pop/remove items out of a python tuple
Question:
I am not sure if I can make myself clear but will try.
I have a tuple in python which I go through as follows (see code below). While going through it, I maintain a counter (let’s call it ‘n’) and ‘pop’ items that meet a certain condition.
Now of course once I pop the first item, the numbering all goes wrong, how can I do what I want to do more elegantly while removing only certain entries of a tuple on the fly?
for x in tupleX:
n=0
if (condition):
tupleX.pop(n)
n=n+1
Answers:
Maybe you want dictionaries?
d = dict( (i,value) for i,value in enumerate(tple))
while d:
bla bla bla
del b[x]
ok I figured out a crude way of doing it.
I store the “n” value in the for loop when condition is satisfied in a list (lets call it delList) then do the following:
for ii in sorted(delList, reverse=True):
tupleX.pop(ii)
Any other suggestions are welcome too.
As DSM mentions, tuple‘s are immutable, but even for lists, a more elegant solution is to use filter:
tupleX = filter(str.isdigit, tupleX)
or, if condition is not a function, use a comprehension:
tupleX = [x for x in tupleX if x > 5]
if you really need tupleX to be a tuple, use a generator expression and pass that to tuple:
tupleX = tuple(x for x in tupleX if condition)
There is a simple but practical solution.
As DSM said, tuples are immutable, but we know Lists are mutable.
So if you change a tuple to a list, it will be mutable. Then you can delete the items by the condition, then after changing the type to a tuple again. That’s it.
Please look at the codes below:
tuplex = list(tuplex)
for x in tuplex:
if (condition):
tuplex.pop(tuplex.index(x))
tuplex = tuple(tuplex)
print(tuplex)
For example, the following procedure will delete all even numbers from a given tuple.
tuplex = (1, 2, 3, 4, 5, 6, 7, 8, 9)
tuplex = list(tuplex)
for x in tuplex:
if (x % 2 == 0):
tuplex.pop(tuplex.index(x))
tuplex = tuple(tuplex)
print(tuplex)
if you test the type of the last tuplex, you will find it is a tuple.
Finally, if you want to define an index counter as you did (i.e., n), you should initialize it before the loop, not in the loop.
Yes we can do it.
First convert the tuple into an list, then delete the element in the list after that again convert back into tuple.
Demo:
my_tuple = (10, 20, 30, 40, 50)
# converting the tuple to the list
my_list = list(my_tuple)
print my_list # output: [10, 20, 30, 40, 50]
# Here i wanna delete second element "20"
my_list.pop(1) # output: [10, 30, 40, 50]
# As you aware that pop(1) indicates second position
# Here i wanna remove the element "50"
my_list.remove(50) # output: [10, 30, 40]
# again converting the my_list back to my_tuple
my_tuple = tuple(my_list)
print my_tuple # output: (10, 30, 40)
Thanks
The best solution is the tuple applied to a list comprehension, but to extract one
item this could work:
def pop_tuple(tuple, n):
return tuple[:n]+tuple[n+1:], tuple[n]
say you have a dict with tuples as keys, e.g: labels = {(1,2,0): 'label_1'} you can modify the elements of the tuple keys as follows:
formatted_labels = {(elem[0],elem[1]):labels[elem] for elem in labels}
Here, we ignore the last elements.
In Python 3 this is no longer an issue, and you really don’t want to use list comprehension, coercion, filters, functions or lambdas for something like this.
Just use
popped = unpopped[:-1]
Remember that it’s an immutable, so you will have to reassign the value if you want it to change
my_tuple = my_tuple[:-1]
Example
>>> foo= 3,5,2,4,78,2,1
>>> foo
(3, 5, 2, 4, 78, 2, 1)
foo[:-1]
(3, 5, 2, 4, 78, 2)
If you want to have the popped value,,
>>> foo= 3,5,2,4,78,2,1
>>> foo
(3, 5, 2, 4, 78, 2, 1)
>>> foo, bit = foo[:-1], foo[-1]
>>> bit
1
>>> foo
(3, 5, 2, 4, 78, 2)
Or, to work with each value of a tuple starting at the back…
foo = 3,5,2,4,78,2,1
for f in reversed(foo):
print(f) # 1; 2; 78; ...
Or, with the count…
foo = 3,5,2,4,78,2,1
for f, i in enumerate(reversed(foo)):
print(i, f) # 0 1; 1 2; 2 78; ...
Or, to coerce into a list..
bar = [*foo]
#or
bar = list(foo)
One solution is to convert to set and bring back to tuple
tupleX = (
"ZAR",
"PAL",
"SEV",
"ALC",
"LPA",
"TFN",)
remove = (
"LPA",
"TFN",)
tuple(set(tupleX) - set(remove))
(‘ZAR’, ‘PAL’, ‘ALC’, ‘SEV’)
A shorter way perhaps:
tup = (0, 1, 2, 3)
new_tup = (*tup[:-2], tup[-1])
print(new_tup) # (0, 1, 3)
I am not sure if I can make myself clear but will try.
I have a tuple in python which I go through as follows (see code below). While going through it, I maintain a counter (let’s call it ‘n’) and ‘pop’ items that meet a certain condition.
Now of course once I pop the first item, the numbering all goes wrong, how can I do what I want to do more elegantly while removing only certain entries of a tuple on the fly?
for x in tupleX:
n=0
if (condition):
tupleX.pop(n)
n=n+1
Maybe you want dictionaries?
d = dict( (i,value) for i,value in enumerate(tple))
while d:
bla bla bla
del b[x]
ok I figured out a crude way of doing it.
I store the “n” value in the for loop when condition is satisfied in a list (lets call it delList) then do the following:
for ii in sorted(delList, reverse=True):
tupleX.pop(ii)
Any other suggestions are welcome too.
As DSM mentions, tuple‘s are immutable, but even for lists, a more elegant solution is to use filter:
tupleX = filter(str.isdigit, tupleX)
or, if condition is not a function, use a comprehension:
tupleX = [x for x in tupleX if x > 5]
if you really need tupleX to be a tuple, use a generator expression and pass that to tuple:
tupleX = tuple(x for x in tupleX if condition)
There is a simple but practical solution.
As DSM said, tuples are immutable, but we know Lists are mutable.
So if you change a tuple to a list, it will be mutable. Then you can delete the items by the condition, then after changing the type to a tuple again. That’s it.
Please look at the codes below:
tuplex = list(tuplex)
for x in tuplex:
if (condition):
tuplex.pop(tuplex.index(x))
tuplex = tuple(tuplex)
print(tuplex)
For example, the following procedure will delete all even numbers from a given tuple.
tuplex = (1, 2, 3, 4, 5, 6, 7, 8, 9)
tuplex = list(tuplex)
for x in tuplex:
if (x % 2 == 0):
tuplex.pop(tuplex.index(x))
tuplex = tuple(tuplex)
print(tuplex)
if you test the type of the last tuplex, you will find it is a tuple.
Finally, if you want to define an index counter as you did (i.e., n), you should initialize it before the loop, not in the loop.
Yes we can do it.
First convert the tuple into an list, then delete the element in the list after that again convert back into tuple.
Demo:
my_tuple = (10, 20, 30, 40, 50)
# converting the tuple to the list
my_list = list(my_tuple)
print my_list # output: [10, 20, 30, 40, 50]
# Here i wanna delete second element "20"
my_list.pop(1) # output: [10, 30, 40, 50]
# As you aware that pop(1) indicates second position
# Here i wanna remove the element "50"
my_list.remove(50) # output: [10, 30, 40]
# again converting the my_list back to my_tuple
my_tuple = tuple(my_list)
print my_tuple # output: (10, 30, 40)
Thanks
The best solution is the tuple applied to a list comprehension, but to extract one
item this could work:
def pop_tuple(tuple, n):
return tuple[:n]+tuple[n+1:], tuple[n]
say you have a dict with tuples as keys, e.g: labels = {(1,2,0): 'label_1'} you can modify the elements of the tuple keys as follows:
formatted_labels = {(elem[0],elem[1]):labels[elem] for elem in labels}
Here, we ignore the last elements.
In Python 3 this is no longer an issue, and you really don’t want to use list comprehension, coercion, filters, functions or lambdas for something like this.
Just use
popped = unpopped[:-1]
Remember that it’s an immutable, so you will have to reassign the value if you want it to change
my_tuple = my_tuple[:-1]
Example
>>> foo= 3,5,2,4,78,2,1
>>> foo
(3, 5, 2, 4, 78, 2, 1)
foo[:-1]
(3, 5, 2, 4, 78, 2)
If you want to have the popped value,,
>>> foo= 3,5,2,4,78,2,1
>>> foo
(3, 5, 2, 4, 78, 2, 1)
>>> foo, bit = foo[:-1], foo[-1]
>>> bit
1
>>> foo
(3, 5, 2, 4, 78, 2)
Or, to work with each value of a tuple starting at the back…
foo = 3,5,2,4,78,2,1
for f in reversed(foo):
print(f) # 1; 2; 78; ...
Or, with the count…
foo = 3,5,2,4,78,2,1
for f, i in enumerate(reversed(foo)):
print(i, f) # 0 1; 1 2; 2 78; ...
Or, to coerce into a list..
bar = [*foo]
#or
bar = list(foo)
One solution is to convert to set and bring back to tuple
tupleX = (
"ZAR",
"PAL",
"SEV",
"ALC",
"LPA",
"TFN",)
remove = (
"LPA",
"TFN",)
tuple(set(tupleX) - set(remove))
(‘ZAR’, ‘PAL’, ‘ALC’, ‘SEV’)
A shorter way perhaps:
tup = (0, 1, 2, 3)
new_tup = (*tup[:-2], tup[-1])
print(new_tup) # (0, 1, 3)