在列表列表中编辑项目时遇到麻烦

Question

I wanted to a make a list of list using pythons and I know two methods to do it.

The first method is like so:

l = []
l_in = []
for j in range(2):
   l_in.append("1")
print l_in
for i in range(2):
   l.append(l_in)
print l

and the second method is:

l = []
for i in range(2):
   l.append([1]*2)
print l

Both these methods creates a list

l = [['1', '1'], ['1', '1']] 

I want to change only on element int this list say the (1,1) element. If I do

l[1][1] = "Something I want to replace with"

the element should be replaced. This works fine for the second method. But if I use the first method both (1,1) and (0,1) changes to "Something I want to replace with" .

Can someone tell me why using the first method gives this output?

Answer 1

It's because in the first method, in the second loop you are appending the same list twice.

for i in range(2):
    l.append(l_in)

Now whenever you're modifying an element in l , you are actually modifying elements in one list. An easy way to understand is to look at the values the builtin id() gives for the content of l .

>>> map(id, l)
[32750120, 32750120]

You can see that both elements in l share the same id (and the same memory address). They both refer to the same list l_in .

As Aशwini चhaudhary suggested, an easy fix is l.append(l_in[:]) where the [:] creates a copy of the list l_in .

l = []
l_in = []
for j in range(2):
   l_in.append("1")
print l_in
for i in range(2):
   l.append(l_in[:])
print l

You can also create lists of lists with list and generator comprehensions. Some examples below:

# empty lists..
>>> [[] for _ in range(4)]
[[], [], [], []]

# generator expression..
>>> list([] for _ in range(4))
[[], [], [], []]

# [1, 1] for each sublist.
>>> list([1, 1] for _ in range(4))
[[1, 1], [1, 1], [1, 1], [1, 1]]

Here's the help() on id() :

Help on built-in function id in module __builtin__:

id(...)
    id(object) -> integer

    Return the identity of an object.  This is guaranteed to be unique among
    simultaneously existing objects.  (Hint: it's the object's memory address.)