Replace "zero-columns" with values from a numpy array

Question

Ok it´s late and i cannot solve the easiest problems anymore:

I have a Matrix with "zero-columns", these columns should be replaced with

a value from another array (same column index) that has the same number of columns:

a=np.array([[2,0,0,0],[1,0,2,0],[1,0,5,0]])
b=np.array([0.3,0.4,0.6,0.8])

result should be:

c=np.array([[2,0.4,0,0.8],[1,0.4,2,0.8],[1,0.4,5,0.8]])

i tried:

#searches for an entire zero-column indexes
wildcard_cols = np.nonzero(a.sum(axis=0) == 0)

i get:

out: wildcard_cols=array([[1, 3]], dtype=int64)# that is right

then i wanted to get a list from this output to iterate over the items in a list

wildcard_cols=np.asarray(wildcard_cols)
wildcard_cols=wildcard_cols.tolist()

but i get a list in a list(?) out=[[1, 3]]

so i cannot do:

for item in wildcard_cols:
    a[:,item]=b[item]
    #this does not work because i want to change every value in the column

i maybe thinking to complicated, but maybe someone finds a quick solution...

Why do you fill with 0.3 and 0.8? Wouldn't it be 0.4 and 0.8 if they're aligned? — DSM
– DSM, Commented Dec 23, 2013 at 2:42

DSM · Accepted Answer · 2013-12-23 02:43:55Z

3

IIUC, how about:

>>> a = np.array([[2,0,0,0],[1,0,2,0],[1,0,5,0]])*1.0
>>> b = np.array([0.3,0.4,0.6,0.8])
>>> wild = (a == 0).all(axis=0)
>>> c = a.copy()
>>> c[:,wild] = b[wild]
>>> c
array([[ 2. ,  0.4,  0. ,  0.8],
       [ 1. ,  0.4,  2. ,  0.8],
       [ 1. ,  0.4,  5. ,  0.8]])

answered Dec 23, 2013 at 2:43

DSM

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2 Comments

DSM Over a year ago

Because if you don't, a will have dtype=int. If you try to put a float into an int array, you won't get what you want.

DSM Over a year ago

@EmanuelePaolini: depends on the context. Here, where it seems that 0 was deliberately being used as a marker? Sure. In other cases, you'd want to use (abs(a) <= eps).all(axis=0) or something.

jtaylor · Accepted Answer · 2013-12-31 18:30:33Z

1

a little shorter and works with higher dimensional arrays as long as they are broadcast-able

np.where((a == 0.).all(axis=0), b, a)

unfortunately as of numpy 1.8 its slower than direct indexing like DSM proposed, a more general variant of that would be:

wild = (a == 0).all(axis=0)
c = a.copy()
c[(slice(None),) + np.nonzero(wild)] = b[wild]

answered Dec 31, 2013 at 18:30

jtaylor

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