查找元素是否在python列表中连续出现n次

Question

myList = [True, True, False, False, True, False, True, False, False]

I want to find if True appears 3 times in a row.

I can find it by doing:

for x0, x1, x2 in zip(myList, myList[1:], myList[2:]):
    if x0 == True and x1 == True and x2 == True:
        print True

Is there a better way?

Answer 1

Use itertools.groupby() to group elements, then count each group. Using the any() function lets you exit the loop early if a match was found:

from itertools import groupby, islice

print any(sum(1 for _ in islice(g, 3)) == 3 for k, g in groupby(myList) if k)

The if k filters the groups to only count the groups of True values.

The itertools.islice() function ensures we only look at the first 3 elements of a group, and ignore the rest of that group. This way you avoid having to count the next however many True values just to determine that you found at least 3.

Demo:

>>> from itertools import groupby, islice
>>> myList = [True, True, False, False, True, False, True, False, False]
>>> [sum(1 for _ in islice(g, 3)) for k, g in groupby(myList) if k]
[2, 1, 1]
>>> any(sum(1 for _ in islice(g, 3)) == 3 for k, g in groupby(myList) if k)
False
>>> myList = [True, True, False, False, True, True, True, True, False, True, False, False]
>>> [sum(1 for _ in islice(g, 3)) for k, g in groupby(myList) if k]
[2, 3, 1]
>>> any(sum(1 for _ in islice(g, 3)) == 3 for k, g in groupby(myList) if k)
True

I used a list comprehension to show the group sizes (counting only True groups) to show why the any() call returns False first, then True ; the second example has a group of 4 consecutive True values.

Answer 2

I like the brevity of the groupby , but I find the following slightly more readable so I thought I'd add an alternative;

needle = 3 * [True]
any(1 for i in range(len(myList)) if myList[i:i+len(needle)] == needle)

Answer 3

Here it is my solution

% cat hsol.py
import itertools

myList = [True, True, False, False, True, False, True, False, False]

def test_sequentiality(l, item, n):

    if n>len(l): return False
    s = 0
    for i in l:
        if i != item:
            s = 0
        else:
            s = s+1
            if s == n: return True

    return False

print test_sequentiality(myList, True, 3)
print test_sequentiality(myList, True, 2)
% python2 hsol.py
False
True
% 

Answer 4

One way would be to reduce the if statements etc in your question, and directly print the boolean value.

for x0, x1, x2 in zip(myList, myList[1:], myList[2:]):
    print x0 == x1 == x2 == True

Using any statement, we can short circuit the same

any(x0 == x1 == x2 == True for (x0, x1, x2) in zip(myList, myList[1:], myList[2:]))

Answer 5

print reduce(
             lambda acc, x: (acc[0]+1 if x else 0, max(acc[0], acc[1])),
             myList+[None],
             (0, 0)
             )[1]

It will accumulate the sequential counts of True with maximal count of sequential True s +[None] is needed in case the longest sequence ends with last element.

Answer 6

A straightforward and (in my opinion) very easy-to-understand way:

myList = [True, True, False, False, True, False, True, False, False]
v = True  # search for this value
n = 3  # number of consecutive appearances of v
vList = [v] * n

for i in range(len(myList) - n):
    if myList[i:i + n] == vList:
        print True
        break
else:
    print False

It's kind of C-style (or Pascal or Fortran, etc.), but simple. The above code assumes you just want a single True or False at the end. You can easily adapt it if you want to print the index of each found occurrence.