Iterating through every other element within list recursively

-1

I have been trying to figure out a clear and concise method to iterate through elements in a list in a form of recursive loop.

For example, if I have a list:

My_list = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]

I would like to extract every other element from the list.

New_list = [2, 6, 10, 14, 18]

I then want to take the discarded values and take every second of those. And so on, recursively.

Since my first run through the list ended at 18, I will skip over 20 and go back to the beginning of the list as I want to extract every other elements.

My_list = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



# Elements not been used after first operation = [4, 8, 12, 16, 20]



New_list = [2, 6, 10, 14, 18, 4, 12, 20, 8, 16]  # desired output

What methods can I use to loop within a list?

edited Nov 12 at 12:44

jpp

90k2052101

asked Nov 12 at 12:15

V Anon

2066

Have you written some code that we can look at?
– L3viathan
Nov 12 at 12:18

My_list[::2] ....
– JBernardo
Nov 12 at 12:18

Do you also need to extract a list for Elements not been used ? Is [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] your desired output?
– jpp
Nov 12 at 12:19

1

@jpp yes I already know how to extract every two elements but it is my goal to loop over the list and extract again from elements not been used. Indeed [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] is my desired output
– V Anon
Nov 12 at 12:24

1

@JBernardo yes I am aware of the method of extracting every other two elements, but would there be a way to loop through the list again?
– V Anon
Nov 12 at 12:25

add a comment |

-1

I have been trying to figure out a clear and concise method to iterate through elements in a list in a form of recursive loop.

For example, if I have a list:

My_list = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]

I would like to extract every other element from the list.

New_list = [2, 6, 10, 14, 18]

I then want to take the discarded values and take every second of those. And so on, recursively.

Since my first run through the list ended at 18, I will skip over 20 and go back to the beginning of the list as I want to extract every other elements.

My_list = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



# Elements not been used after first operation = [4, 8, 12, 16, 20]



New_list = [2, 6, 10, 14, 18, 4, 12, 20, 8, 16]  # desired output

What methods can I use to loop within a list?

edited Nov 12 at 12:44

jpp

90k2052101

asked Nov 12 at 12:15

V Anon

2066

Have you written some code that we can look at?
– L3viathan
Nov 12 at 12:18

My_list[::2] ....
– JBernardo
Nov 12 at 12:18

Do you also need to extract a list for Elements not been used ? Is [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] your desired output?
– jpp
Nov 12 at 12:19

1

@jpp yes I already know how to extract every two elements but it is my goal to loop over the list and extract again from elements not been used. Indeed [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] is my desired output
– V Anon
Nov 12 at 12:24

1

@JBernardo yes I am aware of the method of extracting every other two elements, but would there be a way to loop through the list again?
– V Anon
Nov 12 at 12:25

add a comment |

-1

I have been trying to figure out a clear and concise method to iterate through elements in a list in a form of recursive loop.

For example, if I have a list:

My_list = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]

I would like to extract every other element from the list.

New_list = [2, 6, 10, 14, 18]

I then want to take the discarded values and take every second of those. And so on, recursively.

Since my first run through the list ended at 18, I will skip over 20 and go back to the beginning of the list as I want to extract every other elements.

My_list = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



# Elements not been used after first operation = [4, 8, 12, 16, 20]



New_list = [2, 6, 10, 14, 18, 4, 12, 20, 8, 16]  # desired output

What methods can I use to loop within a list?

edited Nov 12 at 12:44

jpp

90k2052101

asked Nov 12 at 12:15

V Anon

2066

I have been trying to figure out a clear and concise method to iterate through elements in a list in a form of recursive loop.

For example, if I have a list:

My_list = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]

I would like to extract every other element from the list.

New_list = [2, 6, 10, 14, 18]

I then want to take the discarded values and take every second of those. And so on, recursively.

Since my first run through the list ended at 18, I will skip over 20 and go back to the beginning of the list as I want to extract every other elements.

My_list = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



# Elements not been used after first operation = [4, 8, 12, 16, 20]



New_list = [2, 6, 10, 14, 18, 4, 12, 20, 8, 16]  # desired output

What methods can I use to loop within a list?

python python-3.x python-2.7 list recursion

edited Nov 12 at 12:44

jpp

90k2052101

asked Nov 12 at 12:15

V Anon

2066

edited Nov 12 at 12:44

jpp

90k2052101

asked Nov 12 at 12:15

V Anon

2066

edited Nov 12 at 12:44

jpp

90k2052101

edited Nov 12 at 12:44

jpp

90k2052101

edited Nov 12 at 12:44

jpp

90k2052101

asked Nov 12 at 12:15

V Anon

2066

asked Nov 12 at 12:15

V Anon

2066

asked Nov 12 at 12:15

V Anon

2066

Have you written some code that we can look at?
– L3viathan
Nov 12 at 12:18

My_list[::2] ....
– JBernardo
Nov 12 at 12:18

Do you also need to extract a list for Elements not been used ? Is [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] your desired output?
– jpp
Nov 12 at 12:19

1

@jpp yes I already know how to extract every two elements but it is my goal to loop over the list and extract again from elements not been used. Indeed [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] is my desired output
– V Anon
Nov 12 at 12:24

1

@JBernardo yes I am aware of the method of extracting every other two elements, but would there be a way to loop through the list again?
– V Anon
Nov 12 at 12:25

add a comment |

Have you written some code that we can look at?
– L3viathan
Nov 12 at 12:18

My_list[::2] ....
– JBernardo
Nov 12 at 12:18

Do you also need to extract a list for Elements not been used ? Is [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] your desired output?
– jpp
Nov 12 at 12:19

1

@jpp yes I already know how to extract every two elements but it is my goal to loop over the list and extract again from elements not been used. Indeed [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] is my desired output
– V Anon
Nov 12 at 12:24

1

@JBernardo yes I am aware of the method of extracting every other two elements, but would there be a way to loop through the list again?
– V Anon
Nov 12 at 12:25

Have you written some code that we can look at?
– L3viathan
Nov 12 at 12:18

My_list[::2] ....
– JBernardo
Nov 12 at 12:18

Do you also need to extract a list for Elements not been used ? Is [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] your desired output?
– jpp
Nov 12 at 12:19

@jpp yes I already know how to extract every two elements but it is my goal to loop over the list and extract again from elements not been used. Indeed [2, 6, 10, 14, 18, 4, 12, 20, 16, 8] is my desired output
– V Anon
Nov 12 at 12:24

@JBernardo yes I am aware of the method of extracting every other two elements, but would there be a way to loop through the list again?
– V Anon
Nov 12 at 12:25

add a comment |

3 Answers
3

active

oldest

votes

Yes, you can use a while loop with a generator:

L = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



def recursive_odds(x):

    while x:

        yield from x[::2]

        x = x[1::2]



res = list(recursive_odds(L))



[2, 6, 10, 14, 18, 4, 12, 20, 8, 16]

edited Nov 12 at 12:43

answered Nov 12 at 12:33

jpp

90k2052101

add a comment |

you can use python's really cool slicing syntax:

new_list=my_list[::2]

to get every other element.

It means new_list is my_list from the begining to the end with a stride of 2

then your ramainder elements are elements_not_used = [item for item in my_list if item not in new_list] and you can just continue until len(my_list)<2

edited Nov 12 at 12:20

answered Nov 12 at 12:18

vencaslac

1,002217

This doesn't actually answer OP's question.
– jpp
Nov 12 at 12:38

add a comment |

using while loop you can do this:

final=

while len(mylist)>0:

     final.extend(mylist[::2])

     mylist=mylist[1::2]

print(final)

edited Nov 12 at 13:02

vencaslac

1,002217

answered Nov 12 at 12:42

Gautham Mohan

213

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Yes, you can use a while loop with a generator:

L = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



def recursive_odds(x):

    while x:

        yield from x[::2]

        x = x[1::2]



res = list(recursive_odds(L))



[2, 6, 10, 14, 18, 4, 12, 20, 8, 16]

edited Nov 12 at 12:43

answered Nov 12 at 12:33

jpp

90k2052101

add a comment |

Yes, you can use a while loop with a generator:

L = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



def recursive_odds(x):

    while x:

        yield from x[::2]

        x = x[1::2]



res = list(recursive_odds(L))



[2, 6, 10, 14, 18, 4, 12, 20, 8, 16]

edited Nov 12 at 12:43

answered Nov 12 at 12:33

jpp

90k2052101

add a comment |

Yes, you can use a while loop with a generator:

L = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



def recursive_odds(x):

    while x:

        yield from x[::2]

        x = x[1::2]



res = list(recursive_odds(L))



[2, 6, 10, 14, 18, 4, 12, 20, 8, 16]

edited Nov 12 at 12:43

answered Nov 12 at 12:33

jpp

90k2052101

Yes, you can use a while loop with a generator:

L = [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]



def recursive_odds(x):

    while x:

        yield from x[::2]

        x = x[1::2]



res = list(recursive_odds(L))



[2, 6, 10, 14, 18, 4, 12, 20, 8, 16]

edited Nov 12 at 12:43

answered Nov 12 at 12:33

jpp

90k2052101

edited Nov 12 at 12:43

answered Nov 12 at 12:33

jpp

90k2052101

answered Nov 12 at 12:33

jpp

90k2052101

answered Nov 12 at 12:33

jpp

90k2052101

add a comment |

you can use python's really cool slicing syntax:

new_list=my_list[::2]

to get every other element.

It means new_list is my_list from the begining to the end with a stride of 2

then your ramainder elements are elements_not_used = [item for item in my_list if item not in new_list] and you can just continue until len(my_list)<2

edited Nov 12 at 12:20

answered Nov 12 at 12:18

vencaslac

1,002217

This doesn't actually answer OP's question.
– jpp
Nov 12 at 12:38

add a comment |

you can use python's really cool slicing syntax:

new_list=my_list[::2]

to get every other element.

It means new_list is my_list from the begining to the end with a stride of 2

then your ramainder elements are elements_not_used = [item for item in my_list if item not in new_list] and you can just continue until len(my_list)<2

edited Nov 12 at 12:20

answered Nov 12 at 12:18

vencaslac

1,002217

This doesn't actually answer OP's question.
– jpp
Nov 12 at 12:38

add a comment |

you can use python's really cool slicing syntax:

new_list=my_list[::2]

to get every other element.

It means new_list is my_list from the begining to the end with a stride of 2

then your ramainder elements are elements_not_used = [item for item in my_list if item not in new_list] and you can just continue until len(my_list)<2

edited Nov 12 at 12:20

answered Nov 12 at 12:18

vencaslac

1,002217

you can use python's really cool slicing syntax:

new_list=my_list[::2]

to get every other element.

It means new_list is my_list from the begining to the end with a stride of 2

then your ramainder elements are elements_not_used = [item for item in my_list if item not in new_list] and you can just continue until len(my_list)<2

edited Nov 12 at 12:20

answered Nov 12 at 12:18

vencaslac

1,002217

edited Nov 12 at 12:20

answered Nov 12 at 12:18

vencaslac

1,002217

answered Nov 12 at 12:18

vencaslac

1,002217

answered Nov 12 at 12:18

vencaslac

1,002217

This doesn't actually answer OP's question.
– jpp
Nov 12 at 12:38

add a comment |

This doesn't actually answer OP's question.
– jpp
Nov 12 at 12:38

This doesn't actually answer OP's question.
– jpp
Nov 12 at 12:38

add a comment |

using while loop you can do this:

final=

while len(mylist)>0:

     final.extend(mylist[::2])

     mylist=mylist[1::2]

print(final)

edited Nov 12 at 13:02

vencaslac

1,002217

answered Nov 12 at 12:42

Gautham Mohan

213

add a comment |

using while loop you can do this:

final=

while len(mylist)>0:

     final.extend(mylist[::2])

     mylist=mylist[1::2]

print(final)

edited Nov 12 at 13:02

vencaslac

1,002217

answered Nov 12 at 12:42

Gautham Mohan

213

add a comment |

using while loop you can do this:

final=

while len(mylist)>0:

     final.extend(mylist[::2])

     mylist=mylist[1::2]

print(final)

edited Nov 12 at 13:02

vencaslac

1,002217

answered Nov 12 at 12:42

Gautham Mohan

213

using while loop you can do this:

final=

while len(mylist)>0:

     final.extend(mylist[::2])

     mylist=mylist[1::2]

print(final)

edited Nov 12 at 13:02

vencaslac

1,002217

answered Nov 12 at 12:42

Gautham Mohan

213

edited Nov 12 at 13:02

vencaslac

1,002217

edited Nov 12 at 13:02

vencaslac

1,002217

edited Nov 12 at 13:02

vencaslac

1,002217

answered Nov 12 at 12:42

Gautham Mohan

213

answered Nov 12 at 12:42

Gautham Mohan

213

answered Nov 12 at 12:42

Gautham Mohan

213

add a comment |

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