Finding a Run in a List Without using any pre-defined Python Functions

Question

I'm trying to find a run in a list without using pre-defined list functions in Python, such as enumerate, zip, index, etc.

By a run, I mean given a list [1,2,3,6,4], I want to return a new list [1,2,3] as well as returning where the run starts and ends, and the length of the run.

Below is my attempt at this but I can't seem to get it, I seem to always keep running into the problem where I get an index out of range error.

def run_cards (hand_shuffle, hand_sample):

    true_length = len(hand_shuffle) - 1
    y = 0
    r = []
    for x in range(len(hand_shuffle)):
        y = x + 1

        if y <= true_length: #Checks if y is in range with x 
            t = hand_shuffle[y] - hand_shuffle[x] #Subtracts the two numbers to see if they run
            r.append(t)

    lent = len(hand_shuffle)

    if hand_shuffle[lent-1] - hand_shuffle[lent-2] == 1:
        r.append(1)
    h = []

    for i in range(len(r)):

        if r[i] == 1: #If t from above for loop is equal to 1 append it to the new list
            h.append(i)
            h.append(i+1)

    p = []

    for j in h:
        p.append(hand_shuffle[j])

    print (p) 
    return hand_shuffle, hand_sample

Right now, my code runs, but doesn't give me what I am looking for.

`range` and `len` are "predefined functions" and yet you use them. How are they classified differently than `enumerate` and `zip`? — , Nov 14 '13 at 18:35
what if there is more than one run? (eg `[1,2,3,6,4,5,6,7]`) — Joran Beasley, Nov 14 '13 at 18:45
@iCodez functions such as enumerate, zip, index make this task a lot more simple and really doesn't test my knowledge of loops and lists. — l00kitsjake, Nov 14 '13 at 18:47
`break` certainly does test your knowledge of control flow however — Joran Beasley, Nov 14 '13 at 18:49
@JoranBeasley For this question specifically, I only need to worry about the first run. — l00kitsjake, Nov 14 '13 at 18:49

score 1 · Answer 1 · edited May 23 '17 at 11:50

zip and enumerate less version of my previous answer:

def solve(lis):
    run_length = 0
    ind = 0
    for i in range(len(lis)-1):
        x, y = lis[i], lis[i+1]
        if run_length and y-x != 1:
            break
        if y-x == 1:
            if not run_length:
                ind = i
            run_length += 1
    if run_length:
        return run_length+1, ind
    return -1

Demo:

>>> solve([1,2,3,6,4])
(3, 0)                  #run length, index
>>> solve([0, 0, 0, 4, 5, 6])
(3, 3)
>>> solve([4, 5, 5, 1, 8, 3, 1, 6, 2, 7])
(2, 0)
>>> solve([1, 1, 1, 2, 3, 5, 1, 1])
(3, 2)
>>> solve([1, 1, 1, 1])
-1
>>> solve([1, 2, 5, 6, 7] )
(2, 0)
>>> solve([1, 0, -1, -2, -1, 0, 0])
(3, 3)

thanks again, anyway I can do this without using "break"? I will try and figure out myself. — l00kitsjake, Nov 14 '13 at 18:46

score 1 · Accepted Answer · answered Nov 14 '13 at 18:54

>>> runFind(L)
[(4, 0), (2, 3), (2, 4)]
>>> def runFind(L):
...   i = 1
...   start = 0
...   answer = []
...   while i < len(L):
...     if L[i] != L[i-1]+1:
...       answer.append((i-start, start))
...       start = i
...     i += 1
...   answer.append((i-start, start))
...   
...   return answer
... 
>>> L = [1,2,3,6,4]
>>> runFind(L)
[(3, 0), (1, 3), (1, 4)]
>>> L = [0, 0, 0, 4, 5, 6]
>>> runFind(L)
[(1, 0), (1, 1), (1, 2), (3, 3)]
>>> L = [4, 5, 5, 1, 8, 3, 1, 6, 2, 7]
>>> runFind(L)
[(2, 0), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9)]
>>> L = [1, 1, 1, 2, 3, 5, 1, 1]
>>> runFind(L)
[(1, 0), (1, 1), (3, 2), (1, 5), (1, 6), (1, 7)]
>>> L = [1, 1, 1, 1]
>>> runFind(L)
[(1, 0), (1, 1), (1, 2), (1, 3)]
>>> L = [1, 2, 5, 6, 7]
>>> runFind(L)
[(2, 0), (3, 2)]
>>> L = [1, 0, -1, -2, -1, 0, 0]
>>> runFind(L)
[(1, 0), (1, 1), (1, 2), (3, 3), (1, 6)]

I really like this way, this would be great if I need to find more than one run in the list. — l00kitsjake, Nov 14 '13 at 18:57

score 0 · Answer 3 · answered Nov 14 '13 at 18:39

Just have two flags where one flag represents the start and one represents the end.

runs = []
indices = []
for start in range(len(hand)):
    run = [hand[start]]
    for end in range(start+1,len(hand)):
        if hand[end] == hand[end-1]:
            run.append(hand[end])
        else:
            if len(run) > min_run_length:
                runs.append(run)
                indices.append((start,end))
            break

Finding a Run in a List Without using any pre-defined Python Functions

3 Answers3