Sorting Python List Based on the Length of the String

Sorting Python list based on the length of the string

When you pass a lambda to sort, you need to return an integer, not a boolean. So your code should instead read as follows:

xs.sort(lambda x,y: cmp(len(x), len(y)))

Note that cmp is a builtin function such that cmp(x, y) returns -1 if x is less than y, 0 if x is equal to y, and 1 if x is greater than y.

Of course, you can instead use the key parameter:

xs.sort(key=lambda s: len(s))

This tells the sort method to order based on whatever the key function returns.

EDIT: Thanks to balpha and Ruslan below for pointing out that you can just pass len directly as the key parameter to the function, thus eliminating the need for a lambda:

xs.sort(key=len)

And as Ruslan points out below, you can also use the built-in sorted function rather than the list.sort method, which creates a new list rather than sorting the existing one in-place:

print(sorted(xs, key=len))

Sort list of strings in decreasing order (according to length)

You can simply use (-len(x), x) instead of (len(x), x) and reverse=True.

>>> sample_list = ['a', 'bb', 'cc', 'ddd', 'bbc']
>>> sorted(sample_list, key=lambda x: (-len(x), x))
['bbc', 'ddd', 'bb', 'cc', 'a']

This will sort the list with decreasing order of length first (because of -len(x)), and then sort the strings with the same length in a lexicographical order.

How to sort a list by length of string followed by alphabetical order?

You can do it in two steps like this:

the_list.sort() # sorts normally by alphabetical order
the_list.sort(key=len, reverse=True) # sorts by descending length

Python's sort is stable, which means that sorting the list by length leaves the elements in alphabetical order when the length is equal.

You can also do it like this:

the_list.sort(key=lambda item: (-len(item), item))

Generally you never need cmp, it was even removed in Python3. key is much easier to use.

Sort elements in list based on length first and alphabetical order second

use key argument of sorted function. convert each element to a tuple of priority. here (len(s), s), this means len(s) have more priority to s.

proxy_list = ['my', 'list', 'looks', 'like', 'this']
print(sorted(proxy_list, key=lambda s: (len(s), s)))
#>>> ['my', 'like', 'list', 'this', 'looks']

How to sort a list of lists based on length of inner list?

Using sort or sorted with custom key.

Ex:

data = [['A', '1234', 'X'],['B', '12', 'X'],['C', '12345', 'X'],['D', '123', 'X']]
data.sort(key=lambda x: len(x[1]), reverse=True)

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or sorted

data = sorted(data, key=lambda x: len(x[1]), reverse=True)

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print(data)

Sorting list of strings by their length and the last N characters

Use x[::-1].

Ex:

l = [['apple', 'banana', 'grape', 'kiwi', 'pear'], ['airplane', 'bicycle', 'boat', 'car']]
for i in l:
    print(sorted(i, key=lambda x: (x[::-1], len(x))))

Output:

['banana', 'apple', 'grape', 'kiwi', 'pear']
['bicycle', 'airplane', 'car', 'boat']

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output = [sorted(i, key=lambda x: (x[::-1], len(x))) for i in l]
print(output)
--> [['banana', 'apple', 'grape', 'kiwi', 'pear'], ['bicycle', 'airplane', 'car', 'boat']]

Sort list in pandas series by length of string

You can use pd.Series.apply with sorted:

def sorter(x):
    return sorted(x, key=len, reverse=True)

df['serie'] = df['serie'].apply(sorter)

How to sort array based on length and alphabet, where alphabet takes precedence over length python

You can use a sort key with a tuple:

d = ["@note1", "@note2", "=dir1/", "=dir2/", "@dir1/note1", "@dir1/note2"]
new_d = sorted(d, key=lambda x:(x[1:], len(x)))

Output:

['=dir1/', '@dir1/note1', '@dir1/note2', '=dir2/', '@note1', '@note2']

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