Counting the words in a dictionary

Question

I have a dictionary which has fruits that were bought that month.

The dictionary looks like this:

bought = {
January: ['Apple', 'Banana', 'Orange', 'Kiwi', 'Raspberry'],
February: ['Orange', 'Mango', 'Banana'], 
March: ['Apple', 'Starfruit', 'Apricot']
}

Is there any way that I could list how many times I bought each fruit.

I want the output to look something like this

Apple: 2, Banana: 2, Orange: 2, Kiwi: 1, Raspberry: 1, Mango: 1, Starfruit: 1, Apricot: 1

All the other places I have use a dictionary to show how many times a word is used in a list. I am looking for a dictionary that finds out how many times a word is mentioned in another dictionary.

The thing is that I only just started learning how to code so I don't know much. — El numero dos, Feb 23 '19 at 19:31
Go to the python tutorial and look at what you can do with lists. — Bayleef, Feb 23 '19 at 19:32
@Elnumerodos you might want to take a look at this [link](https://stackoverflow.com/questions/23240969/python-count-repeated-elements-in-the-list/23240989) — Hass786123, Feb 23 '19 at 19:38
Thank you @system123456 but the link that you sent me was for a list right? I am trying to find the same thing as that but with a dictionary instead. — El numero dos, Feb 23 '19 at 19:42
See this link for a hint of how to proceed: https://stackoverflow.com/a/5829377/7292937 — liamhawkins, Feb 23 '19 at 19:47

Charles Landau · Accepted Answer · 2019-02-23T20:21:21.447

3

Your first problem is to get the counts, so we'll restructure into a new dictionary called out:

out = {}
for month in bought:
  for fruit in bought[month]:
    out[fruit] = out.get(fruit, 0) + 1

This for-loop just goes through all the keys (months) in bought and iterates through the list it finds there. As it reads fruits in that list, it checks to see if that fruit is already in out. If not, it initializes that fruit in out as 0. Finally, it increments the value by one.

Now we need to print in the format we want:

for k in out:
      print("{}: {}".format(k, out[k]))

String formatting in Python is extremely mature. You can learn all sorts of tricks at https://pyformat.info/ for example. Here we are just inserting the key and then the value to a string template, as we iterate through out.

edited Feb 23 '19 at 20:21

answered Feb 23 '19 at 19:41

Charles Landau

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Thank you so much, This worked perfectly. And thank you for the site too. I'll make sure to check it if I have other problems from now on. – El numero dos Feb 23 '19 at 19:48
That's a really good way to do it @Charles Landau. Have my upvote - in my answer, I never thought about just assigning as I go – PrinceOfCreation Feb 23 '19 at 19:49
I'll test that now :) – PrinceOfCreation Feb 23 '19 at 19:51
They both have a sys time of 0.010s, but mine is faster in real: 0.164s as opposed to your 0.189. Although I suspect for bigger lists, yours will come out on top - I really like the nested loops. _Tests ran 10 times, on an Intel 8300. 8GB RAM (not really needed but sure) and SSD._ – PrinceOfCreation Feb 23 '19 at 19:58
Very interesting @PrinceOfCreation – Charles Landau Feb 23 '19 at 20:00
Yes it is, I would have expected yours to come out faster. – PrinceOfCreation Feb 23 '19 at 20:00
@PrinceOfCreation try it on much larger data if you like – Charles Landau Feb 23 '19 at 20:01
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/188935/discussion-between-princeofcreation-and-charles-landau). – PrinceOfCreation Feb 23 '19 at 20:01
1

Instead of checking if a key exists you could simply do ```out[fruit] = out.get(fruit, 0) + 1``` – Joshua Nixon Feb 23 '19 at 20:18

score 2 · Answer 2 · answered Feb 23 '19 at 19:45

I just quickly threw this together.

Code:

bought = {
"January": ['Apple', 'Banana', 'Orange', 'Kiwi', 'Raspberry'],
"February": ['Orange', 'Mango', 'Banana'], 
"March": ['Apple', 'Starfruit', 'Apricot']
}

numbers = {}
allb = []
for month in bought:
    allb += bought[month]
for item in allb:
    if item in numbers:
        numbers[item] += 1
    else:
        numbers[item] = 1
print(numbers) # Note you can format this however you want: just iterate through the dictionary again

Expected Output:

Apple: 2, Banana: 2, Orange: 2, Kiwi: 1, Raspberry: 1, Mango: 1, Starfruit: 1, Apricot: 1

Actual Output:

{'Apple': 2, 'Banana': 2, 'Orange': 2, 'Kiwi': 1, 'Raspberry': 1, 'Mango': 1, 'Starfruit': 1, 'Apricot': 1}

(As numbers is a dictionary, you can do very powerful things to lookup numbers. It acts like a table in this case.)

The first for loop goes through all the keys in bought (as it is formatted, all the months) and looks up their value. It then adds those values to my list allb (all bought) Next, I iterate through allb and if the item is already in numbers: I add 1 to the count, otherwise I set the count at 1.

Thank you as well. This also worked. I really appreciate it. — El numero dos, Feb 23 '19 at 19:49

score 2 · Answer 3 · answered Feb 24 '19 at 16:45

Using builtin Counter. One liner :-)

from collections import Counter

bought = {
    "January": ['Apple', 'Banana', 'Orange', 'Kiwi', 'Raspberry'],
    "February": ['Orange', 'Mango', 'Banana'],
    "March": ['Apple', 'Starfruit', 'Apricot']
}

count = Counter([y for x in bought.values() for y in x])
print(count)

Mykola Zotko · Answer 4 · 2019-02-23T20:19:25.440

1

You can use Counter:

from collections import Counter
from itertools import chain
from pprint import pprint

c = Counter(chain.from_iterable(bought.values()))

pprint(c)

Output:

Counter({'Apple': 2,
         'Banana': 2,
         'Orange': 2,
         'Kiwi': 1,
         'Raspberry': 1,
         'Mango': 1,
         'Starfruit': 1,
         'Apricot': 1})

edited Feb 23 '19 at 20:19

answered Feb 23 '19 at 20:16

Mykola Zotko

15,583
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Counting the words in a dictionary

4 Answers4

Code:

Expected Output:

Actual Output: