Based on this StackOverflow answer:
NumPy does not support jagged arrays natively. gives an array that may or may not behave as you expect.
A workaround using masked arrays can be as follows:
import numpy as np
import numpy.ma as ma
a = np.array([0, 1])
b = np.array([2, 3, 4, 5])
c = np.array([6, 7, 8, 9, 10, 11])
jagged_array = ma.vstack(
[
ma.array(np.resize(a, c.shape[0]), mask=[False, False, True, True, True, True]),
ma.array(
np.resize(b, c.shape[0]), mask=[False, False, False, False, True, True]
),
c,
]
)
print(jagged_array)
print(jagged_array.ndim)
print(jagged_array.shape)
Your output would look like:
❯ python3 sample.py
[[0 1 -- -- -- --]
[2 3 4 5 -- --]
[6 7 8 9 10 11]]
2
(3, 6)
Answer from user4109800 on Stack OverflowBased on this StackOverflow answer:
NumPy does not support jagged arrays natively. gives an array that may or may not behave as you expect.
A workaround using masked arrays can be as follows:
import numpy as np
import numpy.ma as ma
a = np.array([0, 1])
b = np.array([2, 3, 4, 5])
c = np.array([6, 7, 8, 9, 10, 11])
jagged_array = ma.vstack(
[
ma.array(np.resize(a, c.shape[0]), mask=[False, False, True, True, True, True]),
ma.array(
np.resize(b, c.shape[0]), mask=[False, False, False, False, True, True]
),
c,
]
)
print(jagged_array)
print(jagged_array.ndim)
print(jagged_array.shape)
Your output would look like:
❯ python3 sample.py
[[0 1 -- -- -- --]
[2 3 4 5 -- --]
[6 7 8 9 10 11]]
2
(3, 6)
def ndim(arr):
return len(arr)-1
jagged_array = np.array([[None, None], [None, None, None, None], [None, None, None,None, None, None]])
print(jagged_array)
print(ndim(jagged_array))
print(jagged_array.shape)
RDataFrame -> AsNumpy as jagged arrays
c++ - Jagged Numpy Arrays in Boost Numpy - Stack Overflow
Awkward: Nested, jagged, differentiable, mixed type, GPU-enabled, JIT'd NumPy
python - Convert jagged lists into numpy array - Stack Overflow
Short answer: you can't. NumPy does not support jagged arrays natively.
Long answer:
>>> a = ones((3,))
>>> b = ones((2,))
>>> c = array([a, b])
>>> c
array([[ 1. 1. 1.], [ 1. 1.]], dtype=object)
gives an array that may or may not behave as you expect. E.g. it doesn't support basic methods like sum or reshape, and you should treat this much as you'd treat the ordinary Python list [a, b] (iterate over it to perform operations instead of using vectorized idioms).
Several possible workarounds exist; the easiest is to coerce a and b to a common length, perhaps using masked arrays or NaN to signal that some indices are invalid in some rows. E.g. here's b as a masked array:
>>> ma.array(np.resize(b, a.shape[0]), mask=[False, False, True])
masked_array(data = [1.0 1.0 --],
mask = [False False True],
fill_value = 1e+20)
This can be stacked with a as follows:
>>> ma.vstack([a, ma.array(np.resize(b, a.shape[0]), mask=[False, False, True])])
masked_array(data =
[[1.0 1.0 1.0]
[1.0 1.0 --]],
mask =
[[False False False]
[False False True]],
fill_value = 1e+20)
(For some purposes, scipy.sparse may also be interesting.)
In general, there is an ambiguity in putting together arrays of different length because alignment of data might matter. Pandas has different advanced solutions to deal with that, e.g. to merge series into dataFrames.
If you just want to populate columns starting from first element, what I usually do is build a matrix and populate columns. Of course you need to fill the empty spaces in the matrix with a null value (in this case np.nan)
a = ones((3,))
b = ones((2,))
arraylist=[a,b]
outarr=np.ones((np.max([len(ps) for ps in arraylist]),len(arraylist)))*np.nan #define empty array
for i,c in enumerate(arraylist): #populate columns
outarr[:len(c),i]=c
In [108]: outarr
Out[108]:
array([[ 1., 1.],
[ 1., 1.],
[ 1., nan]])
Your array is 2x2:
In [298]: A
Out[298]:
array([[array([1, 2, 3]), array([4, 5])],
[array([6, 7, 8, 9]), array([10])]], dtype=object)
While A+A works, boolean tests have not been implemented for this kind of array:
In [299]: A>4
...
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
I'm going to flatten A because it makes it easier to compare with list operations:
In [301]: A1=A.flatten()
In [303]: A1+A1
Out[303]:
array([array([2, 4, 6]), array([ 8, 10]), array([12, 14, 16, 18]),
array([20])], dtype=object)
In [304]: [a+a for a in A1]
Out[304]: [array([2, 4, 6]), array([ 8, 10]), array([12, 14, 16, 18]), array([20])]
In [305]: timeit A1+A1
100000 loops, best of 3: 6.85 µs per loop
In [306]: timeit [a+a for a in A1]
100000 loops, best of 3: 9.09 µs per loop
The array operation is a bit faster than a list comprehension. But if I first turn the array into a list:
In [307]: A1l=A1.tolist()
In [308]: A1l
Out[308]: [array([1, 2, 3]), array([4, 5]), array([6, 7, 8, 9]), array([10])]
In [309]: timeit [a+a for a in A1l]
100000 loops, best of 3: 5.2 µs per loop
times improve. This is a good indication that the A1+A1 (or even A+A) is using a similar sort of iteration.
So the straight forward way of performing your A,B calculation is
In [310]: A2=[a[a>4] for a in A1]
In [311]: B=[a+a for a in A2]
In [312]: B
Out[312]: [array([], dtype=int32), array([10]), array([12, 14, 16, 18]), array([20])]
(we can convert to/from arrays and lists as needed).
A numpy array stores its data a flat databuffer, and uses the shape and strides attributes to quickly calculate the location of any element, regardless of the dimensions. The fast array operations use compiled code that rapidly steps though the databuffers of arguments, performing the operations element by element (or some other combination).
A dtype object array also has the flat databuffer, but the elements are pointers to lists or arrays elsewhere. So while it can index individual elements quickly, it still has to perform a Python call(s) to access the arrays. So especially when the array is 1d, it is virtually the same as a flat list with the same pointers.
Multidimensional object arrays are nicer than nested lists. You can reshape them, access elements (A[1,3] v Al[1][3]), transpose them, etc. But when it comes to iterating through all the subarrays they don't offer much of a benefit.
Looking again at your 2d array:
In [315]: timeit A+A
100000 loops, best of 3: 6.93 µs per loop # 6.85 for A1+A1 (above)
In [316]: timeit [[j+j for j in i] for i in A]
100000 loops, best of 3: 17.1 µs per loop
In [317]: Al = A.tolist()
In [318]: timeit [[j+j for j in i] for i in Al]
100000 loops, best of 3: 7.01 µs per loop # 5.2 for A1l flat list
Basically the same time for summing the array and iterating through the equivalent nested list.
The performance of numpy jagged array may not be optimal, but there are enough reasons to believe that it should be much better than using python nested list. As explained in your earlier post:
On principle you should have some performance bonus because every element is a numpy array. So you just need a 2 dimensional loop rather than a 3D loop (if you store every number in nested lists). Also it always saves you lots of memory allocation time to avoid using python list.
Here is a simple test:
import time,sys,random
import numpy as np
rand = np.random.rand
L = np.array([[rand(100), rand(200)],[rand(400), rand(300)]], dtype=object)
L1 = [random.random() for i in range(1000)]
arrFunc = np.vectorize(lambda x:x[x>0.3],otypes=[np.ndarray])
start = time.time()
if sys.argv[1]=='np':
for i in range(100000):
B=i*L
else:
for i in range(100000):
B=[i*x for x in L1]
end = time.time()
print ('Arithmetic Op: ', end-start)
start = time.time()
if sys.argv[1]=='np':
for i in range(100000):
B=arrFunc(L)
else:
for i in range(100000):
B=[x for x in L1 if x<0.3]
end = time.time()
print ('Indexing ', end-start)
Result:
> python testNpJarray.py np
Arithmetic Op: 3.9719998836517334
Indexing 8.079999923706055
> python testNpJarray.py list
Arithmetic Op: 53.289000034332275
Indexing 52.10899996757507
This test may not be quite fare because the outter numpy array is quite small, you are welcome to change the size to fit into your application and tell us the results.
You can transform your initial array of iterable-objects to ndarray by padding them with zeros in a vectorized manner:
import numpy as np
a = np.array([[1, 2, 3, 4],
[2, 3, 4],
[4, 5, 6]])
max_len = len(max(a, key = lambda x: len(x))) # max length of iterable-objects contained in array
cust_func = np.vectorize(pyfunc=lambda x: np.pad(array=x,
pad_width=(0,max_len),
mode='constant',
constant_values=(0,0))[:max_len], otypes=[list])
a_pad = np.stack(cust_func(a))
output:
array([[1, 2, 3, 4],
[2, 3, 4, 0],
[4, 5, 6, 0]])
It depends. Do you know the size of the vectors before or are you appending to a list?
see e.g. http://stackoverflow.com/a/58085045/7919597
You could for example pad the arrays
import numpy as np
a1 = [1, 2, 3, 4]
a2 = [2, 3, 4, np.nan] # pad with nan
a3 = [4, 5, 6, np.nan] # pad with nan
b = np.stack([a1, a2, a3], axis=0)
print(b)
# you can apply the normal numpy operations on
# arrays with nan, they usually just result in a nan
# in a resulting array
c = np.diff(b, axis=-1)
print(c)
Afterwards you can apply a moving window on each row over the columns.
Have a look at https://stackoverflow.com/a/22621523/7919597 which is only 1d, but can give you an idea of how it could work.
It is possible to use a 2d array with only one row as kernel (shape e.g. (1, 3)) with scipy.signal.convolve2d and use the idea above. This is a workaround to get a "row-wise 1D convolution":
from scipy import signal
krnl = np.array([[0, 1, 0]])
d = signal.convolve2d(c, krnl, mode='same')
print(d)