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 = ...