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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Kai Long Chan
Kai Long Chan
3,238 Points
Posted by Kai Long Chan
Kai Long Chan
3,238 Points

I cannot pass the wordcount.py . May I have the test case so that I can test my program?

I cannot pass the wordcount.py . May I have the test case so that I can test my program?

wordcount.py 
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.

def word_count(a_str):
    a_str = a_str.strip()
    a_str = a_str.lower()
    list_word = a_str.split(" ")
    #print(list_word)
    d = {}
    for word in list_word:
        if word not in d.keys():
            d[word] = 1
        else:
            d[word] += 1
    return d

3 Answers

Tonye Jack
seal-mask
.a{fill-rule:evenodd;}techdegree
Tonye Jack
Full Stack JavaScript Techdegree Student 12,469 Points
Tonye Jack
.a{fill-rule:evenodd;}techdegree
Tonye Jack
Full Stack JavaScript Techdegree Student 12,469 Points

Faster way

def word_count(arg):
    words = arg.lower().split()
    keys = set(words)
    ret = dict(zip(keys, [0 for  _ in range(len(keys))]))
    for key in words:
        if ret.get(key):
            ret[key] += 1
    return ret
Kai Long Chan
Kai Long Chan
3,238 Points
Kai Long Chan
Kai Long Chan
3,238 Points

My problem is fixed. It is because I should use "split()" instead of "split(" ")"

carlos florez
carlos florez
9,664 Points
carlos florez
carlos florez
9,664 Points

this worked for me

def word_count(x):
    a = x.lower()
    b = a.split()
    dict1 = {}
    for item in b:
        dict1[item] = b.count(item)
    return dict1
Tonye Jack
.a{fill-rule:evenodd;}techdegree
Tonye Jack
Full Stack JavaScript Techdegree Student 12,469 Points

Set will eliminate duplicate keys but this works but will suggest having 

for item in set(b)

no need to loop through similar values twice.