How to find consecutive items in a list, a set of 5, setting counter = 0 and reset the counter to ch-CodePudding

I have a list in which there are consecutive and non-consecutive values as shown in the code. I want to find 5 consecutive values and then append the 4 values excluding the first one in new list row1. for eg.201,202,203,204,205 this is a set of 5 consecutive values and I want to store 202,203,204,205 in row1 list. Then the new set begins from 206 onwards. If there is a set of 5 then the same approach like above should be followed. I have done the simple coding for it. But the output which I get when I print row1 is-- [202, 203, 204, 205, 206, 231]. I WANT [202, 203, 204, 205, 231]. What "if" condition should I provide to get the desired result? I had used break---if count becomes 4, but it stops the execution of the loop. So I used continue but it doesn't give the required output. Any help is much appreciated.

row1 = []
l = [201, 202, 203, 204, 205, 206, 230, 231]
count = 0
for i in range(len(l) - 1):  
    if l[i 1] == l[i]   1:
            row1.append(l[i 1])
            count  = 1
            if count == 4:  
                continue
    else:
        count = 0
print(row1)

CodePudding user response：

You can use sum of 1st and 5th element and sum of range(1st ele, 5th ele 1) if the sum is equals that means numbers are consecutive

lst = [201, 202, 203, 204, 205, 206, 230, 233, 234, 235, 236, 237, 239]
i, res = 0, []

while (i   4) < len(lst):
    sub = lst[i: i   5]
    if sum(range(sub[0], sub[-1]   1)) == sum(sub):
        res  = lst[i   1:i   5]
        i  = 5
    else:
        i  = 1
print(res) # [202, 203, 204, 205, 234, 235, 236, 237]

CodePudding user response：

following my proposal if i got your point right :

from math import ceil
l = [201, 202, 203, 204, 205, 206, 230, 231, 234, 235, 236, 237, 238]
l = sorted(l)

print([l[i 1:i 5]  for i in range(0,max(len(l), int(ceil(len(l)/5.0)*5)),5) if l[i 1:i 5] !=[]])

Result:

[[202, 203, 204, 205], [230, 231, 234, 235], [237, 238]]

CodePudding user response：

Not completely sure, but it appears to me that you want to find all sequences of 5 elements in a row. If that is the case you need a list of lists as a result. The counter can then be replaced by the length of the list holding the current sequence:

l = [201, 202, 203, 204, 205, 206, 230, 231, 234, 235, 236, 237, 238]

sequences = []
count = 0
current_sequence = []

for i in range(len(l) - 1):  
    if l[i 1] == l[i]   1:
            current_sequence.append(l[i 1])
            if len(current_sequence) == 4:  
                sequences.append(current_sequence)
                current_sequence = []
                continue
    else:
        current_sequence = []
        
print(sequences)

Output for input with two sequences:

[[202, 203, 204, 205], [235, 236, 237, 238]]

A better solution would be this I think as you do not create lists that do not appear in the result:

l = [201, 202, 203, 204, 205, 206, 207, 230, 231, 234, 235, 236, 237, 238]

sequences = []
i=0
while i < len(l) - 4:  
    if l[i] == l[i 1]-1 == l[i 2]-2 == l[i 3]-3 == l[i 4]-4:
        sequences.append(l[i 1:i 5])
        i=i 5
    else:
        i=i 1
            
print(sequences)

CodePudding user response：

This code does exactly what OP wants. I begin at index 0 and append the next four consecutive numbers(if there are that much). After that I move to the element after the last element in the previous consecutive and start all over again.

row1 = []
l = [201, 202, 203, 204, 205, 206, 230, 231]

# Counter to iterate through l
i = 0
n = len(l)

while i < n:
    for j in range(1, 5):
        ind = i   j
        
        # We use try and except so that we can handle any out of bounds errors
        try:
            curr = l[ind]
            prev = l[ind-1]
            if curr == prev 1:
                row1.append(curr)
            else:
                i = ind
                break
        except:
            i = ind
            break
            
        # After reaching the last consecutive in the five, move on to the next element
        if j == 4:
            i = ind   1

Output for this:

[202, 203, 204, 205, 231]