python How to replace the first f(x) values of each x in array(x,n)

Question

I would like to replace the first x values in every row of my array a with ones and to keep all the other values NaN. The first x values however changes in every row and is determined by a list b.

Since I'm not very familiar with arrays I thought I might do this with a for loop as shown below, but this doesn't work (I've got inspiration for the basics of replacement in arrays from How to set first N elements of array to zero?).

In:
a = np.empty((3,4))
a.fill(np.nan)
b = [2,3,1]

for i in range(b):
    a[0:b[i]] = [1] * b[i] 
    a[i:] = np.ones((b[i]))
pass

Out:
line 7:
ValueError: could not broadcast input array from shape (2) into shape (2,4)

Result should be like:

Out:
[[1, 1, nan, nan], 
 [1, 1, 1, nan], 
 [1, nan, nan, nan]] 

Answer 1

In the linked answer, How to set first N elements of array to zero?

the solution for arrays is

y = numpy.array(x)
y[0:n] = 0

In other words if we are filling a slice (range of indices) with the same number we can specify a scalar. It could be an array of the same size, e.g. np.ones(n). But it doesn't have to be.

So we just need to iterate over the rows of a (and elements of b) and perform this indexed assignment

In [368]: a = np.ones((3,4))*np.nan
In [369]: for i in range(3):
     ...:     a[i,:b[i]] = 1
     ...:     
In [370]: a
Out[370]: 
array([[  1.,   1.,  nan,  nan],
       [  1.,   1.,   1.,  nan],
       [  1.,  nan,  nan,  nan]])

There are various ways of 'filling' the original array with nan. np.full does an np.empty followed by a copyto.

A variation on the row iteration is with for i,n in enumerate(a):.

Another good way of iterating in a coordinated sense is with zip.

In [371]: for i,x in zip(b,a):
     ...:     x[:i] = 1

This takes advantage of the fact that iteration on a 2d array iterates on its rows. So x is an 1d view of a and can be changed in-place.

But with a bit of indexing trickery, we don't even have to loop.

In [373]: a = np.full((3,4),np.nan)

In [375]: mask = np.array(b)[:,None]>np.arange(4)
In [376]: mask
Out[376]: 
array([[ True,  True, False, False],
       [ True,  True,  True, False],
       [ True, False, False, False]], dtype=bool)
In [377]: a[mask] = 1
In [378]: a
Out[378]: 
array([[  1.,   1.,  nan,  nan],
       [  1.,   1.,   1.,  nan],
       [  1.,  nan,  nan,  nan]])

This is a favorite of one of the top numpy posters, @Divakar.

Numpy: Fix array with rows of different lengths by filling the empty elements with zeros

It can be used to pad a list of lists. Speaking of padding, itertools has a handy tool, zip_longest (py3 name)

In [380]: np.array(list(itertools.zip_longest(*[np.ones(x).tolist() for x in b],fillvalue=np.nan))).T
Out[380]: 
array([[  1.,   1.,  nan],
       [  1.,   1.,   1.],
       [  1.,  nan,  nan]])

---

Your question should have specified what was wrong; what kinds of errors you got:

for i in w2:
    a[0:b[i]] = [1] * b[i] 
    a[i:] = np.ones((b[i]))

w2 is unspecified, but probably is range(3).

a[0:b[i]] is wrong because it specifies all rows, where as you are working on just one at a time. a[i:] specifies a range of rows as well.

Answer 2

You can do this via a loop. Initialize an array of nan values then loop through the list of first n's and set values to 1 according to the n for each row.

a = np.full((3, 4), np.nan)
b = [2, 3, 1]
for i, x in enumerate(b):
    a[i, :x] = 1

Answer 3

You can initialise you matrix using a list comprehension:

>>> import numpy as np
>>> b = [2, 3, 1]
>>> max_len = 4
>>> gen_array = lambda i: [1] * i + [np.NAN] * (max_len - i)
>>> np.matrix([gen_array(i) for i in b])

With detailed steps:

[1] * N will create an array of length N filled with 1:

>>> [1] * 3
[1, 1, 1]

You can concat array using +:

>>> [1, 2] + [3, 4]
[1, 2, 3, 4]

You just have to combine both [1] * X + [np.NAN] * (N - X) will create an array of N dimension filled with X 1

last one, list-comprehension:

[i for i in b]

is a "shortcut" (not really, but it is easier to understand) for:

a = []
for i in b:
    a.append(i)

Answer 4

import numpy as np
a = np.random.rand(3,4)  #Create matrix with random numbers (you can change this to np.empty or whatever you want.
b = [1, 2, 3] # Your 'b' list
for idr, row in enumerate(a): # Loop through the matrix by row
  a[idr,:b[idr]] = 1  # idr is the row index, here you change the row 'idr' from the column 0 to the column b[idr] that will be 0, 1 and 3
  a[idr,b[idr]:] = 'NaN'  # the values after that receive NaN
print(a) # Outputs matrix
#[[  1.  nan  nan  nan]
 [  1.   1.  nan  nan]
 [  1.   1.   1.  nan]]