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.
Related
This question already has answers here:
Why does += behave unexpectedly on lists?
(9 answers)
Closed last month.
I had learned that n = n + v and n += v are the same. Until this;
def assign_value(n, v):
n += v
print(n)
l1 = [1, 2, 3]
l2 = [4, 5, 6]
assign_value(l1, l2)
print(l1)
The output will be:
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 5, 6]
Now when I use the expanded version:
def assign_value(n, v):
n = n + v
print(n)
l1 = [1, 2, 3]
l2 = [4, 5, 6]
assign_value(l1, l2)
print(l1)
The output will be:
[1, 2, 3, 4, 5, 6]
[1, 2, 3]
Using the += has a different result with the fully expanded operation. What is causing this?
Thats because in the first implementation you are editing the list n itself (and therefore the changes still apply when leaving the function), while on the other implementation you are creating a new temporary list with the same name, so when you leave the function the new list disappears and the variable n is linked to the original list.
the += operator works similarly to x=x+y for immutable objects (since they always create new objects), but for mutable objects such as lists they work differently. x=x+y creats a new object x while x+=y edits the current object.
It may seem counter-intuitive, but they are not always the same. In fact,
a = a + b means a = a.__add__(b), creating a new object
a += b means a = a.__iadd__(b), mutating the object
__iadd__, if absent, defaults to the __add__, but it also can (and it does, in the case of lists) mutate the original object in-place.
This works on how python treats objects and passes variables into functions.
Basically - in first example (with += )
You are passing n and v into function by "pass-by-assignment"
So n gets modified and it will be also modified out of function scope.
In second example - n is reassigned inside of the function to a new list. Which is not seen outside of the function.
In your 1st code. You changes list n itself see the below image..!
In your 2nd code. you just created a temporary list which is cleared when function call ends.. see the below images..!
In the next step when function ends the temporary list clear!!
This is a question out of curiosity rather than trying to use it for a practical purpose.
Consider I have the following simple example where I generate a list through list comprehension:
>>> a = [1, 2, 3]
>>> b = [2 * i for i in a]
>>> b
[2, 4, 6]
>>> b.append(a)
>>> b
[2, 4, 6, [1, 2, 3]]
However if I try and do this all in one action
>>> a = [1, 2, 3]
>>> b = [2 * i for i in a].append(a)
>>> b == None
True
The result returns None. Is there any reason why this is the case?
I would have thought that an action like this would either return an answer like in the first example or throw an error.
For reference I'm using Python 3.6.5
append only works on variables, not list literals, since it updates the list object itself and does not return the resulting list.
As #Tomalak mentioned noted, running a similar operation on a simple list also returns None
>>> [1, 2, 3].append(4) == None
True
You can use concatination + instead of append in list comprehension
In [1]: a = [1, 2, 3]
In [2]: b = [2 * i for i in a] + [a]
In [3]: b
Out[3]: [2, 4, 6, [1, 2, 3]]
#ScottMcC, methods defined on mutable objects like list, dictionary mostly perform operations on calling object and doesn't return anything.
In case of immutable object like string you may see, methods return the modified form(a different object) of the calling object. In case of list, it's different.
You can't expect the below operations on list kind of mutable objects.
s = "hello DJANGO"
s2 = s.upper()
s3 = s.lower()
print(s) # hello DJANGO
print(s2) # HELLO DJANGO
print(s3) # hello django
Now, have a look at the below examples.
list is mutable object.
Calling sort() method on list directly modified the calling object and doesn't return anything (That's why None).
Calling sorted() function doesn't alter the passing list. It creates a separate sorted list based on the passed list. As it is not a method defined on list object, it returns the new sorted list.
append() method appends item on the calling list and doesn't return anything. Once you call it, you are done with updating (appending an item) the list.
# sort() method defined on list updates the calling list
# As it updates current list, it doesn't return anything. That's why None.
a = [5, 8, 1, 2, 7]
n = a.sort()
print (a)
print(n)
print ()
# sorted() function returns a new sorted list
# It doesn't update the calling list a2
a2 = [5, 8, 1, 2, 7];
n = sorted(a2);
print (a2)
print(n)
print()
# append() is method defined on list, it updates calling list so it doesn't return anything (None)
l = []
n = l.append(34)
print(l)
print (n)
Output
[1, 2, 5, 7, 8]
None
[5, 8, 1, 2, 7]
[1, 2, 5, 7, 8]
[34]
None
I have this snippet of code that just sorts a list of numbers that are guaranteed to be between 0 and R-1 (inclusive). The following code does the sort correctly but I don't understand why the input passed in remains unmodified.
def bucket(arr, R):
assert type(arr) is list
for i in arr:
assert i >=0 and i < R
b = [0] * R
for i in arr:
b[i]+=1
arr = []
for ind, v in enumerate(b):
arr = arr + [ind] * v
print(arr)
Why is inp in this example unchanged after the function has been called:
>>> inp
[3, 1, 4, 5, 4, 5, 5, 5, 1, 5]
>>> bucket(inp, 8)
[1, 1, 3, 4, 4, 5, 5, 5, 5, 5]
>>> inp # unchanged, why?
[3, 1, 4, 5, 4, 5, 5, 5, 1, 5]
Because you create a new variable called arr in the line arr = [] and from this point on you operate on a new list. Similarly you always create new lists inside the following for-loop with the arr = arr + [ind] * v operations.
You could simply change it to:
def bucket(arr, R):
assert type(arr) is list
for i in arr:
assert i >= 0 and i < R
b = [0] * R
for i in arr:
b[i] += 1
arr[:] = [] # remove all items from the list (in-place)
for ind, v in enumerate(b):
arr.extend([ind] * v) # extend the list in-place, you could also use "arr += [ind] * v"
print(arr)
Example:
>>> inp = [3, 1, 4, 5, 4, 5, 5, 5, 1, 5]
>>> bucket(inp, 8)
[1, 1, 3, 4, 4, 5, 5, 5, 5, 5]
>>> inp
[1, 1, 3, 4, 4, 5, 5, 5, 5, 5]
By assigning [] to arr you are losing the reference to the existing array, and creating a new one.
To change it, you could use
inp.sort()
More info on sort vs sorted
Python passes by assignment, and the semantics are extremely similar to Java's pass by value.
The confusion arises because you are passing the pointer by value. This means you cannot modify the pointer inside the function, but no one can stop you from modifying what the pointer is pointing to (i.e. the data) So, for example:
x = 3
def change_x(x):
x = 5
return x
change_x(x)
print x
# Outputs 3
The value of x outside the function (in this case, 3) was copied before x entered the function, so the assignment does nothing. This is typically what we expect when we hear pass by value.
x = [1,2,3]
print(id(x))
# Outputs some "address"
def change_x(x):
x.clear()
x.extend([4,5,6])
print(id(x))
# Outputs the SAME address as above
return x
change_x(x)
print(x)
# Outputs [4,5,6]
What the heck. I thought we just said we couldn't change x. The catch is we didn't change x! We just changed the data x is pointing to (there is a subtle, but important difference here). x is still the same pointer. Note that the addresses output by id are the same.
x = [1,2,3]
print(id(x))
# Outputs some "address"
def change_x(x):
x = [4,5,6]
print(id(x))
# Outputs some DIFFERENT "address". The pointer to x was changed (inside this function).
return x
change_x(x)
print(x)
# Outputs [1,2,3]
It copied the value of x before it was passed in (the pointer value, not the data, was copied). So, again, reassignment does nothing. Note that the id function outputs different values this time. So after the function returns, we get the original value of x (the original value of the pointer, that is)
I can't seem to understand following behavior in python:
x = [0, [1,2,3,4,5],[6]]
y = list(x)
y[0] = 10
y[2][0] = 7
print x
print y
It Outputs:
[0, [1, 2, 3, 4, 5], [7]]
[10, [1, 2, 3, 4, 5], [7]]
Why is second index of x and y updated and only the first index of y?
This happens because list(x) creates a shallow copy of the list x. Some of the elements in x are lists themselves. No copies are created for them; they are instead passed as references. In this way x and y end up having a reference to the same list as an element.
If you want to create a deep copy of x (i.e. to also copy the sublists) use:
import copy
y = copy.deepcopy(x)
In Python,sequence are divided into mutable sequence,which can be changed after they are created, and immutable sequence.For immutable sequences(string,Unicode,Tuples),Python make a copy for them.For mutable sequences(Lists,Byte Arrays), Python make a reference for them.
So if you change x ,y will also be changed as as they have a reference to the same list.
The standard type hierarchy
I'm having trouble understanding the odd behaviour in python functions if i pass in a list.
I made the following functions:
def func(x):
y = [4, 5, 6]
x = y
def funcsecond(x):
y = [4, 5, 6]
x[1] = y[1]
x = [1, 2, 3]
When i call func(x) and then print out x , it prints out [1, 2, 3], just the way x was before, it doesnt assign the list y to x. However, if i call funcsecond(x), it assigns 5 to the second position of x. Why is that so? When i assign the whole list, it doesn't do anything, but when i assign only one element, it changes the list i originally called it with.
Thank you very much and i hope you understand what im intending to say , i'm having a hard time expressing myself in English.
The former rebinds the name, the latter mutates the object. Changes to the name only exist in local scope, whereas a mutated object remains mutated after the scope is exited.
this is happening beacuse x points to a object which is mutable.
def func(x): # here x is a local variable which refers to the object[1,2,3]
y = [4, 5, 6]
x = y #now the local variable x refers to the object [4,5,6]
def funcsecond(x): # here x is a local variable which refers to the object[1,2,3]
y = [4, 5, 6]
x[1] = y[1] # it means [1,2,3][1]=5 , means you changed the object x was pointing to
x = [1, 2, 3]