Python快捷方式来填充列表

Question

Java has Arrays.fill(A,1) . For a pre-existing List A , is there a shortcut for filling the list with 1 ? I am writing a function that takes an array and changes the array in some ways. Since arrays are pointers, my function will not return an array. The caller will see the changes after my function returns. The first step in my function is to fill the array with 1s . Doing

def my_work(A):
   A =[1]*len(A) 
   # more work on A

does not seem to change A when my_work is done.

So is my only option

for i in range(len(A)):
  A[i]=1

or is there a shortcut? Mine looks like a workaround.

Answer 1

If you really want to change A in place, the A[:] syntax should work:

>>> A = [1,2,3]
>>> def my_work(A):
...     A[:] = [1]*len(A) 
...     
>>> A
[1, 2, 3]
>>> my_work(A)
>>> A
[1, 1, 1]

And here is the relevant section of the tutorial ("assignment to slices").

Answer 2

A is not changing because your not returning it our of my_work . Once my_work is complete the A you have in there is not referenced.

def my_work(A):
    A =[1]*len(A) 
    return A  #This line is needed

A = my_work(A)

Another way to do this is something like this.

def reset_list(L, x):
     return [x for i in xrange(len(L))]

A = reset_list(A,1)