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Completely loop over a list with a different start index

So I have this list

station_list = [1, 2, 3, 4]

This is an example, the real list may be longer, but always this pattern with increasing integers. As part of my code, if any condition is met, I want to access that list at a different point than 1. Let's say I want to start iterating at 2. How can I iterate over a list using for-loop

starting at 2, but still getting all elements exactly once? Therefore, the iteration order must be 2, 3, 4, 1

.

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5 answers


You can do it:

station_list = [1, 2, 3, 4]
start = 1  # item = 2 is at index position 1!
for item in (station_list[start:] + station_list[:start]):
    # your code here

these outputs:



[2, 3, 4, 1]

You can read more about slicing in python here

And if you want more information about lists in Python, this might help.

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Easy way: create a reordered list using two snippets from the original list:

station_list = [1, 2, 3, 4]
start = 1
for i in station_list[start:] + station_list[:start]:
    print (i)


If you don't want to create a list of reordered stations, you can use index wrapping:

station_list = [1, 2, 3, 4]
start = 1
n = len(station_list)
for i in range(n):
    print (station_list[(i+start)%n])
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This can be achieved with a simple collection.deque and then rotated at any index.

from collections import deque
d = deque([1, 2, 3, 4])
d.rotate(-1)
print (d)

Output:

deque([2, 3, 4, 1])
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station_list = [1, 2, 3, 4]
element_to_find = 2
start = station_list.index(element_to_find) - len(station_list)

for index in range(start, station_list.index(element_to_find)):
    print(station_list[index])
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First, define the visible list where it is # !

. (That is, after defining the list.)

int_list = [1, 2, 3, 4, 5, 6]
seen = [] # {1}

Then define the beginning.

start = 2

This is part of the cycle.

for n in range(start, list_length): # starts at position you desire...
    print(int_list[n])
    seen.append(int_list[n])

If you start at a position not equivalent to the start, this goes back again ...

if int_list != seen: # if there are still some ints at start, go redo the loop until a seen int is found
    for x in range(0, list_length):
        print(int_list[x])
        seen.append(int_list[x])
        if int_list[x+1] in seen:
            break

All code:

int_list = [1, 2, 3, 4, 5, 6]
seen = []

start = 2 

list_length = len(int_list)

for n in range(start, list_length): # starts at position you desire...
    print(int_list[n])
    seen.append(int_list[n])

if int_list != seen:
# if there are still some ints at start, go redo the 
#loop until a seen int is found
    for x in range(0, list_length):
        print(int_list[x])
        seen.append(int_list[x])
        if int_list[x+1] in seen:
            break

Not that compact, but it works.

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