扫描列表连续5个大于x的值

Question

I want to scan a large list for consecutive values that are greater than x. This example x is greater than 1.0.

For example,

my_list = [0.2, 0.1, 0.3, 1.1, 0.7, 0.5, 1.2, 1.3, 1.4, 1.2, 1.9, 1.1, 0.2, 1.3, 1.5, 1.4, 1.2, 1.1, 0.2, 1.3, 0.1., 1.6, 0.2, 0.5, 1.0, 1.1, 0.2]

I can subset this list by

for i in range(0, len(my_list)):
    subset = my_list[i:i+5]

so I get

[0.2, 0.1, 0.3, 1.1, 0.7]
[0.1, 0.3, 1.1, 0.7, 0.5]
[0.3, 1.1, 0.7, 0.5, 1.2]
[1.1, 0.7, 0.5, 1.2, 1.3]
[0.7, 0.5, 1.2, 1.3, 1.4]
[0.5, 1.2, 1.3, 1.4, 1.2]
[1.2, 1.3, 1.4, 1.2, 1.9] <-- values I want
[1.3, 1.4, 1.2, 1.9, 1.1] <-- values I want
[1.4, 1.2, 1.9, 1.1, 0.2]
[1.2, 1.9, 1.1, 0.2, 1.3]
[1.9, 1.1, 0.2, 1.3, 1.5]
[1.1, 0.2, 1.3, 1.5, 1.4]
[0.2, 1.3, 1.5, 1.4, 1.2]
[1.3, 1.5, 1.4, 1.2, 1.1] <-- values I want

What is the best way to do this?

Answer 1

You can do that as follows:

my_list = [0.2, 0.1, 0.3, 1.1, 0.7, 0.5, 1.2, 1.3, 1.4, 1.2, 1.9, 1.1, 0.2, 1.3, 1.5, 1.4, 1.2, 1.1, 0.2, 1.3, 0.1, 1.6, 0.2, 0.5, 1.0, 1.1, 0.2]

x = 1

result = [my_list[i:i+5] for i in range(len(my_list)-4) if all(i > x for i in my_list[i:i+5])]

Answer 2

Here's an itertools based approach that won't need any extra memory and returns results as a generator:

from itertools import tee, islice

def find_consecutive(the_list, threshold, count=5):
    my_iters = tee(the_list, count)
    for i, it in enumerate(my_iters):
        next(islice(it, i, i), None)
    return (f for f in zip(*my_iters) if all(x > threshold for x in f))

my_list = [0.2, 0.1, 0.3, 1.1, 0.7, 0.5, 1.2, 1.3, 1.4, 1.2, 1.9, 1.1, 0.2, 1.3, 1.5, 1.4, 1.2, 1.1, 0.2, 1.3, 0.1, 1.6, 0.2, 0.5, 1.0, 1.1, 0.2]
list(find_consecutive(my_list, 1.0))
# [(1.2, 1.3, 1.4, 1.2, 1.9),
# (1.3, 1.4, 1.2, 1.9, 1.1),
# (1.3, 1.5, 1.4, 1.2, 1.1)]

The function is parameterized by threshold and count so you can look for any N consecutive values. You could even factor out the condition by passing in a function for that instead of just a threshold value.

Answer 3

>>>my_list = [0.2, 0.1, 0.3, 1.1, 0.7, 0.5, 1.2, 1.3, 1.4, 1.2, 1.9,
1.1, 0.2, 1.3, 1.5, 1.4, 1.2, 1.1, 0.2, 1.3, 0.1, 1.6, 0.2, 0.5, 1.0,1.1, 0.2]

>>>x = 1.0
>>>for i in range(0, len(my_list)):
       subset = my_list[i:i+5]
       if(all(item >x for item in subset)):
           print subset

[1.2, 1.3, 1.4, 1.2, 1.9]
[1.3, 1.4, 1.2, 1.9, 1.1]
[1.3, 1.5, 1.4, 1.2, 1.1]

Answer 4

It might save some time to keep a counter so you don't need to keep on checking values that are already checked. Here n=5

def scan_k(arr, val,n):
    counter = 0
    results = set()
    for i in range(len(arr)):
        if arr[i] > val:
            counter +=1
        else:
            counter = 0
            continue
        if counter >= n:
            results.add(arr[(i-n+1):i])
return(results)

Answer 5

Here's a fast solution using a generator function and a variable that keeps track of the number encountered so far:

def find_n_consecutive_greater_than_x(mylist, n, x):
    num_greater_than_x = 0
    for index, val in enumerate(mylist):
        if val > x:
            num_greater_than_x += 1
            if num_greater_than_x == n:
                yield tuple(mylist[index-n+1:index+1])
                num_greater_than_x -= 1
        else:
            num_greater_than_x = 0

mylist = [2]*6
n = 5
x = 1.0

print(list(find_n_consecutive_greater_than_x(mylist, n, x)))
# [(2, 2, 2, 2, 2), (2, 2, 2, 2, 2)]

This will be much faster than any solution that computes all slices of length 5 of a given list because it only processes each element once and avoids the creation of objects, which is painfully slow in most Python implementations.