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Given nested list element, find the one level back list value
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?How do I sort a list of dictionaries by a value of the dictionary?Finding the index of an item given a list containing it in PythonHow to return multiple values from a function?How do I remove an element from a list by index in Python?Find intersection of two nested lists?Getting the last element of a list in Python“Least Astonishment” and the Mutable Default ArgumentHow do I get the number of elements in a list in Python?Is there a simple way to delete a list element by value?Why is [] faster than list()?
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Say i have List=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
i know that if i call List[0][0] i will get [['a', 'b'], ['c', 'd'], ['e', 'f']], and so on.
Is there any built-in or external python function(suppose its func(a)) or way to get the nested list element a one level back?
So if i call func(List[0][1]) those function will return List[0] or when i call func(List[1][0][1]) those function will return List[1][0] but if i call func(List) it will return List since it's already at the root. I've been searching for this kind of problem for hours but still couldn't find the solution.
python list indexing nested
asked Mar 22 at 8:23
FauziFauzi
83
|
show 1 more comment
Say i have List=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
i know that if i call List[0][0] i will get [['a', 'b'], ['c', 'd'], ['e', 'f']], and so on.
Is there any built-in or external python function(suppose its func(a)) or way to get the nested list element a one level back?
So if i call func(List[0][1]) those function will return List[0] or when i call func(List[1][0][1]) those function will return List[1][0] but if i call func(List) it will return List since it's already at the root. I've been searching for this kind of problem for hours but still couldn't find the solution.
python list indexing nested
asked Mar 22 at 8:23
FauziFauzi
83
1
To summarize, if you see your list as a tree, you want a function that, given a subtree, returns the parent node of that subtree? If so, the answer is no (at least with list). There are, however, many tree implementations.
– Guybrush
Mar 22 at 8:25
yes exactly, i want to get back to the parent node of that list. So there really is not any alternative way to get this in python?
– Fauzi
Mar 22 at 8:28
1
The bottom line is, List is a not a bad structure to "build" that kind of logic/abstraction on, but a list simply ISNT itself an implementation of that kind of logic/structure. You would have to use a different structure or build your own. Once you understand that, you can start looking for better alternatives to "store/represent" this relation. Be it graphs, some type of dictionary/tree structure, or so on.
– Paritosh Singh
Mar 22 at 8:29
1
@Fauzi Indeed, lists are not suited to get the "parent", because lists weren't created with that specific use case in mind (obviously). That's why you need something "more". You can either go for trees, or implement your own structure (and both can be based on lists, of course).
– Guybrush
Mar 22 at 8:31
So that's how it is :( Thank you guys for pointing out that list can't be seen as parent-child like node logic.
– Fauzi
Mar 22 at 8:37
|
show 1 more comment
Say i have List=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
i know that if i call List[0][0] i will get [['a', 'b'], ['c', 'd'], ['e', 'f']], and so on.
Is there any built-in or external python function(suppose its func(a)) or way to get the nested list element a one level back?
So if i call func(List[0][1]) those function will return List[0] or when i call func(List[1][0][1]) those function will return List[1][0] but if i call func(List) it will return List since it's already at the root. I've been searching for this kind of problem for hours but still couldn't find the solution.
python list indexing nested
asked Mar 22 at 8:23
FauziFauzi
83
Say i have List=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
i know that if i call List[0][0] i will get [['a', 'b'], ['c', 'd'], ['e', 'f']], and so on.
Is there any built-in or external python function(suppose its func(a)) or way to get the nested list element a one level back?
So if i call func(List[0][1]) those function will return List[0] or when i call func(List[1][0][1]) those function will return List[1][0] but if i call func(List) it will return List since it's already at the root. I've been searching for this kind of problem for hours but still couldn't find the solution.
python list indexing nested
python list indexing nested
asked Mar 22 at 8:23
FauziFauzi
83
asked Mar 22 at 8:23
FauziFauzi
83
edited Mar 22 at 8:26
Fauzi
asked Mar 22 at 8:23
FauziFauzi
83
asked Mar 22 at 8:23
FauziFauzi
83
asked Mar 22 at 8:23
FauziFauzi
83
83
1
To summarize, if you see your list as a tree, you want a function that, given a subtree, returns the parent node of that subtree? If so, the answer is no (at least with list). There are, however, many tree implementations.
– Guybrush
Mar 22 at 8:25
yes exactly, i want to get back to the parent node of that list. So there really is not any alternative way to get this in python?
– Fauzi
Mar 22 at 8:28
1
The bottom line is, List is a not a bad structure to "build" that kind of logic/abstraction on, but a list simply ISNT itself an implementation of that kind of logic/structure. You would have to use a different structure or build your own. Once you understand that, you can start looking for better alternatives to "store/represent" this relation. Be it graphs, some type of dictionary/tree structure, or so on.
– Paritosh Singh
Mar 22 at 8:29
1
@Fauzi Indeed, lists are not suited to get the "parent", because lists weren't created with that specific use case in mind (obviously). That's why you need something "more". You can either go for trees, or implement your own structure (and both can be based on lists, of course).
– Guybrush
Mar 22 at 8:31
So that's how it is :( Thank you guys for pointing out that list can't be seen as parent-child like node logic.
– Fauzi
Mar 22 at 8:37
|
show 1 more comment
1
To summarize, if you see your list as a tree, you want a function that, given a subtree, returns the parent node of that subtree? If so, the answer is no (at least with list). There are, however, many tree implementations.
– Guybrush
Mar 22 at 8:25
yes exactly, i want to get back to the parent node of that list. So there really is not any alternative way to get this in python?
– Fauzi
Mar 22 at 8:28
1
The bottom line is, List is a not a bad structure to "build" that kind of logic/abstraction on, but a list simply ISNT itself an implementation of that kind of logic/structure. You would have to use a different structure or build your own. Once you understand that, you can start looking for better alternatives to "store/represent" this relation. Be it graphs, some type of dictionary/tree structure, or so on.
– Paritosh Singh
Mar 22 at 8:29
1
@Fauzi Indeed, lists are not suited to get the "parent", because lists weren't created with that specific use case in mind (obviously). That's why you need something "more". You can either go for trees, or implement your own structure (and both can be based on lists, of course).
– Guybrush
Mar 22 at 8:31
So that's how it is :( Thank you guys for pointing out that list can't be seen as parent-child like node logic.
– Fauzi
Mar 22 at 8:37
1
1
To summarize, if you see your list as a tree, you want a function that, given a subtree, returns the parent node of that subtree? If so, the answer is no (at least with list). There are, however, many tree implementations.
– Guybrush
Mar 22 at 8:25
To summarize, if you see your list as a tree, you want a function that, given a subtree, returns the parent node of that subtree? If so, the answer is no (at least with list). There are, however, many tree implementations.
– Guybrush
Mar 22 at 8:25
yes exactly, i want to get back to the parent node of that list. So there really is not any alternative way to get this in python?
– Fauzi
Mar 22 at 8:28
yes exactly, i want to get back to the parent node of that list. So there really is not any alternative way to get this in python?
– Fauzi
Mar 22 at 8:28
1
1
The bottom line is, List is a not a bad structure to "build" that kind of logic/abstraction on, but a list simply ISNT itself an implementation of that kind of logic/structure. You would have to use a different structure or build your own. Once you understand that, you can start looking for better alternatives to "store/represent" this relation. Be it graphs, some type of dictionary/tree structure, or so on.
– Paritosh Singh
Mar 22 at 8:29
The bottom line is, List is a not a bad structure to "build" that kind of logic/abstraction on, but a list simply ISNT itself an implementation of that kind of logic/structure. You would have to use a different structure or build your own. Once you understand that, you can start looking for better alternatives to "store/represent" this relation. Be it graphs, some type of dictionary/tree structure, or so on.
– Paritosh Singh
Mar 22 at 8:29
1
1
@Fauzi Indeed, lists are not suited to get the "parent", because lists weren't created with that specific use case in mind (obviously). That's why you need something "more". You can either go for trees, or implement your own structure (and both can be based on lists, of course).
– Guybrush
Mar 22 at 8:31
@Fauzi Indeed, lists are not suited to get the "parent", because lists weren't created with that specific use case in mind (obviously). That's why you need something "more". You can either go for trees, or implement your own structure (and both can be based on lists, of course).
– Guybrush
Mar 22 at 8:31
So that's how it is :( Thank you guys for pointing out that list can't be seen as parent-child like node logic.
– Fauzi
Mar 22 at 8:37
So that's how it is :( Thank you guys for pointing out that list can't be seen as parent-child like node logic.
– Fauzi
Mar 22 at 8:37
|
show 1 more comment
1 Answer
1
active
oldest
votes
You can use the following recursive function:
def get_parent_list(the_elem, the_list):
if (the_elem == the_list):
return (True, the_elem)
elif the_elem in the_list:
return (True, the_list)
else:
for e in the_list:
if (type(e) is list):
(is_found, the_parent) = get_parent_list(the_elem, e)
if (is_found):
return (True, the_parent)
return (False, None)
Testing it out:
my_list=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],
[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
Test Case 1:
the_child = my_list[0][1][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
['3', '4']
[['1', '2'], ['3', '4'], ['5', '6']]
Test Case 2:
the_child = my_list[0][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[['1', '2'], ['3', '4'], ['5', '6']]
[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]
Test Case 3:
the_child = my_list[:]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
Test Case 4:
the_child = my_list[0][1] + ['Non-existent value']
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
False
[['1', '2'], ['3', '4'], ['5', '6'], 'Non-existent value']
None
answered Mar 22 at 9:40
fountainheadfountainhead
1,315313
add a comment |
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1 Answer
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active
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votes
You can use the following recursive function:
def get_parent_list(the_elem, the_list):
if (the_elem == the_list):
return (True, the_elem)
elif the_elem in the_list:
return (True, the_list)
else:
for e in the_list:
if (type(e) is list):
(is_found, the_parent) = get_parent_list(the_elem, e)
if (is_found):
return (True, the_parent)
return (False, None)
Testing it out:
my_list=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],
[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
Test Case 1:
the_child = my_list[0][1][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
['3', '4']
[['1', '2'], ['3', '4'], ['5', '6']]
Test Case 2:
the_child = my_list[0][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[['1', '2'], ['3', '4'], ['5', '6']]
[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]
Test Case 3:
the_child = my_list[:]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
Test Case 4:
the_child = my_list[0][1] + ['Non-existent value']
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
False
[['1', '2'], ['3', '4'], ['5', '6'], 'Non-existent value']
None
answered Mar 22 at 9:40
fountainheadfountainhead
1,315313
add a comment |
You can use the following recursive function:
def get_parent_list(the_elem, the_list):
if (the_elem == the_list):
return (True, the_elem)
elif the_elem in the_list:
return (True, the_list)
else:
for e in the_list:
if (type(e) is list):
(is_found, the_parent) = get_parent_list(the_elem, e)
if (is_found):
return (True, the_parent)
return (False, None)
Testing it out:
my_list=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],
[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
Test Case 1:
the_child = my_list[0][1][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
['3', '4']
[['1', '2'], ['3', '4'], ['5', '6']]
Test Case 2:
the_child = my_list[0][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[['1', '2'], ['3', '4'], ['5', '6']]
[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]
Test Case 3:
the_child = my_list[:]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
Test Case 4:
the_child = my_list[0][1] + ['Non-existent value']
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
False
[['1', '2'], ['3', '4'], ['5', '6'], 'Non-existent value']
None
answered Mar 22 at 9:40
fountainheadfountainhead
1,315313
add a comment |
You can use the following recursive function:
def get_parent_list(the_elem, the_list):
if (the_elem == the_list):
return (True, the_elem)
elif the_elem in the_list:
return (True, the_list)
else:
for e in the_list:
if (type(e) is list):
(is_found, the_parent) = get_parent_list(the_elem, e)
if (is_found):
return (True, the_parent)
return (False, None)
Testing it out:
my_list=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],
[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
Test Case 1:
the_child = my_list[0][1][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
['3', '4']
[['1', '2'], ['3', '4'], ['5', '6']]
Test Case 2:
the_child = my_list[0][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[['1', '2'], ['3', '4'], ['5', '6']]
[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]
Test Case 3:
the_child = my_list[:]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
Test Case 4:
the_child = my_list[0][1] + ['Non-existent value']
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
False
[['1', '2'], ['3', '4'], ['5', '6'], 'Non-existent value']
None
answered Mar 22 at 9:40
fountainheadfountainhead
1,315313
You can use the following recursive function:
def get_parent_list(the_elem, the_list):
if (the_elem == the_list):
return (True, the_elem)
elif the_elem in the_list:
return (True, the_list)
else:
for e in the_list:
if (type(e) is list):
(is_found, the_parent) = get_parent_list(the_elem, e)
if (is_found):
return (True, the_parent)
return (False, None)
Testing it out:
my_list=[[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]],
[[['a','b'],['c','d'],['e','f']],[['1','2'],['3','4'],['5','6']]]]
Test Case 1:
the_child = my_list[0][1][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
['3', '4']
[['1', '2'], ['3', '4'], ['5', '6']]
Test Case 2:
the_child = my_list[0][1]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[['1', '2'], ['3', '4'], ['5', '6']]
[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]
Test Case 3:
the_child = my_list[:]
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
True
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
[[[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]], [[['a', 'b'], ['c', 'd'], ['e', 'f']], [['1', '2'], ['3', '4'], ['5', '6']]]]
Test Case 4:
the_child = my_list[0][1] + ['Non-existent value']
the_flag, the_parent = get_parent_list(the_child, my_list)
print (the_flag)
print (the_child)
print (the_parent)
Result:
False
[['1', '2'], ['3', '4'], ['5', '6'], 'Non-existent value']
None
answered Mar 22 at 9:40
fountainheadfountainhead
1,315313
answered Mar 22 at 9:40
fountainheadfountainhead
1,315313
answered Mar 22 at 9:40
fountainheadfountainhead
1,315313
answered Mar 22 at 9:40
fountainheadfountainhead
1,315313
1,315313
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1
To summarize, if you see your list as a tree, you want a function that, given a subtree, returns the parent node of that subtree? If so, the answer is no (at least with list). There are, however, many tree implementations.
– Guybrush
Mar 22 at 8:25
yes exactly, i want to get back to the parent node of that list. So there really is not any alternative way to get this in python?
– Fauzi
Mar 22 at 8:28
1
The bottom line is, List is a not a bad structure to "build" that kind of logic/abstraction on, but a list simply ISNT itself an implementation of that kind of logic/structure. You would have to use a different structure or build your own. Once you understand that, you can start looking for better alternatives to "store/represent" this relation. Be it graphs, some type of dictionary/tree structure, or so on.
– Paritosh Singh
Mar 22 at 8:29
1
@Fauzi Indeed, lists are not suited to get the "parent", because lists weren't created with that specific use case in mind (obviously). That's why you need something "more". You can either go for trees, or implement your own structure (and both can be based on lists, of course).
– Guybrush
Mar 22 at 8:31
So that's how it is :( Thank you guys for pointing out that list can't be seen as parent-child like node logic.
– Fauzi
Mar 22 at 8:37