Is there something more efficient than the following code to swap two values of a numpy 1D array?
input_seq = arange(64)
ix1 = randint(len(input_seq))
ixs2 = randint(len(input_seq))
temp = input_seq[ix2]
input_seq[ix2] = input_seq[ix1]
input_seq[ix1] = temp
I see you're using numpy arrays. In that case, you can also do this:
input_seq[[ix1, ix2]] = input_seq[[ix2, ix1]]
To explain this a bit:
input_seq[ix1] = 5 # sets value at index `ix1` to `5`
input_seq[ix2] = 42 # sets value at index `ix2` to `42`
You can also do this:
ixs = [ix1, ix2] # create a list of indices to alter
input_seq[ixs] = [5, 42] # sets the two values simultaneously
To even shorten this further, substituting ixs:
input_seq[[ix1, ix2]] = [5, 42] # sets the two values simultaneously in one line
Now Python has a very nice feature in which you can swap two variables like this without a temporary variable:
a, b = b, a # swap the values of a and b
So if you know you can read two variables by index from the array like this:
input_seq[[ix2, ix1]] # reads two values
You can set them as explained in the top of this comment.
You can use tuple unpacking. Tuple unpacking allows you to avoid the use of a temporary variable in your code (in actual fact I believe the Python code itself uses a temp variable behind the scenes but it's at a much lower level and so is much faster).
input_seq[ix1], input_seq[ix2] = input_seq[ix2], input_seq[ix1]
I have flagged this question as a duplicate, the answer in the dupe post has a lot more detail.
A now has 2s all over.
a=np.array([1,2,3]); b=np.array([4,5,6,7]); a[0:3], b[0:3] = b[0:3], a[0:3]. The result will be that the first 3 values in b get moved to a, but the a values don't copy into b. This is likely due to passing pointers (i.e. a gets overwritten by b values and THEN copied into b). It's what makes this question interesting IMHO. Adding .copy() helps, e.g. a, b[0:3] = (b[0:3]).copy(), a works.