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 = [item[0] item in mylist]     return lst2.count(str(n))  print count(result,1) #prints 2 print count(result,2) #prints 1