mode method...
from statistics import mode
new_line = ""
for key in new_dict:
x = mode(new_dict[key])
new_line = new_line + x
However, your question is quite vague, please elaborate more next time.
P.s. I'm a newbie so all you experienced programmers plz don't hate :)

I would use a combination of defaultdict, enumerate, and Counter:
>>> from collections import Counter, defaultdict
>>> data = '''abcd
a5c1
b51d
'''
>>> poscount = defaultdict(Counter)
>>> for line in data.split():
for i, character in enumerate(line):
poscount[i][character] += 1
>>> ''.join([poscount[i].most_common(1)[0][0] for i in sorted(poscount)])
'a5cd'
Here's how it works:
The defaultdict() creates new entries when it sees a new key.
The enumerate() function returns both the character and its position in the line.
The Counter counts the occurences of individual characters
Combining the three makes a defaultdict whose keys are the column positions and whose values are character counters. That gives you one character counter per column.
The most_common() method returns the highest frequency (character, count) pair for that counter.
The [0][0] extracts the character from the list of (character, count) tuples.
The str.join() method combines the results back together.

Getting word frequency of a list of sentence, but not combined. (python)

def getWordFreq(corpus):
wordFreq = []
for sent in corpus:
for word in sent:
wordFreq.append((word, sent.count(word)))
return wordFreq
Wrote this function to get a frequency of each word in a corpus.
To test it, I wrote
cc = [ ['hi','ho'], ['hee','ho']]
getWordFreq(cc)
but this returned
[('hi', 1), ('ho', 1), ('hee', 1), ('ho', 1)]
instead of ('ho', 2).
What am I missing?

Hope this simplest one will be helpful. Here we are using for.
Try this code snippet here
def getWordFreq(corpus):
result = {}
for data in corpus:
for word in data:
if word in result:
result[word] += 1 #adding result in the dictionary
else:
result[word] = 1
return result.items() #returning items
cc = [['hi', 'ho'], ['hee', 'ho']]
print(getWordFreq(cc))
Output: [('hee', 1), ('hi', 1), ('ho', 2)]

You'd be better off using a dictionary for this task:
def getWordFrequency(corpus):
frequencies = {}
for sentence in corpus:
for word in sentence:
if word in frequencies:
frequencies[word] += 1
else:
frequencies[word] = 1
return frequencies
A dictionary preserves a mapping from keys (words) to values (corresponding frequencies). This way, it's easier and faster to keep track of the frequencies, because you don't have to deal with merging words yourself.
Your implementation just adds a tuple consisting of the word and its frequency in the sentence to the result. That won't combine words together and keep track of frequencies for you.
Python's collections module also provides a Counter designed for this type of thing.
from collections import Counter
def getWordFrequency(corpus):
freq = Counter()
for sentence in corpus:
for word in sentence:
freq[word] += 1
return freq
Notice that we don't have to check if the word is already present in the counter, because Counter handles it for us.

You can try this solution:
from collections import Counter
def getWordFreq(corpus):
wordFreq = [j for i in corpus for j in i]
return list(Counter(wordFreq).items())

As I mentioned in my comment, you are counting each word in a sent not in the entire corpus Here is what you need to do
def getWordFreq(corpus):
wordFreq = []
for sent in corpus:
for word in sent:
wordFreq.append((word, sum(map(lambda x: x.count(word), corpus))))
return wordFreq
cc = [ ['hi','ho'], ['hee','ho']]
getWordFreq(cc)
gives
[('hi', 1), ('ho', 2), ('hee', 1), ('ho', 2)]
If you want to print each word only once, change wordFreq to set and use add instead of append
def getWordFreq(corpus):
wordFreq = set()
for sent in corpus:
for word in sent:
wordFreq.add((word, sum(map(lambda x: x.count(word), corpus))))
return wordFreq
cc = [ ['hi','ho'], ['hee','ho']]
getWordFreq(cc)
gives
{('hee', 1), ('hi', 1), ('ho', 2)}

Top Counts...return tuple format of output back to dictionary format

I need the top 10 words and their count, already contained in a dictionary, in the following format:
word count (e.g. hello 10)
I have the following code:
for word in word:
if word not in counts:
counts[word] = 1
else:
counts[word] += 1
for word in counts:
top = sorted(counts.items())
top_10 = top[:9]
print top
the output is a list with tuples inside: [('red', 5), ('blue', 2), ('green', 1), ...]
However, I need it in the format of :
red 5
blue 2
green 1
How can this be done???

First, you can count with less (and more pythonic) code:
for word in words:
count[word] = count.get(word,0) + 1
Second, you can achieve the printing format you want with:
for k in count:
print k,count[k]
If sorting is your problem you can use operator.itemgetter():
from operator import itemgetter
words = ['brown','yellow','red','blue','green','blue','green','blue','green']
count = {}
for word in words:
count[word] = count.get(word,0) + 1
top = sorted(count.iteritems(), key=itemgetter(1))
top_10 = top[-10:]
print top_10

replace
print top
with
for k,v in top: print k, v

you can easily acheive this by using Counter
from collections import Counter
words = ['brown','yellow','red','blue','green','blue','green','blue','green']
c=Counter(words)
for i,j in c.most_common(10):
print i,j