Get Word Length In Dict Python

Example: def test_get_word_len_dict(): text = 'May your coffee be strong and your Monday be short' the_dict = get_word_len_dict(text) print_dict_in_key_order(the_dict)

Solution 1:

A defaultdict of sets should do. First, let's define a function.

from collections import defaultdict

def get_word_counts(text):
    d = defaultdict(set)

    for word in text.split():
        d[len(word)].add(word)    # observe this bit carefully

    return {k : list(v) for k, v in d.items()}

The idea is to find the length of each word, and insert it into the list/set that it belongs to. Once you've defined the function, you may call it as you please.

text = "May your coffee be strong and your Monday be short"print(get_word_counts(text))
{2: ['be'], 3: ['and', 'May'], 4: ['your'], 5: ['short'], 6: ['coffee', 'strong', 'Monday']}

Solution 2:

You can also use itertools.groupby with sorted in order to get similar "lazy" results:

a = 'a long list of words in nice'
x = groupby(sorted(a.split(), key=len), len)  # word countsprint(dict((a, list(b)) for a, b in x))
>>> {1: ['a'], 2: ['of', 'in'], 4: ['long', 'list', 'nice'], 5: ['words']}

By "lazy" I mean that things would not start actually computing (e.g. if you have a really large string), until you start iterating over it. Be careful with iterators returned by groupby() though! That is quite easy to accidentally empty them and then attempt to read second time (and get empty list).

The returned group is itself an iterator that shares the underlying iterable with groupby(). Because the source is shared, when the groupby() object is advanced, the previous group is no longer visible. So, if that data is needed later, it should be stored as a list