I'm sure this can be done, but I have thus far been unsuccessful:
I have a list of strings. I want to create a dictionary with the length of said strings (which can be expressed as a range) as the key and the string itself as the value.
example: Here's something like the list I have: ['foo','bar','help','this','guy']
I'd like to end up with a dictionary like this: {3:['foo','bar','guy], 4:['this','help']}
Using defaultdict so you don't have to check whether or not to create the list for a new key:
from collections import defaultdict
x = ['foo','bar','help','this','guy']
len_dict = defaultdict(list)
for word in x:
len_dict[len(word)].append(word)
len_dict
#
# Out[5]: defaultdict(list, {3: ['foo', 'bar', 'guy'], 4: ['help', 'this']})
You can use a dictionary as a container with setdefault:
lst = ['foo','bar','help','this','guy']
result = {}
for w in lst:
result.setdefault(len(w), []).append(w)
result
# {3: ['foo', 'bar', 'guy'], 4: ['help', 'this']}
You can do it like that:
d={}
lst=['foo','bar','help','this','guy']
for i in lst:
if len(i) in d:
d[len(i)].append(i)
else:
d[len(i)]=[i]
This solution is pythonic, elegant and fast: (by the Famous Raymond Hettinger in one of his many conferences).
dict.setdefault is the dictionary method that initialises a key-value if the key is not found in dict as well as performing dict.get for provided key.
l = ['foo','bar','help','this','guy']
d = {}
for e in l:
key = len(e)
d.setdefault(key, []).append(name)
print(d)
Output:
{3: ['foo', 'bar', 'guy'], 4: ['help', 'this']}
This solution is the modern way of the solution above:
defaultdict from collection is a subclass of dict that automatically initialises value to any given key that is not in the defaultdict.
from collections import defaultdict
l = ['foo','bar','help','this','guy']
d = defaultdict(list)
for e in l:
key = len(e)
d[key].append(e)
print(d)
Output:
defaultdict(<class 'list'>, {3: ['foo', 'bar', 'guy'], 4: ['help', 'this']})
Similar to what have been said, but using the get method of dict class:
the_list=['foo','bar','help','this','guy']
d = {}
for word in the_list:
key = len(word)
d[key] = d.get(key, []) + [word]
print(d)
# {3: ['foo', 'bar', 'guy'], 4: ['help', 'this']}
Another approach:
from collections import defaultdict
given_list=['foo','bar','help','this','guy']
len_words=[len(i) for i in given_list]
d=defaultdict(list)
for i,j in list(zip(len_words,given_list)):
d[i].append(j)
