ChildishJack - 5 months ago 10

Python Question

So, I have a snippet of a script:

`lol = []`

latv1 = 0

latv2 = 0

latv3 = 0

#Loop a

for a in range(100):

#Refresh latv2 after each iteration of loop a

latv2 = 0

#Loop b

for b in range(100):

#Refresh latv3 after each iteration of loop b

latv3 = 0

#Loop c

for c in range(100):

#Make 4 value list according to iteration and append to lol

midl2 = [latv1,latv2,latv3,0]

lol.append(midl2)

#Iterate after loop

latv3 = latv3 + 1

latv2 = latv2 + 1

latv1 = latv1 + 1

Which will do what I want it to do.... but very slowly. It gives:

`[[0,0,0,0]`

[0,0,1,0]

...

[0,1,0,0]

[0,1,1,0]

...

[9,9,8,0]

[9,9,9,0]]

I've read about numpy and its speed and optimization. I cannot figure out how to implement with numpy what I have above. I've learned how to make an array of zeroes with numpy via the manuals:

`numpy_array = np.zeroes((100,4))`

To give:

`[[ 0. 0. 0. 0.]`

[ 0. 0. 0. 0.]

[ 0. 0. 0. 0.]

...,

[ 0. 0. 0. 0.]

[ 0. 0. 0. 0.]

[ 0. 0. 0. 0.]]

and can change the values of each column with:

`numpA = np.arange(0,100,1)`

numpB = np.arange(0,100,1

numpC = np.arange(0,100,1)

numArr[:,0] = numpA

numArr[:,1] = numpB

numArr[:,2] = numpC

giving:

`[[ 0. 0. 0. 0.]`

[ 1. 1. 1. 0.]

[ 2. 2. 2. 0.]

...,

[ 997. 997. 997. 0.]

[ 998. 998. 998. 0.]

[ 999. 999. 999. 0.]]

but I cannot create a numpy array 1000000 lines long and have the columns increment like the original example did. If I call the zero array creation with 1000000 instead of 100 the column substitution does not work, which makes sense as the length of the array and the substitution are unequal - but I am not sure how to correctly iterate the substitution arrays to work.

How can I replicate the original scripts output via numpy arrays?

Note: This is a python 2.7 machine, but it's 64 bit at least. I know RAM use is an issue, but I should be able to change the dtype of the array to fit my needs.

Answer

**Approach #1**

To create the NumPy equivalent of the posted code and have NumPy array as output, you could additionally make use of `itertools`

, like so -

```
from itertools import product
out = np.zeros((N**3,4),dtype=int)
out[:,:3] = list(product(np.arange(N), repeat=3))
```

Please note that it would be `N = 100`

to make it equivalent to the posted code.

**Approach #2**

Another potentially faster approach based on purely NumPy and using it's vectorized `broadcasting`

capabilities could be suggested like so -

```
out = np.zeros((N**3,4),dtype=int)
out[:,:3] = (np.arange(N**3)[:,None]/[N**2,N,1])%N
```

I would think this to be faster than the previous `itertools`

based one, because that created a list of tuples that are to be set into a NumPy array. We will test this theory out in the next section.

**Runtime test**

```
In [111]: def itertools_based(N):
...: out = np.zeros((N**3,4),dtype=int)
...: out[:,:3] = list(product(np.arange(N), repeat=3))
...: return out
...:
...: def broadcasting_based(N):
...: out = np.zeros((N**3,4),dtype=int)
...: out[:,:3] = (np.arange(N**3)[:,None]/[N**2,N,1])%N
...: return out
In [112]: N = 20
In [113]: np.allclose(itertools_based(N),broadcasting_based(N)) # Verify results
Out[113]: True
In [114]: %timeit itertools_based(N)
100 loops, best of 3: 7.42 ms per loop
In [115]: %timeit broadcasting_based(N)
1000 loops, best of 3: 1.23 ms per loop
```

Now, let's time just the creation of list of tuples of those iterated elements and put it against the NumPy based one -

```
In [116]: %timeit list(product(np.arange(N), repeat=3))
1000 loops, best of 3: 746 µs per loop
In [117]: %timeit (np.arange(N**3)[:,None]/[N**2,N,1])%N
1000 loops, best of 3: 1.09 ms per loop
```

Well, so the creation part for the `itertools-based`

one is faster now, as predicted/thought out earlier! So, if you are happy with the first three columns as output and them being list of tuples, then go with `itertools`

.

Source (Stackoverflow)

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