问题
In my code, at some point I try to modify a value of a masked array, yet python seems to ignore this. I'm thinking this has to do with the way memory is stored in arrays, as if I were modifying a copy of the value and not the value itself, but I'm not well versed enough in this to have any clue how to resolve it.
Here is a simplified version of what I'm trying to do :
x = np.zeros((2,5)) # create 2D array of zeroes
x[0][1:3] = 5 # replace some values along 1st dimension with 5
mask = (x[0] > 0) # create a mask to only deal with the non negative values
x[0][mask][1] = 10 # change one of the values that is non negative
print x[0][mask][1] # value isn't changed in the original array
the output of this is :
5.0
when it should be 10.
Any help would be greatly appreciated, ideally this need to be scalable (meaning I don't necessarily know the shape of x, or where the values are non-negative, or which one I will need to modify).
I'm working with numpy 1.11.0, on python 2.7.12 on Ubuntu 16.04.2
Thanks !
回答1:
Let's generalize your problem a bit:
In [164]: x=np.zeros((2,5))
In [165]: x[0, [1, 3]] = 5 # index with a list, not a slice
In [166]: x
Out[166]:
array([[ 0., 5., 0., 5., 0.],
[ 0., 0., 0., 0., 0.]])
When the indexing occurs right before the =, it's part of a __setitem__ and acts on the original array. This is true whether the indexing uses slices, a list or a boolean mask.
But a selection with the list or mask produces a copy. Further indexed assignment affects only that copy, not the original.
In [167]: x[0, [1, 3]]
Out[167]: array([ 5., 5.])
In [168]: x[0, [1, 3]][1] = 6
In [169]: x
Out[169]:
array([[ 0., 5., 0., 5., 0.],
[ 0., 0., 0., 0., 0.]])
The best way around this is to modify the mask itself:
In [170]: x[0, np.array([1,3])[1]] = 6
In [171]: x
Out[171]:
array([[ 0., 5., 0., 6., 0.],
[ 0., 0., 0., 0., 0.]])
If the mask is boolean, you may need to convert it to indexing array
In [174]: mask = x[0]>0
In [175]: mask
Out[175]: array([False, True, False, True, False], dtype=bool)
In [176]: idx = np.where(mask)[0]
In [177]: idx
Out[177]: array([1, 3], dtype=int32)
In [178]: x[0, idx[1]]
Out[178]: 6.0
Or you can tweak the boolean values directly
In [179]: mask[1]=False
In [180]: x[0,mask]
Out[180]: array([ 6.])
So in your big problem you need to be aware of when indexing produces a view and it is a copy. And you need to be comfortable with index with lists, arrays and booleans, and understand how to switch between them.
回答2:
It's not really a masked array what you've created:
x = np.zeros((2,5))
x[0][1:3] = 5
mask = (x[0] > 0)
mask
Out[14]: array([False, True, True, False, False], dtype=bool)
So, this is just a boolean array. To create a masked array you should use numpy.ma module:
masked_x = np.ma.array(x[0], mask=~(x[0] > 0)) # let's mask first row as you did
masked_x
Out[15]:
masked_array(data = [-- 5.0 5.0 -- --],
mask = [ True False False True True],
fill_value = 1e+20)
Now you can change your masked array, and accordingly the main array:
masked_x[1] = 10.
masked_x
Out[36]:
masked_array(data = [-- 10.0 5.0 -- --],
mask = [ True False False True True],
fill_value = 1e+20)
x
Out[37]:
array([[ 0., 10., 5., 0., 0.],
[ 0., 0., 0., 0., 0.]])
And notice that in masked arrays invalid entries marked as True.
回答3:
To understand what's going on I suggest reading this http://scipy-cookbook.readthedocs.io/items/ViewsVsCopies.html
This boils down to the misleading use of fancy indexing. The following statements are the same and as you can see it's directly setting to 10 the elements of x using mask.
x[0][mask] = 10
x[0,mask] = 10
x.__setitem__((0, mask), 10)
What you're doing on the other hand is the following
x[0][mask][1] = 10
x[0,mask][1] = 10
x[0,mask].__setitem__(1, 10)
x.__getitem__((0, mask)).__setitem__(1, 10)
Which is creating a copy with __getitem__()
In conclusion you need to rethink how to modify that single number with a different mask __setitem()__
来源：https://stackoverflow.com/questions/43921510/how-can-i-change-the-value-of-a-masked-array-in-numpy