How to fill an nd array with values from a 1d-array?

Question

The following is a real-world problem in numPy reduced to the essentials, just with smaller dimensions.

Let's say I want to create an n-dimensional array all with dimensions (10, 10, 100):

all = np.empty((10, 10, 100))

I also have a 1d array data, simulated here as

data = np.arange(0, 100)

for all i, j I now want to achieve that

all[i,j]=data

So I do:

all[:, :]=data

Of course that works.

But now I want to import data to all2 with shape (100, 10, 10). I could do that with

all2 = np.empty((100, 10, 10)) # new target to be populated
for i in range(100):
  for j in range(10):
    for k in range(10):
      all2[i, j, k]=data[i]

But is there an easier way to do this without looping? I would be surprised if it couldn't be done more elegantly, but I don't see how.

I'm not big on NumPy but you should have a look into the reshape functions, this may help! — Victor
– Victor, Commented Apr 18, 2024 at 8:41
all2[:] = data[:, None,None]. This puts a (100, 1,1) shape into the (100,10,10). — hpaulj
– hpaulj, Commented Apr 18, 2024 at 10:30

Julien · Accepted Answer · 2024-04-18 08:47:46Z

1

You can use transpose: all2.T[:] = data (note the second , : insn't necessary)

answered Apr 18, 2024 at 8:47

Julien

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6 Comments

MichaelW Over a year ago

most easy. I would never have thought of that

Julien Over a year ago

You can also define directly all = np.tile(data, (10,10,1)) and all2 = np.tile(data, (10,10,1)).T although using all as variable name is a terrible idea...

MichaelW Over a year ago

as I told - this is only a boiled-down version of my original problem

user2357112 Over a year ago

@MichaelW: NumPy automatically fills in as many trailing :s as necessary.

hpaulj Over a year ago

Assignment to the transposed view is faster than the tile (or repeat) approach. But what about all3 with a (10,100,10) shape?

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