Is it possible to pass a slice of a list into a function and modify the list via the slice?
This doesn't seem to work:
def foo(a_list):
a_list[0]='abc'
x=[1,2,3,4]
foo(x[0:2])
I want x to now be x=['abc',2,3,4]
No. A "list slice" in the sense you describe is nothing but a new list. When you do x[0:2], the resulting list does not "know" that it was created as a slice of another list. It's just a new list.
What you could do is pass in the slice object itself, separately from the list:
def foo(a_list, a_slice):
a_list[a_slice]='abc'
>>> foo(x, slice(0, 2))
>>> x
['a', 'b', 'c', 3, 4]
No! Slices create copies of lists (see the documentation). You can do this:
>>> x = [1, 2, 3, 4]
>>> x[1:3] = [7, 8, 9]
>>> X
[1, 7, 8, 9, 4]
But, when you get a new list from a slicing operation, it's a copy, and thus changes to it won't affect the original:
>>> x = [1, 2, 3, 4]
>>> y = x[1:3]
>>> y[0] = 5
>>> y
[5, 3]
>>> x
[1, 2, 3, 4]
def foo(a_list):
a_list[0]='abc'
return a_list
x=[1,2,3,4]
foo(x) #it returns x=['abc',2,3,4]
Assumably you're trying to pass in x[0:2], rather than x[0,2], but the reason it doesn't work is because when you create a slice you are creating a subarray copy of x.
You are not operating on the same instance of that array, what you are doing is passing an entirely new array. Passing in 'x' alone would work, but passing in x[0:2] would not unless you specifically wrote it as x[0:2] = foo(x[0:2]) and had foo() return a_list.
As brenns10 explained, slices create a copy (even in python 3.0) of your origional data.
You could do something like the following:
def foo(x):
x[0] = 'abc'
return x
x = [0, 1, 2, 3]
x[0:2] = foo(x[0:2]) # x = ['abc', 1, 2, 3]
While this gives you the desired outcome, it doesn't exactly work as you would want. This could be problematic if needing to perform large slices as you'd have to perform a lot of copying.
Related
consider the following code:
>>> x = y = [1, 2, 3, 4]
>>> x += [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4, 4]
and then consider this:
>>> x = y = [1, 2, 3, 4]
>>> x = x + [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4]
Why is there a difference these two?
(And yes, I tried searching for this).
__iadd__ mutates the list, whereas __add__ returns a new list, as demonstrated.
An expression of x += y first tries to call __iadd__ and, failing that, calls __add__ followed an assignment (see Sven's comment for a minor correction). Since list has __iadd__ then it does this little bit of mutation magic.
The first mutates the list, and the second rebinds the name.
1)'+=' calls in-place add i.e iadd method. This method takes two parameters, but makes the change in-place, modifying the contents of the first parameter (i.e x is modified). Since both x and y point to same Pyobject they both are same.
2)Whereas x = x + [4] calls the add mehtod(x.add([4])) and instead of changing or adding values in-place it creates a new list to which a points to now and y still pointing to the old_list.
I define this function to do: [1,2,3] --> [2,3,1]
def shift_to_left(p):
p.append(p[0])
return p[1:]
When I check like this, results are ok:
p1 = [1,2,3]
print p1
p1 = shift_to_left(p1)
print p1
The result:
[1, 2, 3]
[2, 3, 1]
However, when I introduce another list and concatenate as I go the result is different:
ss = []
p1 = [1,2,3]
ss.append(p1)
p1 = shift_to_left(p1)
ss.append(p1)
print ss
The result
[[1, 2, 3, 1], [2, 3, 1]]
But I want:
[1,2,3]
[2,3,1]
why is it happening?
Thanks very much,
In Python, most arguments are taken by reference.
Your function, shift_to_left, actually mutates its argument (through the use of append), but then returns a slice (which is a shallow copy of the list).
When you replace your original variable with the output of shift_to_left, this behaviour is hidden:
In [1]: def shift_to_left(p):
...: p.append(p[0])
...: return p[1:]
...:
In [2]: xs = [1, 2, 3]
In [3]: xs = shift_to_left(xs)
In [4]: xs
Out[4]: [2, 3, 1]
But if we instead assign the result into a new variable, we can see that the original list has indeed been changed:
In [5]: ys = shift_to_left(xs)
In [6]: ys
Out[6]: [3, 1, 2]
In [7]: xs
Out[7]: [2, 3, 1, 2]
Our result, ys, is the slice of xs from the second element onwards. That's what you expected.
But xs itself has also been changed by the call to append: it's now one element longer than before.
This is what you're experiencing in your second example.
If you do not want this behaviour, one way of avoiding it is by passing a copy of your list to shift_to_left:
In [8]: zs = shift_to_left(ys[:])
In [9]: zs
Out[9]: [1, 2, 3]
In [10]: ys
Out[10]: [3, 1, 2]
Here, you can see that the original list ys has not been modified, as shift_to_left was given a copy of it, not the object itself. (This is still passing by reference, of course; it's just not a reference to ys).
Alternatively, and probably more reasonably, you could change shift_to_left itself, so that it does not modify its arguments:
def shift_to_left(xs):
return xs[1:] + xs[0] # build a new list in the return statement
The big problem with both these approaches is that they create lots of copies of lists, which can be incredibly slow (and use a lot of memory) when the lists are large.
Of course, as #Marcin points out, if this is more than an academic exercise, you should probably use one of the built-in data structures such as deque.
If you want to shift/rotate elements in a list, I think better would be to use a deque, rather than reinvent the wheel. For example:
from collections import deque
d = deque([1,2,3])
d.rotate(-1)
print(d)
#[2, 3, 1]
If you run your code here, you can notice that ss remains pointing to the original (mutated in your shift function because of p.append(p[0])) copy of p1, where as p1 points to a knew list all together when it gets reassigned, resulting in the behavior. (Step 10 out of 11)
(p becomes mutated, and ss[0] = p)
(p1 gets assigned to a new list altogether, which is latter appended to ss)
why is it happening?
return p[1:] is "non-destructive": it creates a new list. However, p.append(p[0]) is "destructive": it changes p itself.
First you append p1 to ss. This makes [[1, 2, 3]], where [1, 2, 3] is p1.
Then you do your shift_to_left, which changes p1 to [1, 2, 3, 1] and returns [2, 3, 1]. Because p1 is contained in ss, your ss becomes [[1, 2, 3, 1]], and then you append the new p1 to form [[1, 2, 3, 1], [2, 3, 1]].
A better implementation would be purely non-destructive:
def shift_to_left(p):
return p[1:] + [p[0]]
Try this:
p1 = [1,2,3]
p1 = shift_to_left(p1)
ss = []
ss.extend(p1)
print ss
That prints [2, 3, 1]. Use extend() instead because append() will create an array in an array. Also you had an extra call to ss.append(p1).
consider the following code:
>>> x = y = [1, 2, 3, 4]
>>> x += [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4, 4]
and then consider this:
>>> x = y = [1, 2, 3, 4]
>>> x = x + [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4]
Why is there a difference these two?
(And yes, I tried searching for this).
__iadd__ mutates the list, whereas __add__ returns a new list, as demonstrated.
An expression of x += y first tries to call __iadd__ and, failing that, calls __add__ followed an assignment (see Sven's comment for a minor correction). Since list has __iadd__ then it does this little bit of mutation magic.
The first mutates the list, and the second rebinds the name.
1)'+=' calls in-place add i.e iadd method. This method takes two parameters, but makes the change in-place, modifying the contents of the first parameter (i.e x is modified). Since both x and y point to same Pyobject they both are same.
2)Whereas x = x + [4] calls the add mehtod(x.add([4])) and instead of changing or adding values in-place it creates a new list to which a points to now and y still pointing to the old_list.
The del statement in python, when applied to a list, does something very strange. It not only deletes the entry in the list in question, but also "backtracks" to delete the same element in whatever list the variable was derived from.
For instance:
>>> x
[1, 2, 3, 4, 5]
>>> z=x
>>> del z[1]
>>> z
[1, 3, 4, 5]
>>> x
[1, 3, 4, 5]
In other words, even though I apply delete to z, the deletion is also applied to x. This "backtracking" doesn't come up with any other commands or functions, e.g.
>>> x=[1,2,3,4,5]
>>> z=x
>>> z+[6]
[1, 2, 3, 4, 5, 6]
>>> x
[1, 2, 3, 4, 5]
applying the add function to z leaves x unaffected, as it very well should.
This is causing me all sorts of headaches with the delete 'NA' scrip that I'm working on. Do you know what's going on with the del statement, and how can I get around this problem so that if I set z=x and then apply del to z, x remains unchanged?
z = x does not make a copy of the list. z and x now point to the same list. Any modification of that list will be visible through both names, z and x.
With z + [6], you did not assign the result back to z. Try printing z after doing that. It will be unchanged, just like x.
z += [6] would be necessary to assign the result back to z and it would have the same behavior as your first example.
You can copy the list using one of these methods:
z = x[:]
z = list(x)
consider the following code:
>>> x = y = [1, 2, 3, 4]
>>> x += [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4, 4]
and then consider this:
>>> x = y = [1, 2, 3, 4]
>>> x = x + [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4]
Why is there a difference these two?
(And yes, I tried searching for this).
__iadd__ mutates the list, whereas __add__ returns a new list, as demonstrated.
An expression of x += y first tries to call __iadd__ and, failing that, calls __add__ followed an assignment (see Sven's comment for a minor correction). Since list has __iadd__ then it does this little bit of mutation magic.
The first mutates the list, and the second rebinds the name.
1)'+=' calls in-place add i.e iadd method. This method takes two parameters, but makes the change in-place, modifying the contents of the first parameter (i.e x is modified). Since both x and y point to same Pyobject they both are same.
2)Whereas x = x + [4] calls the add mehtod(x.add([4])) and instead of changing or adding values in-place it creates a new list to which a points to now and y still pointing to the old_list.