python - Count the number of lists that starts with the same item in a list -
so have list like: result = [["1", "1", "a", 8.2],["1", "2", "c", 6.2],["2", "1", "a", 8.2]] i want function returns count of number of lists starts (index[0]) variable "n" . so, if n = '1' in case 2, if n = '2' i 1 . edit: i've tried few things this, can't work. def count(list,n): result = [] value = 0 in list: if str(i[0]) == n: value = value + 1 sum.append[value] return len(sum) print count(result,1) result = [["1", "1", "a", 8.2],["1", "2", "c", 6.2],["2", "1", "a", 8.2]] lst2 = [item[0] item in result] lst2.count("1") result = [["1", "1", "a", 8.2],["1", "2", "c", 6.2],["2", "1", "a", 8.2]] def count(mylist,n): lst2 = ...