Replace zero array with new values one by one NumPy

Question

I stuck with a simple question in NumPy. I have an array of zero values. Once I generate a new value I would like to add it one by one.

arr=array([0,0,0])
# something like this
l=[1,5,10]
for x in l:
     arr.append(x)   # from python logic

so I would like to add one by one x into array, so I would get: 1st iteration arr=([1,0,0]); 2d iteration arr=([1,5,0]); 3rd arr=([1,5,10]);

Basically I need to substitute zeros with new values one by one in NumPy (I am learning NumPy!!!!!!). I checked many of NumPy options like np.append (it adds to existing values new values), but can't find the right.

thank you

mtzl · Accepted Answer · 2016-04-21 13:00:56Z

1

There are a few things to pick up with numpy:

you can generate the array full of zeros with
```
>>> np.zeros(3)
array([ 0.,  0.,  0.])
```

You can get/set array elements with indexing as with lists etc:

arr[2] = 7

for i, val in enumerate([1, 5, 10]):
    arr[i] = val

Or, if you want to fill with array with something like a list, you can directly use:
```
>>> np.array([1, 5, 10])
array([ 1,  5, 10])
```
Also, numpy's signature for appending stuff to an array is a bit different:
```
arr = np.append(arr, 7)
```

Having said that, you should just consider diving into Numpy's own userguide.

edited Apr 21, 2016 at 13:00

answered Apr 21, 2016 at 12:49

mtzl

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