python word frequencies for dictionary with values and key

Question

I am a first-time learner of python, I understand how to use word frequencies to count the number of each unique variable of a list, like this

sentence = ['hello', 'people', 'are', 'the', 'most', 'common', 'word', 'people', 'use', 'for', 'language ', 'learning']

words_freq ={} #dictionary for the counts
for word in sentence:
    if word not in words_freq:

        words_freq[word] =1
    else:

        words_freq[word] +=1

print (words_freq)

however, I wonder how could a word frequencies do on the dictionary by using a double for loop?

for example, I have a dictionary like this

Food = {
2015: ["Apple", "Milk", "Cookie", "Banana", "Orange" ],
2016: ["Potato", "Orange", "Chocolate", "Milk", "Mango"],
2017: ["Fish", "Potato", "Orange", "Mango", "Banana"],
2018: ["Beef", "Pork", "Fish", "Apple", "Cookie"],
2019: ["Pork", "Orange", "Apple", "Mango", "Chocolate"]
}

how to do a word frequencies/count and print something like this? Or store the highest value in list form ? apple : 3 milk : 2 orange : 3 .. .. ..

You have the right idea. You need two for loops. I suggest you take a stab at it and figure out as much as you can on your own. You should even consider creating a function to wrap your existing code that counts word frequencies in a list. — Code-Apprentice
– Code-Apprentice, Commented Feb 17, 2020 at 21:33
the desired output is list out as all type of food and follows with their number in total, and store it in a new dictionary. So I can print it out — Calvin Mela
– Calvin Mela, Commented Feb 17, 2020 at 21:42
Your result (a dictionary grouped by count) is off! There are lots of items that appear more than once. That cannot happen in a real dictionary. If you print it, you will see it contains a lot less items. While not an error, it still makes helping you (with, apparently per the already provided answers, a ready-made piece of code...) a lot harder. In your preface you mention a word count; your example output can never be made with such a counting script. Please adjust the desired output. — Jongware
– Jongware, Commented Feb 17, 2020 at 21:57
(There seems to be some misunderstanding about your question. Please edit and clarify. My impression was that from your (existing) word count list you wanted to create a dictionary such as the Food one you show. If that is not the case, please reword.) — Jongware
– Jongware, Commented Feb 17, 2020 at 22:02

Nadav · Accepted Answer · 2020-02-17 21:48:58Z

1

defaultdict would be pretty elegant for your use case it create a default dictionary for a given type for int - the default value is 0 for int value, it enable you to write less code

https://docs.python.org/3.3/library/collections.html#collections.defaultdict

from collections import defaultdict

def get_freq(food_dict: dict) -> dict:
    freq = defaultdict(int)
    for year, lst in food_dict.items():
        for elem in lst:
            freq[elem] += 1
    return freq

edited Feb 17, 2020 at 21:48

answered Feb 17, 2020 at 21:41

Nadav

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1 Comment

Calvin Mela Over a year ago

thanks a lot ! I will try my best to understand this !

Sash Sinha · Accepted Answer · 2020-02-17 21:53:47Z

1

Using a standard python dictionary you can make use of get(key[, default]) to deal with the case when the key you want to increment does not exist in the dictionary.

get(key[, default])

Return the value for key if key is in the dictionary, else default. If default is not given, it defaults to None, so that this method never raises a KeyError.

food = {
    2015: ["Apple", "Milk", "Cookie", "Banana", "Orange"],
    2016: ["Potato", "Orange", "Chocolate", "Milk", "Mango"],
    2017: ["Fish", "Potato", "Orange", "Mango", "Banana"],
    2018: ["Beef", "Porn", "Fish", "Apple", "Cookie"],
    2019: ["Pork", "Orange", "Apple", "Mango", "Chocolate"]
}

counts = {}
for year in food:
    for item in food[year]:
        counts[item] = counts.get(item, 0) + 1

print(counts)

Output:

{'Apple': 3, 'Milk': 2, 'Cookie': 2, 'Banana': 2, 'Orange': 4, 'Potato': 2, 'Chocolate': 2, 'Mango': 3, 'Fish': 2, 'Beef': 1, 'Porn': 1, 'Pork': 1}

edited Feb 17, 2020 at 21:53

answered Feb 17, 2020 at 21:42

Sash Sinha

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1 Comment

Calvin Mela Over a year ago

thanks a lot, I just wondering why I got a key error from the second for loop. This helps a lot

CC7052 · Accepted Answer · 2020-02-17 21:35:10Z

0

It helps you:

Food = {
2015: ["Apple", "Milk", "Cookie", "Banana", "Orange" ],
2016: ["Potato", "Orange", "Chocolate", "Milk", "Mango"],
2017: ["Fish", "Potato", "Orange", "Mango", "Banana"],
2018: ["Beef", "Porn", "Fish", "Apple", "Cookie"],
2019: ["Pork", "Orange", "Apple", "Mango", "Chocolate"]
}

words_freq ={} #dictionary for the counts
for word1 in Food.values():
    for word in word1:
        if word not in words_freq:

            words_freq[word] =1
        else:

            words_freq[word] +=1

print (words_freq)

answered Feb 17, 2020 at 21:35

CC7052

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5 Comments

CC7052 Over a year ago

@usr2564301 user shows codes and ask to expand it to nested dict, so I modified, We know that there are many way to do this. In your opinion I should comment behind any answer It will not.

Jongware Over a year ago

No, what OP is asking is "how to do a word frequencies/count and print something like this?", with that shown dictionary Food as output. (Unless I am very mistaken. OP might want to clarify the question ...)

CC7052 Over a year ago

@usr2564301 No, you didn't understood the question, and If you check the answer that CALVIN accepted, has the same output as mine !

Jongware Over a year ago

I've added a comment for the OP for clarification. The whole "word count" preamble and sample code seems to have nothing to do with the Food output.

Tomerikoo Over a year ago

@usr2564301 Food is not the output. It is another dictionary, asking (quite unclearly) how to do a word count across all of its values. The example output (again, quite vaguely) is right in the end of the post

Rob · Accepted Answer · 2020-02-17 21:45:04Z

0

You don't need a double loop, but you can do something like this:

food = {
    2015: ["Apple", "Milk", "Cookie", "Banana", "Orange" ],
    2016: ["Potato", "Orange", "Chocolate", "Milk", "Mango"],
    2017: ["Fish", "Potato", "Orange", "Mango", "Banana"],
    2018: ["Beef", "Porn", "Fish", "Apple", "Cookie"],
    2019: ["Pork", "Orange", "Apple", "Mango", "Chocolate"]
}

words_freq = {}
for year in food.keys():
    for fruit in food[year]:

        if fruit in words_freq.keys():
            words_freq[fruit] += 1
        else:
            words_freq[fruit] = 1

answered Feb 17, 2020 at 21:45

Rob

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Comments

David Collins · Accepted Answer · 2020-02-18 02:04:01Z

0

Another way you could do this is to create a single list, and then count frequencies in that list:

foodList = []

#creates a single list
for idx, f in enumerate(Food):
    foodList = foodList + list(Food.values())[idx]

#makes a count in that list
words_freq ={}
for word in foodList:
    if word not in words_freq:
        words_freq[word] =1
    else:
        words_freq[word] +=1

which yields the correct answer:

{'Apple': 3, 'Milk': 2, 'Cookie': 2, 'Banana': 2, 'Orange': 4, 'Potato': 2, 'Chocolate': 2, 'Mango': 3, 'Fish': 2, 'Beef': 1, 'Pork': 2}

answered Feb 18, 2020 at 2:04

David Collins

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