how to get the result of ［1,1］.insert(1,3)(［1,3,1］) without using b=［1,1］b.insert(1,3) ?

Question

because the latter one b will keep changing in a loop where i can insert many elements in b ,i don't want that,i want b to be ［1,1］all the time during a loop where i can insert other element to it like a pattern.Anyone could help with this?

Answer 1

## else u can do this....if u wanna incorporate "insert" in your code in each step anyway

a=[1, 1, 1, 1]
temp=a.copy()
for i in range(5):
a.insert(2,i)
print(a)
a=[i for i in temp]

Answer 2

You mean
[1,1].insert(1,3) doesn't work?

Answer 3

we don't have a variant to store the result ［1,3,1］using ［1，1］.insert()

Answer 4

Imaging this:
def add_pattern(n):
a=［1,1,1,1］
for i in range(n):
a.insert(2,i)
print a

if n=3,print ［1,1,0,1,1］,［1,1,1,0,1,1] etc.We lose the original pattern [1,1,1,1] which can be replaced by many other patterns.
but what i want is ［1,1,0,1,1］,［1,1,1,1,1］,［1,1,2,1,1］etc.Maybe it is just not the way to do that.Many other way to do that,but i just want to try it anyway.

thanks for your answer.still very helpful😀

Answer 5

yep, We just need to transfer the changed a into the unchanged a(pattern) at the end of the loop, although it seem not to be the best way.Thank 😀you so much

Answer 6

python dont work like that....

in c ; java u need to declair a var furst then initialise...
but in python...
any initialisation is itsekf declairisation..
like that...
when u are doing
a=[1, 1]
a.insert(1, 3)
it makes a [1,3,1] and then returns a

but when u directly do

[1, 1].insert(1, 3)

## think it off like that##

python virtually creats a var
copies [1, 1] in it
changes that var into [1, 3, 1]
returns the list in var as[1, 3, 1] ##destroying the var

Answer 7

the thing is
what u wanna do has nothing to do with insert except 1st move--------<>

a=[1, 1, 1, 1]
a.insert(2,0)
print(a)
for i in range(1,5):
a[2]=i
print(a)