Creating a dictionary where the key is an integer and the value is the length of a random sentence

Question

Super new to to python here, I've been struggling with this code for a while now. Basically the function returns a dictionary with the integers as keys and the values are all the words where the length of the word corresponds with each key.

So far I'm able to create a dictionary where the values are the total number of each word but not the actual words themselves.

So passing the following text

"the faith that he had had had had an affect on his life"

to the function

def get_word_len_dict(text):
    result_dict = {'1':0, '2':0, '3':0, '4':0, '5':0, '6' :0}
    for word in text.split():
        if str(len(word)) in result_dict:
            result_dict[str(len(word))] += 1
    return result_dict

returns

1 - 0
2 - 3
3 - 6
4 - 2
5 - 1
6 - 1

Where I need the output to be:

2 - ['an', 'he', 'on']
3 - ['had', 'his', 'the']
4 - ['life', 'that']
5 - ['faith']
6 - ['affect']

I think I need to have to return the values as a list. But I'm not sure how to approach it.

Solution 1

the problem here is you are counting the word by length, instead you want to group them. You can achieve this by storing a list instead of a int:

def get_word_len_dict(text):
    result_dict = {}
    for word in text.split():
        if len(word) in result_dict:
            result_dict[len(word)].add(word)
        else:
            result_dict[len(word)] = {word} #using a set instead of list to avoid duplicates
    return result_dict

Other improvements:

• don't hardcode the key in the initialized dict but let it empty instead. Let the code add the new keys dynamically when necessary
• you can use int as keys instead of strings, it will save you the conversion
• use sets to avoid repetitions

---

Using groupby

Well, I'll try to propose something different: you can group by length using groupby from the python standard library

import itertools
def get_word_len_dict(text):
    # split and group by length (you get a list if tuple(key, list of values)
    groups = itertools.groupby(sorted(text.split(), key=lambda x: len(x)), lambda x: len(x))
    # convert to a dictionary with sets 
    return {l: set(words) for l, words in groups}

Solution 2

Fixing Sabian's answer so that duplicates aren't added to the list:

def get_word_len_dict(text):
    result_dict = {1:[], 2:[], 3:[], 4:[], 5:[], 6 :[]}
    for word in text.split():
        n = len(word)
        if n in result_dict and word not in result_dict[n]:
            result_dict[n].append(word)
    return result_dict

Solution 3

I think that what you want is a dic of lists.

result_dict = {'1':[], '2':[], '3':[], '4':[], '5':[], '6' :[]}
for word in text.split():
    if str(len(word)) in result_dict:
        result_dict[str(len(word))].append(word)
return result_dict

Solution 4

Instead of defining the default value as 0, assign it as set() and within if condition do, result_dict[str(len(word))].add(word).

Also, instead of preassigning result_dict, you should use collections.defaultdict.

Since you need non-repetitive words, I am using set as value instead of list.

Hence, your final code should be:

from collections import defaultdict
def get_word_len_dict(text):
    result_dict = defaultdict(set)
    for word in text.split():
        result_dict[str(len(word))].add(word)
    return result_dict

In case it is must that you want list as values (I think set should suffice your requirement), you need to further iterate it as:

for key, value in result_dict.items():
    result_dict[key] = list(value)

Solution 5

You say you want the keys to be integers but then you convert them to strings before storing them as a key. There is no need to do this in Python; integers can be dictionary keys.

Regarding your question, simply initialize the values of the keys to empty lists instead of the number 0. Then, in the loop, append the word to the list stored under the appropriate key (the length of the word), like this:

string = "the faith that he had had had had an affect on his life"

def get_word_len_dict(text):
    result_dict = {i : [] for i in range(1, 7)}
    for word in text.split():
        length = len(word)
        if length in result_dict:
            result_dict[length].append(word)
    return result_dict      

This results in the following:

>>> get_word_len_dict(string)
{1: [], 2: ['he', 'an', 'on'], 3: ['the', 'had', 'had', 'had', 'had', 'his'], 4: ['that', 'life'], 5: ['faith'], 6: ['affect']}

If you, as you mentioned, wish to remove the duplicate words when collecting your input string, it seems elegant to use a set and convert to a list as a final processing step, if this is needed. Also note the use of defaultdict so you don't have to manually initialize the dictionary keys and values as a default value set() (i.e. the empty set) gets inserted for each key that we try to access but not others:

from collections import defaultdict

string = "the faith that he had had had had an affect on his life"

def get_word_len_dict(text):
    result_dict = defaultdict(set)
    for word in text.split():
        length = len(word)
        result_dict[length].add(word)
    return {k : list(v) for k, v in result_dict.items()}

This gives the following output:

>>> get_word_len_dict(string)
{2: ['he', 'on', 'an'], 3: ['his', 'had', 'the'], 4: ['life', 'that'], 5: ['faith'], 6: ['affect']}

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