How to change the first element in a tuple that is a key in a dictionary?
I need to set the first element of a tuple which is a key in a dictionary to be current.
{('a', '0'): 'c',
('b', '0'): 'd',}
I need to set 'a' as current variable before looping the code.
How can I do this?
python dictionary tuples
edited Nov 22 '18 at 18:23
Eugene Yarmash
83.7k22177260
asked Nov 22 '18 at 16:02
pollywogzpollywogz
175
add a comment |
I need to set the first element of a tuple which is a key in a dictionary to be current.
{('a', '0'): 'c',
('b', '0'): 'd',}
I need to set 'a' as current variable before looping the code.
How can I do this?
python dictionary tuples
edited Nov 22 '18 at 18:23
Eugene Yarmash
83.7k22177260
asked Nov 22 '18 at 16:02
pollywogzpollywogz
175
3
Can you elaborate on what you are doing? Are you trying to use 'a' as a variable name, or just remember that the last time you checked, 'a' was valid, or what?
– Austin Hastings
Nov 22 '18 at 16:09
add a comment |
I need to set the first element of a tuple which is a key in a dictionary to be current.
{('a', '0'): 'c',
('b', '0'): 'd',}
I need to set 'a' as current variable before looping the code.
How can I do this?
python dictionary tuples
edited Nov 22 '18 at 18:23
Eugene Yarmash
83.7k22177260
asked Nov 22 '18 at 16:02
pollywogzpollywogz
175
I need to set the first element of a tuple which is a key in a dictionary to be current.
{('a', '0'): 'c',
('b', '0'): 'd',}
I need to set 'a' as current variable before looping the code.
How can I do this?
python dictionary tuples
python dictionary tuples
edited Nov 22 '18 at 18:23
Eugene Yarmash
83.7k22177260
asked Nov 22 '18 at 16:02
pollywogzpollywogz
175
edited Nov 22 '18 at 18:23
Eugene Yarmash
83.7k22177260
asked Nov 22 '18 at 16:02
pollywogzpollywogz
175
edited Nov 22 '18 at 18:23
Eugene Yarmash
83.7k22177260
edited Nov 22 '18 at 18:23
Eugene Yarmash
83.7k22177260
edited Nov 22 '18 at 18:23
Eugene Yarmash
83.7k22177260
83.7k22177260
asked Nov 22 '18 at 16:02
pollywogzpollywogz
175
asked Nov 22 '18 at 16:02
pollywogzpollywogz
175
asked Nov 22 '18 at 16:02
pollywogzpollywogz
175
175
3
Can you elaborate on what you are doing? Are you trying to use 'a' as a variable name, or just remember that the last time you checked, 'a' was valid, or what?
– Austin Hastings
Nov 22 '18 at 16:09
add a comment |
3
Can you elaborate on what you are doing? Are you trying to use 'a' as a variable name, or just remember that the last time you checked, 'a' was valid, or what?
– Austin Hastings
Nov 22 '18 at 16:09
3
3
Can you elaborate on what you are doing? Are you trying to use 'a' as a variable name, or just remember that the last time you checked, 'a' was valid, or what?
– Austin Hastings
Nov 22 '18 at 16:09
Can you elaborate on what you are doing? Are you trying to use 'a' as a variable name, or just remember that the last time you checked, 'a' was valid, or what?
– Austin Hastings
Nov 22 '18 at 16:09
add a comment |
3 Answers
3
active
oldest
votes
You can't modify the key itself because it's an immutable object. Instead, replace the dictionary key with a new one, e.g.:
key1, key2 = ('a', '0'), (current, 0)
d[key2] = d[key1]
del d[key1]
or in one step:
d[key2] = d.pop(key1)
answered Nov 22 '18 at 16:05
Eugene YarmashEugene Yarmash
83.7k22177260
add a comment |
If you want to loop through all your dictionaries to update them only by the first item in your tuple key.
d = {('a', '0'): 'c', ('b', '0'): 'd',}
for key in d:
new_key = key[0]
d[new_key] = d.pop(key)
d
>>{'a': 'c', 'b': 'd'}
Essentially here you are deleting the old key and replacing it with a new one where you take index 0 of your tuple to be the new key.
answered Nov 22 '18 at 16:07
BernardLBernardL
2,3581929
add a comment |
Are you trying to access 'a' from the dictionary?
If you are then you can do the following:
list(d.keys())[0][0]
>>> a
However; if the above is not what is required from you but the below is.
{('a', '0'): 'a', ('b', '0'): 'd'}
You can accomplish this with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
print(d)
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
Or are you looking to achieve something as this:
{('b', '0'): 'd', ('c', '0'): 'a'}
Again, with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
answered Nov 22 '18 at 19:13
Victor SVictor S
1668
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
You can't modify the key itself because it's an immutable object. Instead, replace the dictionary key with a new one, e.g.:
key1, key2 = ('a', '0'), (current, 0)
d[key2] = d[key1]
del d[key1]
or in one step:
d[key2] = d.pop(key1)
answered Nov 22 '18 at 16:05
Eugene YarmashEugene Yarmash
83.7k22177260
add a comment |
You can't modify the key itself because it's an immutable object. Instead, replace the dictionary key with a new one, e.g.:
key1, key2 = ('a', '0'), (current, 0)
d[key2] = d[key1]
del d[key1]
or in one step:
d[key2] = d.pop(key1)
answered Nov 22 '18 at 16:05
Eugene YarmashEugene Yarmash
83.7k22177260
add a comment |
You can't modify the key itself because it's an immutable object. Instead, replace the dictionary key with a new one, e.g.:
key1, key2 = ('a', '0'), (current, 0)
d[key2] = d[key1]
del d[key1]
or in one step:
d[key2] = d.pop(key1)
answered Nov 22 '18 at 16:05
Eugene YarmashEugene Yarmash
83.7k22177260
You can't modify the key itself because it's an immutable object. Instead, replace the dictionary key with a new one, e.g.:
key1, key2 = ('a', '0'), (current, 0)
d[key2] = d[key1]
del d[key1]
or in one step:
d[key2] = d.pop(key1)
answered Nov 22 '18 at 16:05
Eugene YarmashEugene Yarmash
83.7k22177260
edited Dec 12 '18 at 13:41
answered Nov 22 '18 at 16:05
Eugene YarmashEugene Yarmash
83.7k22177260
answered Nov 22 '18 at 16:05
Eugene YarmashEugene Yarmash
83.7k22177260
answered Nov 22 '18 at 16:05
Eugene YarmashEugene Yarmash
83.7k22177260
83.7k22177260
add a comment |
add a comment |
If you want to loop through all your dictionaries to update them only by the first item in your tuple key.
d = {('a', '0'): 'c', ('b', '0'): 'd',}
for key in d:
new_key = key[0]
d[new_key] = d.pop(key)
d
>>{'a': 'c', 'b': 'd'}
Essentially here you are deleting the old key and replacing it with a new one where you take index 0 of your tuple to be the new key.
answered Nov 22 '18 at 16:07
BernardLBernardL
2,3581929
add a comment |
If you want to loop through all your dictionaries to update them only by the first item in your tuple key.
d = {('a', '0'): 'c', ('b', '0'): 'd',}
for key in d:
new_key = key[0]
d[new_key] = d.pop(key)
d
>>{'a': 'c', 'b': 'd'}
Essentially here you are deleting the old key and replacing it with a new one where you take index 0 of your tuple to be the new key.
answered Nov 22 '18 at 16:07
BernardLBernardL
2,3581929
add a comment |
If you want to loop through all your dictionaries to update them only by the first item in your tuple key.
d = {('a', '0'): 'c', ('b', '0'): 'd',}
for key in d:
new_key = key[0]
d[new_key] = d.pop(key)
d
>>{'a': 'c', 'b': 'd'}
Essentially here you are deleting the old key and replacing it with a new one where you take index 0 of your tuple to be the new key.
answered Nov 22 '18 at 16:07
BernardLBernardL
2,3581929
If you want to loop through all your dictionaries to update them only by the first item in your tuple key.
d = {('a', '0'): 'c', ('b', '0'): 'd',}
for key in d:
new_key = key[0]
d[new_key] = d.pop(key)
d
>>{'a': 'c', 'b': 'd'}
Essentially here you are deleting the old key and replacing it with a new one where you take index 0 of your tuple to be the new key.
answered Nov 22 '18 at 16:07
BernardLBernardL
2,3581929
answered Nov 22 '18 at 16:07
BernardLBernardL
2,3581929
answered Nov 22 '18 at 16:07
BernardLBernardL
2,3581929
answered Nov 22 '18 at 16:07
BernardLBernardL
2,3581929
2,3581929
add a comment |
add a comment |
Are you trying to access 'a' from the dictionary?
If you are then you can do the following:
list(d.keys())[0][0]
>>> a
However; if the above is not what is required from you but the below is.
{('a', '0'): 'a', ('b', '0'): 'd'}
You can accomplish this with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
print(d)
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
Or are you looking to achieve something as this:
{('b', '0'): 'd', ('c', '0'): 'a'}
Again, with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
answered Nov 22 '18 at 19:13
Victor SVictor S
1668
add a comment |
Are you trying to access 'a' from the dictionary?
If you are then you can do the following:
list(d.keys())[0][0]
>>> a
However; if the above is not what is required from you but the below is.
{('a', '0'): 'a', ('b', '0'): 'd'}
You can accomplish this with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
print(d)
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
Or are you looking to achieve something as this:
{('b', '0'): 'd', ('c', '0'): 'a'}
Again, with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
answered Nov 22 '18 at 19:13
Victor SVictor S
1668
add a comment |
Are you trying to access 'a' from the dictionary?
If you are then you can do the following:
list(d.keys())[0][0]
>>> a
However; if the above is not what is required from you but the below is.
{('a', '0'): 'a', ('b', '0'): 'd'}
You can accomplish this with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
print(d)
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
Or are you looking to achieve something as this:
{('b', '0'): 'd', ('c', '0'): 'a'}
Again, with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
answered Nov 22 '18 at 19:13
Victor SVictor S
1668
Are you trying to access 'a' from the dictionary?
If you are then you can do the following:
list(d.keys())[0][0]
>>> a
However; if the above is not what is required from you but the below is.
{('a', '0'): 'a', ('b', '0'): 'd'}
You can accomplish this with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
print(d)
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
>>> {('a', '0'): 'a', ('b', '0'): 'd'}
Or are you looking to achieve something as this:
{('b', '0'): 'd', ('c', '0'): 'a'}
Again, with lambda or with list comprehension:
# Lambda
d[('a', '0')] = list(map(lambda k: k[0], d))[0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
# list comprehension
d[('a', '0')] = [k[0] for k in d][0]
d[('c', '0')] = d.pop(('a', '0'))
print(d)
>>> {('b', '0'): 'd', ('c', '0'): 'a'}
answered Nov 22 '18 at 19:13
Victor SVictor S
1668
edited Nov 23 '18 at 10:44
answered Nov 22 '18 at 19:13
Victor SVictor S
1668
answered Nov 22 '18 at 19:13
Victor SVictor S
1668
answered Nov 22 '18 at 19:13
Victor SVictor S
1668
1668
add a comment |
add a comment |
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3
Can you elaborate on what you are doing? Are you trying to use 'a' as a variable name, or just remember that the last time you checked, 'a' was valid, or what?
– Austin Hastings
Nov 22 '18 at 16:09