CODEWITHSUNDEEP
python
olcesi
olcesi

Reputation: 17

Lists append in a list in for loop

In Python, I cannot create a list in which every item is a different list. This is an example:

a = [1,2,3,4,5,6,7,8,9]
b = []
c = []

for i in a:
  b.append(i)
  c.append(b)
c

the result is:

[[1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9]]

instead, what I would reach is:

[[1],
 [1, 2],
 [1, 2, 3],
 [1, 2, 3, 4],
 [1, 2, 3, 4, 5],
 [1, 2, 3, 4, 5, 6],
 [1, 2, 3, 4, 5, 6, 7],
 [1, 2, 3, 4, 5, 6, 7, 8],
 [1, 2, 3, 4, 5, 6, 7, 8, 9]]

May you please help me?

Upvotes: 1

Views: 159

Answers (2)

abhiarora
abhiarora

Reputation: 10450

In Python, variables holds references to the Objects. When you append your list b to another list c, you basically copy the reference of b to your list c (NOT THE OBJECT'S CONTENT). Since, list are mutuable, when you modify your list b (after appending it to list c), it's updated value will also be reflected in c.

Try this code to learn more:

a = [10]
c = a
a.append(100)
print(c)

Outputs:

[10, 100]

You can either do:

c.append(b[:])

OR

c.append(list(b))

OR

You can also use deepcopy in Python.

import copy
a = [1,2,3,4,5,6,7,8,9]
b = []
c = []

for i in a:
    b.append(i)
    c.append(copy.deepcopy(b))
print(c)

Upvotes: 1

azro
azro

Reputation: 54168

By doing c.append(b) you're putting the b instance, so b is everywhere in c, and as you fill b you see it in all boxes of c, you need to make a copy with on these ways

c.append(list(b))
c.append(b[:])

Regarding the task itself, I'd propose another way to do it:

for end in a:
    c.append(list(range(1, end + 1)))

Which corresponds to c = [list(range(1, end + 1)) for end in a] in list comprehension

Upvotes: 2

Related Questions