计算列表中重复的前导整数数量的最快方法

Question

example of list being looked at (list will always be sorted). Size of list is 4^12. [1,1,1,1, ...... , 1,1,4]

At the moment I have a for loop that iterates through the list with a counter until it finds the first non 1 and then breaks and returns the counter. This is quicker than using the count function since the list is already sorted. I am now implementing binary search and I was wondering if anyone can think of a better way to count the number of leading ones in a ordered list.

Answer 1

You can use itertools.groupby :

import itertools
s = [1,1,1,1,1,1,4]
runs = [len(list(b)) for a, b in itertools.groupby(s)][0]

Output:

6

Answer 2

Bisect is built in and seems comparably fast to binary sort;

import bisect
import time

import itertools

test = [1] * 100000000
test[len(test) - 1] = 4

start = time.time()
print(bisect.bisect_right(test,1))
print(time.time() - start)

start = time.time()
runs = [len(list(b)) for a, b in itertools.groupby(test)][0]
print(runs)
print(time.time() - start)

start = time.time()
length= len(test)-test[::-1].index(1)
print(length)
print(time.time() - start)


# Bisect
99999999
3.719329833984375e-05

# GroupBy
99999999
1.2406058311462402

# Length
99999999
0.3920767307281494

Answer 3

If you only want to find the count of the first value then you can use this:

a = [1,1,1,...,1,4]
first_value = a[0]
length= len(a)-a[::-1].index(first_value) #get index starting from the end