CODEWITHSUNDEEP
python
Leonardo Echeverria
Leonardo Echeverria

Reputation: 45

Counting capitalized or non-capitalize in dictionary

So I have this dictionary, 

word_counts = {'Two':2, 'one':3, 'Forks.':1, 'knife.':2, 
'glasses.':2, 'plate.':1, 'Naptkin.':1, 'his':2}

And i need to compute how to how many are capitalized and non-capitalized. I know I need to do this by getting the keys of the dictionary and loop through them all, but I'm having trouble. Please help. Thank you!

Tried to use counter variables with for loops and isupper() and islower() but its not working. If you haver better ways please let me know!

#What ive done so far
word_counts = {'Two':2, 'one':3, 'Forks.':1, 'knife.':2, 
    'glasses.':2, 'plate.':1, 'Naptkin.':1, 'his':2}

for word, occurence in words:
        upper_counter = 0
        lower_counter = 0
        for word in word_counts.items():
            if word.isupper():
                upper_counter += 1
            elif word_islower():
                lower_counter += 1

Upvotes: 3

Views: 168

Answers (4)

Dilip Kharel
Dilip Kharel

Reputation: 1

Here's a simplest approach using 1. list comprehension, 2. string formatting.

word_counts = {'Two':2, 'one':3, 'Forks.':1, 'knife.':2,
'glasses.':2, 'plate.':1, 'Naptkin.':1, 'his':2}

Create lists of caps and non-caps words.

caps = [i for i in word_counts.keys() if i[0].isupper()]
non_caps = [i for i in word_counts.keys() if i not in caps]

print("There are {} capitalized words. 
They are:{}{}".format(len(caps), "\n", caps))

print("There are {} non-capitalized words. 
They are:{}{}".format(len(non_caps), "\n", non_caps))

Upvotes: 0

user6172474
user6172474

Reputation:

Your purpose is counting all uppercase from keys, or count number of keys that contain uppercase.

Count all uppercase from keys:

all_keys = ''.join(list(word_counts.keys()))
count = len([w for w in all_keys if w.isupper()])

Count all words contain upper

count = len([word for word in word_couhnts.keys() if
                              any([w for w in word if w.isupper()])])

Upvotes: 0

Iain Shelvington
Iain Shelvington

Reputation: 32244

# The counters should be initialised before the loop
upper_counter = 0
lower_counter = 0

for word in word_counts:  # Looping over a dictionary gives the keys
    if word[0].isupper():  # Just have to check if the first character is upper case
        upper_counter += 1
    else:
        lower_counter += 1

Upvotes: 2

Ajax1234
Ajax1234

Reputation: 71461

You can use re:

import re
cap, non_cap = 0, 0
word_counts = {'Two':2, 'one':3, 'Forks.':1, 'knife.':2, 'glasses.':2, 'plate.':1, 'Naptkin.':1, 'his':2}
for a in word_counts:
   if re.findall('^[A-Z]\w+', a):
      cap += 1
   else:
      non_cap += 1

Output:

3 #cap
5 #non_cap

Upvotes: 0

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