# To sort the list in place… ut.sort(key=lambda x: x.count, reverse=True) # To return a new list, use the sorted() built-in function… newlist = sorted(ut, key=lambda x: x.count, reverse=True) More on sorting by keys.
str.count(sub[, start[, end]]) Return the number of non-overlapping occurrences of substring sub in the range [start, end]. Optional arguments start and end are interpreted as in slice notation. >>> sentence=”Mary had a little lamb” >>> sentence.count(‘a’) 4
To do this in any ES5-compatible environment, such as Node.js, Chrome, Internet Explorer 9+, Firefox 4+, or Safari 5+: Object.keys(obj).length Browser compatibility Object.keys documentation (includes a method you can add to non-ES5 browsers)
If you only want a single item’s count, use the count method: >>> [1, 2, 3, 4, 1, 4, 1].count(1) 3 Important: this is very slow if you are counting multiple different items Each count call goes over the entire list of n elements. Calling count in a loop n times means n * n … Read more