Change the values of a NumPy array that are NOT in a list of indices
Question:
I have a NumPy array like:
a = np.arange(30)
I know that I can replace the values located at positions indices=[2,3,4] using for instance fancy indexing:
a[indices] = 999
But how to replace the values at the positions that are not in indices? Would be something like below?
a[ not in indices ] = 888
Answers:
Obviously there is no general not operator for sets. Your choices are:
- Subtracting your
indices set from a universal set of indices (depends on the shape of a), but that will be a bit difficult to implement and read.
- Some kind of iteration (probably the
for-loop is your best bet since you definitely want to use the fact that your indices are sorted).
-
Creating a new array filled with new value, and selectively copying indices from the old one.
b = np.repeat(888, a.shape)
b[indices] = a[indices]
I don’t know of a clean way to do something like this:
mask = np.ones(a.shape,dtype=bool) #np.ones_like(a,dtype=bool)
mask[indices] = False
a[~mask] = 999
a[mask] = 888
Of course, if you prefer to use the numpy data-type, you could use dtype=np.bool_ — There won’t be any difference in the output. it’s just a matter of preference really.
Only works for 1d arrays:
a = np.arange(30)
indices = [2, 3, 4]
ia = np.indices(a.shape)
not_indices = np.setxor1d(ia, indices)
a[not_indices] = 888
Just overcome similar situation, solved this way:
a = np.arange(30)
indices=[2,3,4]
a[indices] = 999
not_in_indices = [x for x in range(len(a)) if x not in indices]
a[not_in_indices] = 888
I have a NumPy array like:
a = np.arange(30)
I know that I can replace the values located at positions indices=[2,3,4] using for instance fancy indexing:
a[indices] = 999
But how to replace the values at the positions that are not in indices? Would be something like below?
a[ not in indices ] = 888
Obviously there is no general not operator for sets. Your choices are:
- Subtracting your
indicesset from a universal set of indices (depends on the shape ofa), but that will be a bit difficult to implement and read. - Some kind of iteration (probably the
for-loop is your best bet since you definitely want to use the fact that your indices are sorted). -
Creating a new array filled with new value, and selectively copying indices from the old one.
b = np.repeat(888, a.shape) b[indices] = a[indices]
I don’t know of a clean way to do something like this:
mask = np.ones(a.shape,dtype=bool) #np.ones_like(a,dtype=bool)
mask[indices] = False
a[~mask] = 999
a[mask] = 888
Of course, if you prefer to use the numpy data-type, you could use dtype=np.bool_ — There won’t be any difference in the output. it’s just a matter of preference really.
Only works for 1d arrays:
a = np.arange(30)
indices = [2, 3, 4]
ia = np.indices(a.shape)
not_indices = np.setxor1d(ia, indices)
a[not_indices] = 888
Just overcome similar situation, solved this way:
a = np.arange(30)
indices=[2,3,4]
a[indices] = 999
not_in_indices = [x for x in range(len(a)) if x not in indices]
a[not_in_indices] = 888