how can I return the address of the value in the list I wish to change?
I can change a value in a list in this manner:
a = [1, 2, 3]
a[0] = 0
but how can I change a value that a helper function choose?
helper = lambda x: x[0]
a = [1, 2, 3]
helper(a) = 0
will produce a runtime error type syntax error (because python interpret it like 1 = 0)
my question is how can I build a function that chooses the index for me?
functions that return only the index are not helpful:
example:
helper = lambda x: 0
a = [1, 2, 3]
a[helper(a)] = 0
will not solve the problem for (for example) 2d lists:
helper_1 = lambda x: 0, 0
a = [[1, 2, 3]]
a[helper(a)[0]][helper(a)[1]] = 0
as you can see this makes the code very ugly very fast.
I will provide two simple examples for this issue:
first example code, 1d list:
lst = [1, 2, 3]
for x in lst:
x *= 2
# will not change the list
if there were pointers:
for &x in lst:
*x *= 2
second example code, 2d list:
def rotate_90(image, direction):
rows, columns = len(image), len(image[0])
rotated = empty_2d(columns, rows) # allocate 2d list
rotated_place = lambda row, col: rotated[col][-1- row] \
if direction =='R' else \
lambda row, col: rotated[-1-col][row]
for row in range(rows):
for col in range(columns):
rotated_place(row, col) = image[row][col] # error
return rotated
# will produce an error
if there were pointers:
def rotate_90(image, direction):
rows, columns = len(image), len(image[0])
rotated = empty_2d(columns, rows) # aloccate memory
rotated_place = lambda row, col: &rotated[col][-1- row] \
if direction =='R' else \
lambda row, col: &rotated[-1-col][row]
for row in range(rows):
for col in range(columns):
*rotated_place(row, col) = image[row][col]
return rotated
CodePudding user response：
Changing a bit for clarity, you want this to print [42, 2, 3]:
first = ...
a = [1, 2, 3]
first(a) = 42
print(a)
You can make it work with first[a] instead of first(a):
class First:
def __setitem__(_, lst, value):
lst[0] = value
first = First()
a = [1, 2, 3]
first[a] = 42
print(a)