Iterating through every other element within list recursively
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 '18 at 12:44
jpp
92.2k2053103
asked Nov 12 '18 at 12:15
V Anon
2066
add a comment |
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 '18 at 12:44
jpp
92.2k2053103
asked Nov 12 '18 at 12:15
V Anon
2066
Have you written some code that we can look at?
– L3viathan
Nov 12 '18 at 12:18
My_list[::2]....
– JBernardo
Nov 12 '18 at 12:18
Do you also need to extract a list forElements not been used? Is[2, 6, 10, 14, 18, 4, 12, 20, 16, 8]your desired output?
– jpp
Nov 12 '18 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 '18 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 '18 at 12:25
add a comment |
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 '18 at 12:44
jpp
92.2k2053103
asked Nov 12 '18 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
python python-3.x python-2.7 list recursion
edited Nov 12 '18 at 12:44
jpp
92.2k2053103
asked Nov 12 '18 at 12:15
V Anon
2066
edited Nov 12 '18 at 12:44
jpp
92.2k2053103
asked Nov 12 '18 at 12:15
V Anon
2066
edited Nov 12 '18 at 12:44
jpp
92.2k2053103
edited Nov 12 '18 at 12:44
jpp
92.2k2053103
edited Nov 12 '18 at 12:44
jpp
92.2k2053103
92.2k2053103
asked Nov 12 '18 at 12:15
V Anon
2066
asked Nov 12 '18 at 12:15
V Anon
2066
asked Nov 12 '18 at 12:15
V Anon
2066
2066
Have you written some code that we can look at?
– L3viathan
Nov 12 '18 at 12:18
My_list[::2]....
– JBernardo
Nov 12 '18 at 12:18
Do you also need to extract a list forElements not been used? Is[2, 6, 10, 14, 18, 4, 12, 20, 16, 8]your desired output?
– jpp
Nov 12 '18 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 '18 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 '18 at 12:25
add a comment |
Have you written some code that we can look at?
– L3viathan
Nov 12 '18 at 12:18
My_list[::2]....
– JBernardo
Nov 12 '18 at 12:18
Do you also need to extract a list forElements not been used? Is[2, 6, 10, 14, 18, 4, 12, 20, 16, 8]your desired output?
– jpp
Nov 12 '18 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 '18 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 '18 at 12:25
Have you written some code that we can look at?
– L3viathan
Nov 12 '18 at 12:18
Have you written some code that we can look at?
– L3viathan
Nov 12 '18 at 12:18
My_list[::2] ....– JBernardo
Nov 12 '18 at 12:18
My_list[::2] ....– JBernardo
Nov 12 '18 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 '18 at 12:19
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 '18 at 12:19
1
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 '18 at 12:24
@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 '18 at 12:24
1
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 '18 at 12:25
@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 '18 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]
answered Nov 12 '18 at 12:33
jpp
92.2k2053103
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
answered Nov 12 '18 at 12:18
vencaslac
1,002217
This doesn't actually answer OP's question.
– jpp
Nov 12 '18 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 '18 at 13:02
vencaslac
1,002217
answered Nov 12 '18 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
active
oldest
votes
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]
answered Nov 12 '18 at 12:33
jpp
92.2k2053103
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]
answered Nov 12 '18 at 12:33
jpp
92.2k2053103
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]
answered Nov 12 '18 at 12:33
jpp
92.2k2053103
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]
answered Nov 12 '18 at 12:33
jpp
92.2k2053103
edited Nov 12 '18 at 12:43
answered Nov 12 '18 at 12:33
jpp
92.2k2053103
answered Nov 12 '18 at 12:33
jpp
92.2k2053103
answered Nov 12 '18 at 12:33
jpp
92.2k2053103
92.2k2053103
add a comment |
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
answered Nov 12 '18 at 12:18
vencaslac
1,002217
This doesn't actually answer OP's question.
– jpp
Nov 12 '18 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
answered Nov 12 '18 at 12:18
vencaslac
1,002217
This doesn't actually answer OP's question.
– jpp
Nov 12 '18 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
answered Nov 12 '18 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
answered Nov 12 '18 at 12:18
vencaslac
1,002217
edited Nov 12 '18 at 12:20
answered Nov 12 '18 at 12:18
vencaslac
1,002217
answered Nov 12 '18 at 12:18
vencaslac
1,002217
answered Nov 12 '18 at 12:18
vencaslac
1,002217
1,002217
This doesn't actually answer OP's question.
– jpp
Nov 12 '18 at 12:38
add a comment |
This doesn't actually answer OP's question.
– jpp
Nov 12 '18 at 12:38
This doesn't actually answer OP's question.
– jpp
Nov 12 '18 at 12:38
This doesn't actually answer OP's question.
– jpp
Nov 12 '18 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 '18 at 13:02
vencaslac
1,002217
answered Nov 12 '18 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 '18 at 13:02
vencaslac
1,002217
answered Nov 12 '18 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 '18 at 13:02
vencaslac
1,002217
answered Nov 12 '18 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 '18 at 13:02
vencaslac
1,002217
answered Nov 12 '18 at 12:42
Gautham Mohan
213
edited Nov 12 '18 at 13:02
vencaslac
1,002217
edited Nov 12 '18 at 13:02
vencaslac
1,002217
edited Nov 12 '18 at 13:02
vencaslac
1,002217
1,002217
answered Nov 12 '18 at 12:42
Gautham Mohan
213
answered Nov 12 '18 at 12:42
Gautham Mohan
213
answered Nov 12 '18 at 12:42
Gautham Mohan
213
213
add a comment |
add a comment |
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Have you written some code that we can look at?
– L3viathan
Nov 12 '18 at 12:18
My_list[::2]....– JBernardo
Nov 12 '18 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 '18 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 '18 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 '18 at 12:25