Python 检查列表中不定数量的连续元素是否满足条件

Question

Basically I am playing around with nested lists, and would like for a given input n , to find if n consecutive elements in the list satisfy a predetermined condition.

For example:

n = 3
lst = [[1,2,3],[6,7,8],[4,5,6],[1,3,4,5]]

And I would like to know if there are n consecutive sublists within lst that have length greater than or equal to 3 (Or some other condition). What is the most efficient way to find firstly:

1. Is the condition satisfied
2. The range of indices for when condition is satisfied n times

An example output for my illustrative example would be:

True
index_range = (0,2) #Only need the first instance when n consec achieved

Any help would be greatly appreciated, preferably code which doesn't rely too heavily on inbuilt Python libraries such as itertools or numpy as I am new to Python and would like to better understand how such a process works. Thanks in advance!

Answer 1

different conditionals, one loop, constant memory usage, early exiting

def func(lst, n, condition):
    n_consec = 0
    if len(lst) < n:
        return (False, None)
    for i,sub in enumerate(lst):
        if condition(sub):
            n_consec += 1
            if n_consec == n:
                return (True, (i-n+1, i))
        else:
            n_consec = 0
            if len(lst) - (i + 1) < n:
                return (False, None)
    return (False, None)

print(func(lst,n, lambda sub: len(sub) >= n))

Answer 2

lst = [[1,2,3],[6,7,8],[4,5,6],[1,3,4,5]]

def check_cond_func(lst, n):
    count = 0
    range = []
    for i, l in enumerate(lst):
        if count == n:
            break
        if len(l) != n:
            count = 0
        if len(l) == n:
            count += 1
            range.append(i)
    if count == n:
        return (True, (min(range), max(range)))
    else:
        return False

yay_nay, index_range = check_cond_func(lst, 3)

print(yay_nay)    
print(index_range)

Output

True
(0, 2)

Answer 3

Allows setting different conditionals

def check(lst, conditional, n):
  " Checks for consective sublist satisfying conditional "

  for i in range(len(lst)):
    # Starting with i-th sublist
    for j in range(i, i+n):
      # checking for n consecutive sublists
      # where conditional(lst[j]) is True
      if not conditional(lst[j]):
        break
    else:
      return True, i, i+n-1
  return False

def condition(sublist):
  " Condiiton on length "
  return len(sublist) >= 3

lst = [[1,2,3],[6,7,8],[4,5,6],[1,3,4,5]]
print(check(lst, condition, 3))

# Out: (True, 0, 2)