How do you sort a dictionary by word length?

Question

Currently I am trying to get words such as "The quick brown fox jumped over the lazy brown dog’s back" read from a text file and organized by word length and by word count.

So the output should be:

1 The

1 fox

1 the

1 back

1 lazy

1 over

2 brown

1 dog’s

1 quick

1 jumped

I did check quite a lot of stackoverflow questions like how to sort by length of string followed by alphabetical order?, and I'm going to guess I missed it, or I don't understand how to use it. I'm a beginner with python.

This is what I have so far:

 from collections import Counter
 file = open("text.txt","r")

 #read the file & split words 
 wordcount =Counter(file.read().split())

 #printing word count 
 for item in wordcount.items():

     print ("{}\t{}".format(*item))

Could someone help me know what i'm doing wrong?

Answer 1

Try something like-

from collections import Counter
import re
#Identify each word using regex
words = re.findall(r'\w+', open(r"D:\test.txt").read())
#Find counts
data= Counter(words).most_common()
data = sorted(data,key=lambda x:x[0])
print data

Prints-

[('The', 1), ('back', 1), ('brown', 2), ('dog', 1), ('fox', 1), ('jumped', 1), ('lazy', 1), ('over', 1), ('quick', 1), ('s', 1), ('the', 1)]

Or try word by split-

from collections import Counter
import re
words=open(r"D:\test.txt").read().split(" ")
data= Counter(words).most_common()
data = sorted(data,key=lambda x:x[1])
print data

Prints-

[('lazy', 1), ('jumped', 1), ('over', 1), ('fox', 1), ('back', 1), ('quick', 1), ('The', 1), ('the', 1), ('dog's', 1), ('brown', 2)]

Answer 2

as my comment says, you can't sort a dict because dicts are not ordered (it has to do with how the key/value pairs are hashed to allow for O(1) value getting).

You can instead iterate through a sorted dict.items() since .items() returns a list of tuples and lists ARE ordered.

>>> s = "The quick brown fox jumped over the lazy brown dog’s back"

>>> from collections import Counter
>>> wordcount = Counter(s.split())
>>> wordcount
Counter({'brown': 2, 'back': 1, 'quick': 1, 'The': 1, 'over': 1, 'dog’s': 1, 'jumped': 1, 'fox': 1, 'the': 1, 'lazy': 1})
>>> for key,val in sorted(wordcount.items(),key = lambda pair: len(pair[0])):
    print(str(val),key)


1 The
1 fox
1 the
1 back
1 over
1 lazy
1 quick
2 brown
1 dog’s
1 jumped

using the builtin sorted(list,key=somefunction) function, you can sort the list that is returned by wordcount.items() by the length of the key (which is accessed by pair[0] since pair == (key,value)

Answer 3

First, the dictionary has to be converted to a list of tuples, then sort it and print/return:

#shaffled words dict
words = {"The": 1,
"fox": 1,
"dog's": 1,
"quick": 1,
"jumped": 1,
"over": 1,         
"the": 1,
"brown": 1,
"back": 1,
"lazy": 1}

#convert dict to list of tuples
def toList(d1):
    l1 = []
    for k in d1:
        l1.append((k, d1[k]))
    return l1

#sort the list by length and alfabet
output = sorted(toList(words), key = lambda w: (len(w[0]), w[0]))

    for o in output:
        print str(o[1]) + " " + str(o[0])
"""
expected output is:
1 The
1 fox
1 the
1 back
1 lazy
1 over
1 brown
1 dog's
1 quick
1 jumped
"""

Answer 4

[EDIT] I reread the post and figured out that was not exactly what you wanted. See answers of others.

A dictionary is something similar to a list, but instead of integers as indecies you use strings. They are useful if you want to store data with key-value structure like "Mom":39, "Kevin":12, "Sally":14. Dictionarys are not sortable.

For what you need, a simple list of strings will do. (You can sort it afterwords by just calling sort() on the list:

words = file.read().split() #that is a list
words.sort()