How to remove the inner units in a compound list in Python?
x = [['a','b'],['c','d','g'],['e','f','h','i','j'].......['zzy','xxx']]
If I got a compound list like this (a large list) in Python, how can I elegantly remove only, say, the element 'c' without removing the whole element ['c','d','g'] together?
Obviously merely list.remove() doesn't work for this, and implementing a for loop works
for i in x:
for j in i:
if j == 'c':
i = i.remove(j)
but is computationally expensive since it's a very long list...
thank you
python
asked Nov 20 at 21:34
D.Chain
33
add a comment |
x = [['a','b'],['c','d','g'],['e','f','h','i','j'].......['zzy','xxx']]
If I got a compound list like this (a large list) in Python, how can I elegantly remove only, say, the element 'c' without removing the whole element ['c','d','g'] together?
Obviously merely list.remove() doesn't work for this, and implementing a for loop works
for i in x:
for j in i:
if j == 'c':
i = i.remove(j)
but is computationally expensive since it's a very long list...
thank you
python
asked Nov 20 at 21:34
D.Chain
33
3
What have you tried so far?
– G. Anderson
Nov 20 at 21:34
5
your question leaves a lot to be interpreted by us. Do you want this to be applicable only to this small list? why is 'a' a smaller unit than 'b'? 'a' and 'b' are strings. You also didn't provide anything you have tried so far.
– d_kennetz
Nov 20 at 21:36
Is the list always 2-d or can it be arbitrarily deep?
– slider
Nov 20 at 21:38
My apology. I'm new to Python and am not fully used to asking here. I've edited it a little and if you could have some look i'd appreciate it.
– D.Chain
Nov 20 at 21:56
add a comment |
x = [['a','b'],['c','d','g'],['e','f','h','i','j'].......['zzy','xxx']]
If I got a compound list like this (a large list) in Python, how can I elegantly remove only, say, the element 'c' without removing the whole element ['c','d','g'] together?
Obviously merely list.remove() doesn't work for this, and implementing a for loop works
for i in x:
for j in i:
if j == 'c':
i = i.remove(j)
but is computationally expensive since it's a very long list...
thank you
python
asked Nov 20 at 21:34
D.Chain
33
x = [['a','b'],['c','d','g'],['e','f','h','i','j'].......['zzy','xxx']]
If I got a compound list like this (a large list) in Python, how can I elegantly remove only, say, the element 'c' without removing the whole element ['c','d','g'] together?
Obviously merely list.remove() doesn't work for this, and implementing a for loop works
for i in x:
for j in i:
if j == 'c':
i = i.remove(j)
but is computationally expensive since it's a very long list...
thank you
python
python
asked Nov 20 at 21:34
D.Chain
33
asked Nov 20 at 21:34
D.Chain
33
edited Nov 20 at 22:15
asked Nov 20 at 21:34
D.Chain
33
asked Nov 20 at 21:34
D.Chain
33
asked Nov 20 at 21:34
D.Chain
33
33
3
What have you tried so far?
– G. Anderson
Nov 20 at 21:34
5
your question leaves a lot to be interpreted by us. Do you want this to be applicable only to this small list? why is 'a' a smaller unit than 'b'? 'a' and 'b' are strings. You also didn't provide anything you have tried so far.
– d_kennetz
Nov 20 at 21:36
Is the list always 2-d or can it be arbitrarily deep?
– slider
Nov 20 at 21:38
My apology. I'm new to Python and am not fully used to asking here. I've edited it a little and if you could have some look i'd appreciate it.
– D.Chain
Nov 20 at 21:56
add a comment |
3
What have you tried so far?
– G. Anderson
Nov 20 at 21:34
5
your question leaves a lot to be interpreted by us. Do you want this to be applicable only to this small list? why is 'a' a smaller unit than 'b'? 'a' and 'b' are strings. You also didn't provide anything you have tried so far.
– d_kennetz
Nov 20 at 21:36
Is the list always 2-d or can it be arbitrarily deep?
– slider
Nov 20 at 21:38
My apology. I'm new to Python and am not fully used to asking here. I've edited it a little and if you could have some look i'd appreciate it.
– D.Chain
Nov 20 at 21:56
3
3
What have you tried so far?
– G. Anderson
Nov 20 at 21:34
What have you tried so far?
– G. Anderson
Nov 20 at 21:34
5
5
your question leaves a lot to be interpreted by us. Do you want this to be applicable only to this small list? why is 'a' a smaller unit than 'b'? 'a' and 'b' are strings. You also didn't provide anything you have tried so far.
– d_kennetz
Nov 20 at 21:36
your question leaves a lot to be interpreted by us. Do you want this to be applicable only to this small list? why is 'a' a smaller unit than 'b'? 'a' and 'b' are strings. You also didn't provide anything you have tried so far.
– d_kennetz
Nov 20 at 21:36
Is the list always 2-d or can it be arbitrarily deep?
– slider
Nov 20 at 21:38
Is the list always 2-d or can it be arbitrarily deep?
– slider
Nov 20 at 21:38
My apology. I'm new to Python and am not fully used to asking here. I've edited it a little and if you could have some look i'd appreciate it.
– D.Chain
Nov 20 at 21:56
My apology. I'm new to Python and am not fully used to asking here. I've edited it a little and if you could have some look i'd appreciate it.
– D.Chain
Nov 20 at 21:56
add a comment |
2 Answers
2
active
oldest
votes
The only improvement I could find to your code is:
x = [['a','b'],['c','d','g'],['e','f','h','i','j'],['zzy','xxx']]
def remove(compound_list, elem):
for lst in compound_list:
for ix, item in enumerate(lst):
if item == elem:
del lst[ix]
remove(x, 'c')
It is slow O(n^2) but it is faster that first solution which counting i.remove(j) it's O(n^3)
answered Nov 21 at 1:01
edilio
69656
add a comment |
You can find the "smallest" element using min (provide a custom key function if you want to change the definition of smallest).
An O(n) solution (where n is the total number of elements in all the lists) can be to rebuild each list by omitting the smallest element:
x = [['a','b'],['b','b'],['c','c']]
smallest = min(min(l) for l in x)
x = [[e for e in l if e != smallest] for l in x]
print(x)
# [['b'], ['b', 'b'], ['c', 'c']]
answered Nov 20 at 21:50
slider
8,0151129
thanks a lot for your answer! but it's actually a very long list so using loop may cost much time..i've edited the question to make it clearer
– D.Chain
Nov 20 at 21:58
@D.Chain I've updated the answer to not use remove since, as you rightly pointed out, it is computationally expensive to call it on each list. A computationally (in terms of time) more efficient alternative can be to rebuild the lists.
– slider
Nov 20 at 22:03
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The only improvement I could find to your code is:
x = [['a','b'],['c','d','g'],['e','f','h','i','j'],['zzy','xxx']]
def remove(compound_list, elem):
for lst in compound_list:
for ix, item in enumerate(lst):
if item == elem:
del lst[ix]
remove(x, 'c')
It is slow O(n^2) but it is faster that first solution which counting i.remove(j) it's O(n^3)
answered Nov 21 at 1:01
edilio
69656
add a comment |
The only improvement I could find to your code is:
x = [['a','b'],['c','d','g'],['e','f','h','i','j'],['zzy','xxx']]
def remove(compound_list, elem):
for lst in compound_list:
for ix, item in enumerate(lst):
if item == elem:
del lst[ix]
remove(x, 'c')
It is slow O(n^2) but it is faster that first solution which counting i.remove(j) it's O(n^3)
answered Nov 21 at 1:01
edilio
69656
add a comment |
The only improvement I could find to your code is:
x = [['a','b'],['c','d','g'],['e','f','h','i','j'],['zzy','xxx']]
def remove(compound_list, elem):
for lst in compound_list:
for ix, item in enumerate(lst):
if item == elem:
del lst[ix]
remove(x, 'c')
It is slow O(n^2) but it is faster that first solution which counting i.remove(j) it's O(n^3)
answered Nov 21 at 1:01
edilio
69656
The only improvement I could find to your code is:
x = [['a','b'],['c','d','g'],['e','f','h','i','j'],['zzy','xxx']]
def remove(compound_list, elem):
for lst in compound_list:
for ix, item in enumerate(lst):
if item == elem:
del lst[ix]
remove(x, 'c')
It is slow O(n^2) but it is faster that first solution which counting i.remove(j) it's O(n^3)
answered Nov 21 at 1:01
edilio
69656
answered Nov 21 at 1:01
edilio
69656
answered Nov 21 at 1:01
edilio
69656
answered Nov 21 at 1:01
edilio
69656
69656
add a comment |
add a comment |
You can find the "smallest" element using min (provide a custom key function if you want to change the definition of smallest).
An O(n) solution (where n is the total number of elements in all the lists) can be to rebuild each list by omitting the smallest element:
x = [['a','b'],['b','b'],['c','c']]
smallest = min(min(l) for l in x)
x = [[e for e in l if e != smallest] for l in x]
print(x)
# [['b'], ['b', 'b'], ['c', 'c']]
answered Nov 20 at 21:50
slider
8,0151129
thanks a lot for your answer! but it's actually a very long list so using loop may cost much time..i've edited the question to make it clearer
– D.Chain
Nov 20 at 21:58
@D.Chain I've updated the answer to not use remove since, as you rightly pointed out, it is computationally expensive to call it on each list. A computationally (in terms of time) more efficient alternative can be to rebuild the lists.
– slider
Nov 20 at 22:03
add a comment |
You can find the "smallest" element using min (provide a custom key function if you want to change the definition of smallest).
An O(n) solution (where n is the total number of elements in all the lists) can be to rebuild each list by omitting the smallest element:
x = [['a','b'],['b','b'],['c','c']]
smallest = min(min(l) for l in x)
x = [[e for e in l if e != smallest] for l in x]
print(x)
# [['b'], ['b', 'b'], ['c', 'c']]
answered Nov 20 at 21:50
slider
8,0151129
thanks a lot for your answer! but it's actually a very long list so using loop may cost much time..i've edited the question to make it clearer
– D.Chain
Nov 20 at 21:58
@D.Chain I've updated the answer to not use remove since, as you rightly pointed out, it is computationally expensive to call it on each list. A computationally (in terms of time) more efficient alternative can be to rebuild the lists.
– slider
Nov 20 at 22:03
add a comment |
You can find the "smallest" element using min (provide a custom key function if you want to change the definition of smallest).
An O(n) solution (where n is the total number of elements in all the lists) can be to rebuild each list by omitting the smallest element:
x = [['a','b'],['b','b'],['c','c']]
smallest = min(min(l) for l in x)
x = [[e for e in l if e != smallest] for l in x]
print(x)
# [['b'], ['b', 'b'], ['c', 'c']]
answered Nov 20 at 21:50
slider
8,0151129
You can find the "smallest" element using min (provide a custom key function if you want to change the definition of smallest).
An O(n) solution (where n is the total number of elements in all the lists) can be to rebuild each list by omitting the smallest element:
x = [['a','b'],['b','b'],['c','c']]
smallest = min(min(l) for l in x)
x = [[e for e in l if e != smallest] for l in x]
print(x)
# [['b'], ['b', 'b'], ['c', 'c']]
answered Nov 20 at 21:50
slider
8,0151129
edited Nov 20 at 22:02
answered Nov 20 at 21:50
slider
8,0151129
answered Nov 20 at 21:50
slider
8,0151129
answered Nov 20 at 21:50
slider
8,0151129
8,0151129
thanks a lot for your answer! but it's actually a very long list so using loop may cost much time..i've edited the question to make it clearer
– D.Chain
Nov 20 at 21:58
@D.Chain I've updated the answer to not use remove since, as you rightly pointed out, it is computationally expensive to call it on each list. A computationally (in terms of time) more efficient alternative can be to rebuild the lists.
– slider
Nov 20 at 22:03
add a comment |
thanks a lot for your answer! but it's actually a very long list so using loop may cost much time..i've edited the question to make it clearer
– D.Chain
Nov 20 at 21:58
@D.Chain I've updated the answer to not use remove since, as you rightly pointed out, it is computationally expensive to call it on each list. A computationally (in terms of time) more efficient alternative can be to rebuild the lists.
– slider
Nov 20 at 22:03
thanks a lot for your answer! but it's actually a very long list so using loop may cost much time..i've edited the question to make it clearer
– D.Chain
Nov 20 at 21:58
thanks a lot for your answer! but it's actually a very long list so using loop may cost much time..i've edited the question to make it clearer
– D.Chain
Nov 20 at 21:58
@D.Chain I've updated the answer to not use remove since, as you rightly pointed out, it is computationally expensive to call it on each list. A computationally (in terms of time) more efficient alternative can be to rebuild the lists.
– slider
Nov 20 at 22:03
@D.Chain I've updated the answer to not use remove since, as you rightly pointed out, it is computationally expensive to call it on each list. A computationally (in terms of time) more efficient alternative can be to rebuild the lists.
– slider
Nov 20 at 22:03
add a comment |
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3
What have you tried so far?
– G. Anderson
Nov 20 at 21:34
5
your question leaves a lot to be interpreted by us. Do you want this to be applicable only to this small list? why is 'a' a smaller unit than 'b'? 'a' and 'b' are strings. You also didn't provide anything you have tried so far.
– d_kennetz
Nov 20 at 21:36
Is the list always 2-d or can it be arbitrarily deep?
– slider
Nov 20 at 21:38
My apology. I'm new to Python and am not fully used to asking here. I've edited it a little and if you could have some look i'd appreciate it.
– D.Chain
Nov 20 at 21:56