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Let suppose I have a list l and I make new list temp from l. When I try to delete element from temp it also deletes element from l too. This is so strange. How can I only delete element from temp?
l = [1,2,3]
temp = l
temp.remove(3)
print(l, temp)
Output: ([1, 2], [1, 2])
Actually both l and temp are names that refer to the same object list in memory. When you do something with any of them, you are doing it to the underlying object list that is pointed by those two names.
You can always create a new object by using the copy method:
temp = l.copy()
And that way, you will be having two names pointing to different objects.
When you create a list using temp=l, python doesn't create new object.
It refers to the same object as temp so whatever changes you make will reflect in original list as well. Use copy method to avoid this
l = [1,2,3]
temp = l.copy()
temp.remove(3)
print(l, temp)
Output:
[1, 2, 3] [1, 2]
If you do this:
l = [1,2,3]
temp = l
You haven't created a new list, but instead you have merely created a new reference to the same list. Changing the list through either variable changes it for both.
To copy the list, you can do:
temp = list(l)
Another way is to do temp = l[:], which is equivalent. In Python 3 you can also do temp = l.copy(), but that won't work in Python 2.
Then you can change either list without affecting the other.
temp = l is simply assigning temp to a reference of l. So whatever you do to one list, the other is affected.
We can check this with the id() function:
print(id(l))
# 1739514774400
print(id(temp))
# 1739514774400
Which returns the same unique address for each object in memory.
We can see this in the documentation:
Return the “identity” of an object. This is an integer which is guaranteed to be unique and constant for this object during its lifetime. Two objects with non-overlapping lifetimes may have the same id() value.
CPython implementation detail: This is the address of the object in memory.
In this case both l and temp are the same, so they have the same identities.
We can fix this by assigning a copy instead:
l = [1,2,3]
temp = l[:]
Which now gives different identities and only removes 2 from temp:
print(id(l))
# 1822815159168
print(id(temp))
# 1822815159104
temp.remove(3)
print(l, temp)
# [1, 2, 3] [1, 2]
For example, why is a not equal to b?
a = [1]
a.append(2)
print(a) # [1, 2]
b = [1].append(2)
print(b) # None
The syntax for b doesn't look wrong to me, but it is. I want to write one-liners to define a list (e.g. using a generator expression) and then append elements, but all I get is None.
It's because:
append, extend, sort and more list function are all "in-place".
What does "in-place" mean? it means it modifies the original variable directly, some things you would need:
l = sorted(l)
To modify the list, but append already does that, so:
l.append(3)
Will modify l already, don't need:
l = l.append(3)
If you do:
l = [1].append(2)
Yes it will modify the list of [1], but it would be lost in memory somewhere inaccessible, whereas l will become None as we discovered above.
To make it not "in-place", without using append either do:
l = l + [2]
Or:
l = [*l, 2]
The one-liner for b does these steps:
Defines a list [1]
Appends 2 to the list in-place
Append has no return, so b = None
The same is true for all list methods that alter the list in-place without a return. These are all None:
c = [1].extend([2])
d = [2, 1].sort()
e = [1].insert(1, 2)
...
If you wanted a one-liner that is similar to your define and extend, you could do
c2 = [1, *[2]]
which you could use to combine two generator expressions.
All built-in methods under class 'List' in Python are just modifying the list 'in situ'. They only change the original list and return nothing.
The advantage is, you don't need to pass the object to the original variable every time you modify it. Meanwhile, you can't accumulatively call its methods in one line of code such as what is used in Javascript. Because Javascript always turns its objects into DOM, but Python not.
In python objects such as lists are passed by reference. Assignment with the = operator assigns by reference. So this function:
def modify_list(A):
A = [1,2,3,4]
Takes a reference to list and labels it A, but then sets the local variable A to a new reference; the list passed by the calling scope is not modified.
test = []
modify_list(test)
print(test)
prints []
However I could do this:
def modify_list(A):
A += [1,2,3,4]
test = []
modify_list(test)
print(test)
Prints [1,2,3,4]
How can I assign a list passed by reference to contain the values of another list? What I am looking for is something functionally equivelant to the following, but simpler:
def modify_list(A):
list_values = [1,2,3,4]
for i in range(min(len(A), len(list_values))):
A[i] = list_values[i]
for i in range(len(list_values), len(A)):
del A[i]
for i in range(len(A), len(list_values)):
A += [list_values[i]]
And yes, I know that this is not a good way to do <whatever I want to do>, I am just asking out of curiosity not necessity.
You can do a slice assignment:
>>> def mod_list(A, new_A):
... A[:]=new_A
...
>>> liA=[1,2,3]
>>> new=[3,4,5,6,7]
>>> mod_list(liA, new)
>>> liA
[3, 4, 5, 6, 7]
The simplest solution is to use:
def modify_list(A):
A[::] = [1, 2, 3, 4]
To overwrite the contents of a list with another list (or an arbitrary iterable), you can use the slice-assignment syntax:
A = B = [1,2,3]
A[:] = [4,5,6,7]
print(A) # [4,5,6,7]
print(A is B) # True
Slice assignment is implemented on most of the mutable built-in types. The above assignment is essentially the same the following:
A.__setitem__(slice(None, None, None), [4,5,6,7])
So the same magic function (__setitem__) is called when a regular item assignment happens, only that the item index is now a slice object, which represents the item range to be overwritten. Based on this example you can even support slice assignment in your own types.
This question already has answers here:
Different ways of deleting lists
(6 answers)
Closed 7 years ago.
Please what is the most efficient way of emptying a list?
I have a list called a = [1,2,3]. To delete the content of the list I usually write a = [ ]. I came across a function in python called del. I want to know if there is a difference between del a [:] and what I use.
There is a difference, and it has to do with whether that list is referenced from multiple places/names.
>>> a = [1, 2, 3]
>>> b = a
>>> del a[:]
>>> print(b)
[]
>>> a = [1, 2, 3]
>>> b = a
>>> a = []
>>> print(b)
[1, 2, 3]
Using del a[:] clears the existing list, which means anywhere it's referenced will become an empty list.
Using a = [] sets a to point to a new empty list, which means that other places the original list is referenced will remain non-empty.
The key to understanding here is to realize that when you assign something to a variable, it just makes that name point to a thing. Things can have multiple names, and changing what a name points to doesn't change the thing itself.
This can probably best be shown:
>>> a = [1, 2, 3]
>>> id(a)
45556280
>>> del a[:]
>>> id(a)
45556280
>>> b = [4, 5, 6]
>>> id(b)
45556680
>>> b = []
>>> id(b)
45556320
When you do a[:] you are referring to all elements within the list "assigned" to a. The del statement removes references to objects. So, doing del a[:] is saying "remove all references to objects from within the list assigned to a". The list itself has not changed. We can see this with the id function, which gives us a number representing an object in memory. The id of the list before using del and after remains the same, indicating the same list object is assigned to a.
On the other hand, when we assign a non-empty list to b and then assign a new empty list to b, the id changes. This is because we have actually moved the b reference from the existing [4, 5, 6] list to the new [] list.
Beyond just the identity of the objects you are dealing with, there are other things to be aware of:
>>> a = [1, 2, 3]
>>> b = a
>>> del a[:]
>>> print a
[]
>>> print b
[]
Both b and a refer to the same list. Removing the elements from the a list without changing the list itself mutates the list in place. As b references the same object, we see the same result there. If you did a = [] instead, then a will refer to a new empty list while b continues to reference the [1, 2, 3] list.
>>> list1 = [1,2,3,4,5]
>>> list2 = list1
To get a better understanding, let us see with the help of pictures what happens internally.
>>> list1 = [1,2,3,4,5]
This creates a list object and assigns it to list1.
>>> list2 = list1
The list object which list1 was referring to is also assigned to list2.
Now, lets look at the methods to empty an list and what actually happens internally.
METHOD-1: Set to empty list [] :
>>> list1 = []
>>> list2
[1,2,3,4,5]
This does not delete the elements of the list but deletes the reference to the list. So, list1 now points to an empty list but all other references will have access to that old list1.
This method just creates a new list object and assigns it to list1. Any other references will remain.
METHOD-2: Delete using slice operator[:] :
>>> del list1[:]
>>> list2
[]
When we use the slice operator to delete all the elements of the list, then all the places where it is referenced, it becomes an empty list. So list2 also becomes an empty list.
Well, del uses just a little less space in the computer as the person above me implied. The computer still accepts the variable as the same code, except with a different value. However, when you variable is assigned something else, the computer assigns a completely different code ID to it in order to account for the change in memory required.
I sometimes get across this way of printing or returning a list - someList[:].
I don't see why people use it, as it returns the full list.
Why not simply write someList, whithout the [:] part?
[:] creates a slice, usually used to get just a part of a list. Without any minimum/maximum index given, it creates a copy of the entire list. Here's a Python session demonstrating it:
>>> a = [1,2,3]
>>> b1 = a
>>> b2 = a[:]
>>> b1.append(50)
>>> b2.append(51)
>>> a
[1, 2, 3, 50]
>>> b1
[1, 2, 3, 50]
>>> b2
[1, 2, 3, 51]
Note how appending to b1 also appended the value to a. Appending to b2 however did not modify a, i.e. b2 is a copy.
In python, when you do a = b, a doesn't take the value of b, but references the same value referenced by b. To see this, make:
>>> a = {'Test': 42}
>>> b = a
>>> b['Test'] = 24
What is now the value of a?
>>> a['Test']
24
It's similar with lists, so we must find a way to really copy a list, and not make a reference to it. One way could be to recreate the list copy = list(list1), or use the functions of the copy module. But, after all, the easiest way, the prettiest, the best way ( ;) ) for doing this, is to copy each value of the first list to the other, by doing copy = list1[:]. It uses the slices, here list1 is sliced from index 0 to index len(list1), so the whole list1 is returned!
Moreover, the slice method is slightly faster: using the time.clock() method to measure the mean execution time of 1000 assignment of lists, each one containing 10000 random integers, with slices, constructor and deepcopy, the results show that the slices are 15% faster than the constructor method, and deepcopy is 4 times slower. However, this gain of time is negligible while using small lists: thus, using copy = list(list_to_copy) or copy = list_to_copy[:] is up to the developer's preferences.
Finally, we often forget the list.copy method, which seems to be the faster! In fact, it's even 13% faster than the slice method!
To create a copy of a list instead of passing by reference, as Python does. Use next two example to understand the difference.
Example:
# Passing by reference
SomeListA = [1, 2, 3]
SomeListB = [2, 3, 4]
SomeListB = SomeListA
SomeListA[2] = 5
print SomeListB
print SomeListA
# Using slice
SomeListA = [1, 2, 3]
SomeListB = [2, 3, 4]
SomeListB = SomeListA[:]
SomeListA[2] = 5
print SomeListB
print SomeListA
When you need to modify the list and you don't want to change the list and create another list you use
y = [1,2,3]
x = y[:]
and you can do a lot of changes to the list but the origin list will be in (y) and the modified in (x)
#Here is a simpler way for beginners to understand:
list16 = [1,2,3,4,5,6]
list17 = list16[:]
#^Identifying that 'list17' is a copy of 'list16' and not list16 directly
list17[0] = 10
#^ Making an alteration in the new copied list
print(list17)
print(list16)
= [10,2,3,4,5,6]
= [1,2,3,4,5,6]
#Printing the lists so that you can see what is happening. I created a copy of the list and altered it without changing the original list at all.
There are 2 copies available. 1) Deep Copy 2) Shallow Copy.
1) Deep Copy is you just copy the values
list = ['abc',123,'xyz']
list1 = copy.deepcopy(list) or list1 = list[:]
2) Shallow Copy is you just reference to the varible
list2 = copy.copy(list) or list2 = list
When you modify something on list2 it get effected in list also as it is referenced.
list1.append(456)
list2.append('789')
print "list: %s" %list
print "list1: %s" %list1
print "list2: %s" %list2
ans:
list : ['abc',123,'xyz','789']
list1 : ['abc',123,'xyz',456]
list2 : ['abc',123,'xyz','789']