python : problem with insertion in a list

Question

I have written the following code that works and the output is what it is expected :

myList1=[[1,2],[3,4],[5,6]]
myList2=["abc","def","ghi"]
myList3=[]
for i in range(3):
    temp=myList1[i]
    temp.insert(0,myList2[i])
    myList3.append(temp)
print(myList3)

But when I try to write the same code without the temp variable, it does not work.

myList1=[[1,2],[3,4],[5,6]]
myList2=["abc","def","ghi"]
myList3=[]
for i in range(3):
    myList3.append(myList1[i].insert(0,myList2[i]))
print(myList3)

And this code, also, does not work.

myList1=[[1,2],[3,4],[5,6]]
myList2=["abc","def","ghi"]
myList3=[]
for i in range(3):
    temp=myList1[i].insert(0,myList2[i])
    myList3.append(temp)
print(myList3)

I think there are some python fundamentals that I have not understood.

Any comments and help will be appreciated.

Thanks for answer

.insert returns None so you get None,None,None

user15801675
– user15801675

2021-07-22 11:49:33 +00:00
Commented Jul 22, 2021 at 11:49 — user15801675
– user15801675, Commented Jul 22, 2021 at 11:49
Yes. I know that but why it returns None.

Stef1611
– Stef1611

2021-07-22 11:58:20 +00:00
Commented Jul 22, 2021 at 11:58 — Stef1611
– Stef1611, Commented Jul 22, 2021 at 11:58

user15801675 · Accepted Answer · 2021-07-22 11:52:08Z

2

.insert is an inplace function. It returns None or nothing. If you don't want to use temp, you can use zip()

myList1=[[1,2],[3,4],[5,6]]
myList2=["abc","def","ghi"]
myList3=[]
for i,j in zip(myList1,myList2):
    i.insert(0,j)
    myList3.append(i)
print(myList3)

A compact solution:

myList1=[[1,2],[3,4],[5,6]]
myList2=["abc","def","ghi"]

myList3=[[j]+i for i,j in zip(myList1,myList2)]
print(myList3)

answered Jul 22, 2021 at 11:52

user15801675

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2 Comments

Stef1611 Over a year ago

Thanks for the keyword : inplace function

Stef1611 Over a year ago

The sighingnow comment makes me in trouble (see its answer and comments). As I am a python beginner. I am going to read some docs about inplace function. I let you discuss with Sighingnow.

sighingnow · Accepted Answer · 2021-07-22 11:52:39Z

1

.insert of list returns None, rather than the result list.

To avoid the usage of temporary variable tmp, you could have

myList1=[[1,2],[3,4],[5,6]]
myList2=["abc","def","ghi"]
myList3=[]
for i in range(3):
    myList3.append(myList2[i:i+1] + myList1[i])
print(myList3)

Or

myList1=[[1,2],[3,4],[5,6]]
myList2=["abc","def","ghi"]
myList3=[]
for i in range(3):
    myList3.append([myList2[i]] + myList1[i])
print(myList3)

answered Jul 22, 2021 at 11:52

sighingnow

8416 silver badges11 bronze badges

5 Comments

Stef1611 Over a year ago

I see that. But why None is returned

sighingnow Over a year ago

This is how list.insert is implemented. And has noting to do with "inplace function". You could define a inplace insert function as python def insert(xs, idx, x): xs.insert(idx, x) return xs It is a inplace function, but it doesn't return None.

Stef1611 Over a year ago

Your comment makes me in trouble. As I am a python beginner. I am going to read some docs about inplace function. I let you discuss with Sujay.

sighingnow Over a year ago

Inplace function means the function operates on its argument, but inplace function do could have meaningful returns.

Stef1611 Over a year ago

Do you have link / blog / discussion to advise me ?

clemens · Accepted Answer · 2021-07-22 12:00:55Z

1

Take a look at your last example: The line: temp=myList1[i].insert(0,myList2[i]) does change your list "myList1"! The problem is, that the result of the insert will return "None".

So something like this would work, but not be pretty:

myList1=[[1,2],[3,4],[5,6]]
myList2=["abc","def","ghi"]
myList3=[]
for i in range(3):
    myList1[i].insert(0,myList2[i])
    myList3.append(myList1[i])
print(myList3)

answered Jul 22, 2021 at 12:00

clemens

1158 bronze badges

1 Comment

Stef1611 Over a year ago

Thanks for answer. Sujay give me the answer. insert is a inplace function

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