reverse range vs normal range when trying to remove items from a list using a for loop

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For this for loop, this code:

lst = ['NORTH', 'SOUTH', 'EAST', 'EAST']



for num in range(0, len(lst) - 1):

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

I know it doesn't work because by deleting the items in the list, the len(lst) changes, however, why does this work:

for num in range(len(lst) - 1 , -1 , -1):

    if lst[num] == 'SOUTH' and lst[num - 1] == 'NORTH':

        del lst[num]

        del lst[num - 1]

for this, aren't we also changing the list length? Why does the list length not matter in a reverse range?

edited Jan 3 at 7:34

asked Jan 3 at 7:29

tzyyyy

1

By moving forward and removing list items, you are changing the indexes of the items that you have not processed yet. By moving backward and removing list items, you are changing the indexes of the items that you have processed.

– DYZ
Jan 3 at 7:32

1

@DYZ By deleting two items at a time, the second code is also subject to IndexError.

– user9522315
Jan 3 at 7:40

@iBot In theory, yes. But not in this case: deletion takes place only if the current item is SOUTH and the previous is NORTH; if the condition is true for num, it is not true for num-1.

– DYZ
Jan 3 at 7:42

1

@DYZ If in the first iteration, two elements are deleted, then in the second iteration lst[num] will fail (where num = len - 2). See my answer for an example.

– user9522315
Jan 3 at 7:44

@iBot Fair enough.

– DYZ
Jan 3 at 7:45

add a comment |

For this for loop, this code:

lst = ['NORTH', 'SOUTH', 'EAST', 'EAST']



for num in range(0, len(lst) - 1):

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

I know it doesn't work because by deleting the items in the list, the len(lst) changes, however, why does this work:

for num in range(len(lst) - 1 , -1 , -1):

    if lst[num] == 'SOUTH' and lst[num - 1] == 'NORTH':

        del lst[num]

        del lst[num - 1]

for this, aren't we also changing the list length? Why does the list length not matter in a reverse range?

edited Jan 3 at 7:34

asked Jan 3 at 7:29

tzyyyy

1

By moving forward and removing list items, you are changing the indexes of the items that you have not processed yet. By moving backward and removing list items, you are changing the indexes of the items that you have processed.

– DYZ
Jan 3 at 7:32

1

@DYZ By deleting two items at a time, the second code is also subject to IndexError.

– user9522315
Jan 3 at 7:40

@iBot In theory, yes. But not in this case: deletion takes place only if the current item is SOUTH and the previous is NORTH; if the condition is true for num, it is not true for num-1.

– DYZ
Jan 3 at 7:42

1

@DYZ If in the first iteration, two elements are deleted, then in the second iteration lst[num] will fail (where num = len - 2). See my answer for an example.

– user9522315
Jan 3 at 7:44

@iBot Fair enough.

– DYZ
Jan 3 at 7:45

add a comment |

For this for loop, this code:

lst = ['NORTH', 'SOUTH', 'EAST', 'EAST']



for num in range(0, len(lst) - 1):

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

I know it doesn't work because by deleting the items in the list, the len(lst) changes, however, why does this work:

for num in range(len(lst) - 1 , -1 , -1):

    if lst[num] == 'SOUTH' and lst[num - 1] == 'NORTH':

        del lst[num]

        del lst[num - 1]

for this, aren't we also changing the list length? Why does the list length not matter in a reverse range?

edited Jan 3 at 7:34

asked Jan 3 at 7:29

tzyyyy

For this for loop, this code:

lst = ['NORTH', 'SOUTH', 'EAST', 'EAST']



for num in range(0, len(lst) - 1):

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

I know it doesn't work because by deleting the items in the list, the len(lst) changes, however, why does this work:

for num in range(len(lst) - 1 , -1 , -1):

    if lst[num] == 'SOUTH' and lst[num - 1] == 'NORTH':

        del lst[num]

        del lst[num - 1]

for this, aren't we also changing the list length? Why does the list length not matter in a reverse range?

python python-3.x for-loop

edited Jan 3 at 7:34

asked Jan 3 at 7:29

tzyyyy

edited Jan 3 at 7:34

asked Jan 3 at 7:29

tzyyyy

edited Jan 3 at 7:34

asked Jan 3 at 7:29

tzyyyy

asked Jan 3 at 7:29

tzyyyy

asked Jan 3 at 7:29

tzyyyy

1

By moving forward and removing list items, you are changing the indexes of the items that you have not processed yet. By moving backward and removing list items, you are changing the indexes of the items that you have processed.

– DYZ
Jan 3 at 7:32

1

@DYZ By deleting two items at a time, the second code is also subject to IndexError.

– user9522315
Jan 3 at 7:40

@iBot In theory, yes. But not in this case: deletion takes place only if the current item is SOUTH and the previous is NORTH; if the condition is true for num, it is not true for num-1.

– DYZ
Jan 3 at 7:42

1

@DYZ If in the first iteration, two elements are deleted, then in the second iteration lst[num] will fail (where num = len - 2). See my answer for an example.

– user9522315
Jan 3 at 7:44

@iBot Fair enough.

– DYZ
Jan 3 at 7:45

add a comment |

1

By moving forward and removing list items, you are changing the indexes of the items that you have not processed yet. By moving backward and removing list items, you are changing the indexes of the items that you have processed.

– DYZ
Jan 3 at 7:32

1

@DYZ By deleting two items at a time, the second code is also subject to IndexError.

– user9522315
Jan 3 at 7:40

@iBot In theory, yes. But not in this case: deletion takes place only if the current item is SOUTH and the previous is NORTH; if the condition is true for num, it is not true for num-1.

– DYZ
Jan 3 at 7:42

1

@DYZ If in the first iteration, two elements are deleted, then in the second iteration lst[num] will fail (where num = len - 2). See my answer for an example.

– user9522315
Jan 3 at 7:44

@iBot Fair enough.

– DYZ
Jan 3 at 7:45

By moving forward and removing list items, you are changing the indexes of the items that you have not processed yet. By moving backward and removing list items, you are changing the indexes of the items that you have processed.

– DYZ
Jan 3 at 7:32

@DYZ By deleting two items at a time, the second code is also subject to IndexError.

– user9522315
Jan 3 at 7:40

@iBot In theory, yes. But not in this case: deletion takes place only if the current item is SOUTH and the previous is NORTH; if the condition is true for num, it is not true for num-1.

– DYZ
Jan 3 at 7:42

@DYZ If in the first iteration, two elements are deleted, then in the second iteration lst[num] will fail (where num = len - 2). See my answer for an example.

– user9522315
Jan 3 at 7:44

@iBot Fair enough.

– DYZ
Jan 3 at 7:45

add a comment |

1 Answer
1

active

oldest

votes

As DYZ said in comments, with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.

However, it seems you're lucky with the input, since you're deleting two items at once. If you change the input list to this one, it will still fail with an IndexError.

lst = ['EAST', 'EAST', 'NORTH', 'SOUTH']

If you want to perform list manipulation like this, you're better off with a while loop.

num = 0

while num < len(lst) - 1:

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

    else:

        num += 1

answered Jan 3 at 7:42

user9522315

Hi! Thanks for the answer. However, is it possible to elaborate further on this line? 'with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.' according to my list, by moving forward, if i delete index 0 and index 1, i understand the remaining list won't work because i left 2 items in the list which are now 0 and 1. But moving backwards, if i delete the same 2 items, won't the result be the same? The index of the remaining 2 will be 0 and 1?

– tzyyyy
Jan 5 at 1:38

Also, why would the list fail if it were 2 items compared to 1? In both cases, the index of the remaining items would change wouldn't it?

– tzyyyy
Jan 5 at 1:41

@tzyyyy If you do it in the reverse way, you're processing, say for example, 5,4,3,2,1,0. Assume you deleted 3 and move on, the previous 5 and 4 become 4 and 3, but you're going to process 2,1,0 which are unaffected. Deleting items always reduce the index of "later items" (in the list), so if you're moving from big index to small index, it isn't much of a problem.

– iBug
Jan 6 at 5:53

add a comment |

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1 Answer
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As DYZ said in comments, with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.

However, it seems you're lucky with the input, since you're deleting two items at once. If you change the input list to this one, it will still fail with an IndexError.

lst = ['EAST', 'EAST', 'NORTH', 'SOUTH']

If you want to perform list manipulation like this, you're better off with a while loop.

num = 0

while num < len(lst) - 1:

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

    else:

        num += 1

answered Jan 3 at 7:42

user9522315

Hi! Thanks for the answer. However, is it possible to elaborate further on this line? 'with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.' according to my list, by moving forward, if i delete index 0 and index 1, i understand the remaining list won't work because i left 2 items in the list which are now 0 and 1. But moving backwards, if i delete the same 2 items, won't the result be the same? The index of the remaining 2 will be 0 and 1?

– tzyyyy
Jan 5 at 1:38

Also, why would the list fail if it were 2 items compared to 1? In both cases, the index of the remaining items would change wouldn't it?

– tzyyyy
Jan 5 at 1:41

@tzyyyy If you do it in the reverse way, you're processing, say for example, 5,4,3,2,1,0. Assume you deleted 3 and move on, the previous 5 and 4 become 4 and 3, but you're going to process 2,1,0 which are unaffected. Deleting items always reduce the index of "later items" (in the list), so if you're moving from big index to small index, it isn't much of a problem.

– iBug
Jan 6 at 5:53

add a comment |

As DYZ said in comments, with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.

However, it seems you're lucky with the input, since you're deleting two items at once. If you change the input list to this one, it will still fail with an IndexError.

lst = ['EAST', 'EAST', 'NORTH', 'SOUTH']

If you want to perform list manipulation like this, you're better off with a while loop.

num = 0

while num < len(lst) - 1:

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

    else:

        num += 1

answered Jan 3 at 7:42

user9522315

Hi! Thanks for the answer. However, is it possible to elaborate further on this line? 'with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.' according to my list, by moving forward, if i delete index 0 and index 1, i understand the remaining list won't work because i left 2 items in the list which are now 0 and 1. But moving backwards, if i delete the same 2 items, won't the result be the same? The index of the remaining 2 will be 0 and 1?

– tzyyyy
Jan 5 at 1:38

Also, why would the list fail if it were 2 items compared to 1? In both cases, the index of the remaining items would change wouldn't it?

– tzyyyy
Jan 5 at 1:41

@tzyyyy If you do it in the reverse way, you're processing, say for example, 5,4,3,2,1,0. Assume you deleted 3 and move on, the previous 5 and 4 become 4 and 3, but you're going to process 2,1,0 which are unaffected. Deleting items always reduce the index of "later items" (in the list), so if you're moving from big index to small index, it isn't much of a problem.

– iBug
Jan 6 at 5:53

add a comment |

As DYZ said in comments, with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.

However, it seems you're lucky with the input, since you're deleting two items at once. If you change the input list to this one, it will still fail with an IndexError.

lst = ['EAST', 'EAST', 'NORTH', 'SOUTH']

If you want to perform list manipulation like this, you're better off with a while loop.

num = 0

while num < len(lst) - 1:

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

    else:

        num += 1

answered Jan 3 at 7:42

user9522315

As DYZ said in comments, with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.

However, it seems you're lucky with the input, since you're deleting two items at once. If you change the input list to this one, it will still fail with an IndexError.

lst = ['EAST', 'EAST', 'NORTH', 'SOUTH']

If you want to perform list manipulation like this, you're better off with a while loop.

num = 0

while num < len(lst) - 1:

    if lst[num] == 'NORTH' and lst[num + 1] == 'SOUTH':

        del lst[num]

        del lst[num - 1]

    else:

        num += 1

answered Jan 3 at 7:42

user9522315

answered Jan 3 at 7:42

user9522315

answered Jan 3 at 7:42

user9522315

answered Jan 3 at 7:42

user9522315

Hi! Thanks for the answer. However, is it possible to elaborate further on this line? 'with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.' according to my list, by moving forward, if i delete index 0 and index 1, i understand the remaining list won't work because i left 2 items in the list which are now 0 and 1. But moving backwards, if i delete the same 2 items, won't the result be the same? The index of the remaining 2 will be 0 and 1?

– tzyyyy
Jan 5 at 1:38

Also, why would the list fail if it were 2 items compared to 1? In both cases, the index of the remaining items would change wouldn't it?

– tzyyyy
Jan 5 at 1:41

@tzyyyy If you do it in the reverse way, you're processing, say for example, 5,4,3,2,1,0. Assume you deleted 3 and move on, the previous 5 and 4 become 4 and 3, but you're going to process 2,1,0 which are unaffected. Deleting items always reduce the index of "later items" (in the list), so if you're moving from big index to small index, it isn't much of a problem.

– iBug
Jan 6 at 5:53

add a comment |

Hi! Thanks for the answer. However, is it possible to elaborate further on this line? 'with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.' according to my list, by moving forward, if i delete index 0 and index 1, i understand the remaining list won't work because i left 2 items in the list which are now 0 and 1. But moving backwards, if i delete the same 2 items, won't the result be the same? The index of the remaining 2 will be 0 and 1?

– tzyyyy
Jan 5 at 1:38

Also, why would the list fail if it were 2 items compared to 1? In both cases, the index of the remaining items would change wouldn't it?

– tzyyyy
Jan 5 at 1:41

@tzyyyy If you do it in the reverse way, you're processing, say for example, 5,4,3,2,1,0. Assume you deleted 3 and move on, the previous 5 and 4 become 4 and 3, but you're going to process 2,1,0 which are unaffected. Deleting items always reduce the index of "later items" (in the list), so if you're moving from big index to small index, it isn't much of a problem.

– iBug
Jan 6 at 5:53

Hi! Thanks for the answer. However, is it possible to elaborate further on this line? 'with reverse indexing, you're changing the indices of items that you have already processed, so the code won't fail when it continues.' according to my list, by moving forward, if i delete index 0 and index 1, i understand the remaining list won't work because i left 2 items in the list which are now 0 and 1. But moving backwards, if i delete the same 2 items, won't the result be the same? The index of the remaining 2 will be 0 and 1?

– tzyyyy
Jan 5 at 1:38

Also, why would the list fail if it were 2 items compared to 1? In both cases, the index of the remaining items would change wouldn't it?

– tzyyyy
Jan 5 at 1:41

@tzyyyy If you do it in the reverse way, you're processing, say for example, 5,4,3,2,1,0. Assume you deleted 3 and move on, the previous 5 and 4 become 4 and 3, but you're going to process 2,1,0 which are unaffected. Deleting items always reduce the index of "later items" (in the list), so if you're moving from big index to small index, it isn't much of a problem.

– iBug
Jan 6 at 5:53

add a comment |

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