Is there an effective way in which to compare all three numpy arrays at once?
For example, if the given value to check is 5, wherever the value is not 5, it should be not 5 for all three arrays.
The only way I've thought of how to do this would be checking that occurrences that arr1 != 5 & arr2 == 5 is 0. However this only checks one direction between the two arrays, and then I need to also incorporate arr3. This seems inefficient and might end up with some logical hole.
This should pass:
arr1 = numpy.array([[1, 7, 3],
[4, 5, 6],
[4, 5, 2]])
arr2 = numpy.array([[1, 2, 3],
[4, 5, 6],
[8, 5, 6]])
arr3 = numpy.array([[1, 1, 3],
[4, 5, 6],
[9, 5, 6]])
However this should fail due to arr2 having a 3 where other arrays have 5s
arr1 = numpy.array([[1, 2, 3],
[8, 5, 6],
[4, 5, 6]])
arr2 = numpy.array([[1, 2, 3],
[2, 3, 1],
[2, 5, 6]])
arr3 = numpy.array([[1, 2, 3],
[4, 5, 6],
[4, 5, 3]])
There is a general solution (regardless number of arrays). And it's quite educational:
import numpy as np #a recommended way of import
arr = np.array([arr1, arr2, arr3])
is_valid = np.all(arr==5, axis=0) == np.any(arr==5, axis=0) #introduce axis
out = np.all(is_valid)
#True for the first case, False for the second one
Is this a valid solution?
numpy.logical_and(((arr1==5)==(arr2==5)).all(), ((arr2==5)==(arr3==5)).all())
You could AND all comparisons to 5 and compare to any one of the comparisons:
A = (arr1==5)
(A==(A&(arr2==5)&(arr3==5))).all()
Output: True for the first example, False for the second
NB. This works for any number of arrays
If i have a nested array lets say:
arr = [[1, 2, 3, 4], [5, 6, 7, 8]]
I want to divide element wise so my output would be:
[5/1, 6/2, 7/3, 8/4]
Just using fractions to be clear on what i'm asking. Thank you
Try to use the zip() function:
d=[] #This is done to avoid name 'd' is not defined
arr = [[1, 2, 3, 4], [5, 6, 7, 8]]
zipped = zip(arr[1], arr[0])
for i1,i2 in zipped:
d.append(i1/i2)
You can easily do this with numpy.
Extract the second row, and divide it by the first row element wise:
arr = np.array([[1, 2, 3, 4], [5, 6, 7, 8]])
np.array(arr[1, :] / arr[0, :])
# [5. 3. 2.33333333 2. ]
If instead you want to do it with a for loop:
[arr[1][i] / arr[0][i] for i in range(len(arr[0]))]
Basically looking for a similar solution as posted here, but then in python. R: How to subtract every n-th column from the ones before it in a matrix/data-frame?
Some data
import numpy as np
m = np.matrix([[1, 2, 3, 5], [3, 4, 5, 2], [5, 6, 7, 2]])
I want to subtract the value in every 2 columns from the value of the column before. So the solution that I want to end up with two columns. Each row in the first column containing -1 and the second column as [-2, 3, 5]
Thanks in advance!
You can do :
import numpy as np
m = np.matrix([[1, 2, 3, 5], [3, 4, 5, 2], [5, 6, 7, 2]])
a = m[:,0] - m[:,1]
b = m[:,2] - m[:,3]
m2 = np.concatenate((a, b), axis=1)
print(m2)
And for n column :
import numpy as np
m = np.matrix([[1, 2, 3, 5, 5 , 6], [3, 4, 5, 2, 3, 2], [5, 6, 7, 2, 7, 1]])
shape = np.shape(m)
print(shape)
result = []
if(shape[1] % 2 == 0):
for i in range(0, shape[1], 2):
print(i)
result.append(m[:,i] - m[:,i+1])
m2 = result[0]
for i in range(1, len(result)):
m2 = np.concatenate((m2, result[i]), axis=1)
print(m2)
I have numpy 2d array having duplicate values.
I am searching the array like this.
In [104]: import numpy as np
In [105]: array = np.array
In [106]: a = array([[1, 2, 3],
...: [1, 2, 3],
...: [2, 5, 6],
...: [3, 8, 9],
...: [4, 8, 9],
...: [4, 2, 3],
...: [5, 2, 3])
In [107]: num_list = [1, 4, 5]
In [108]: for i in num_list :
...: print(a[np.where(a[:,0] == num_list)])
...:
[[1 2 3]
[1 2 3]]
[[4 8 9]
[4 2 3]]
[[5 2 3]]
The input is list having number similar to column 0 values.
The end result I want is the resulting rows in any format like array, list or tuple for example
array([[1, 2, 3],
[1, 2, 3],
[4, 8, 9],
[4, 2, 3],
[5, 2, 3]])
My code works fine but doesn't seem pythonic. Is there any better searching strategy with multiple values?
like a[np.where(a[:,0] == l)] where only one time lookup is done to get all the values.
my real array is large
Approach #1 : Using np.in1d -
a[np.in1d(a[:,0], num_list)]
Approach #2 : Using np.searchsorted -
num_arr = np.sort(num_list) # Sort num_list and get as array
# Get indices of occurrences of first column in num_list
idx = np.searchsorted(num_arr, a[:,0])
# Take care of out of bounds cases
idx[idx==len(num_arr)] = 0
out = a[a[:,0] == num_arr[idx]]
You can do
a[numpy.in1d(a[:, 0], num_list), :]
I have an array of values that I want to replace with from an array of choices based on which choice is linearly closest.
The catch is the size of the choices is defined at runtime.
import numpy as np
a = np.array([[0, 0, 0], [4, 4, 4], [9, 9, 9]])
choices = np.array([1, 5, 10])
If choices was static in size, I would simply use np.where
d = np.where(np.abs(a - choices[0]) > np.abs(a - choices[1]),
np.where(np.abs(a - choices[0]) > np.abs(a - choices[2]), choices[0], choices[2]),
np.where(np.abs(a - choices[1]) > np.abs(a - choices[2]), choices[1], choices[2]))
To get the output:
>>d
>>[[1, 1, 1], [5, 5, 5], [10, 10, 10]]
Is there a way to do this more dynamically while still preserving the vectorization.
Subtract choices from a, find the index of the minimum of the result, substitute.
a = np.array([[0, 0, 0], [4, 4, 4], [9, 9, 9]])
choices = np.array([1, 5, 10])
b = a[:,:,None] - choices
np.absolute(b,b)
i = np.argmin(b, axis = -1)
a = choices[i]
print a
>>>
[[ 1 1 1]
[ 5 5 5]
[10 10 10]]
a = np.array([[0, 3, 0], [4, 8, 4], [9, 1, 9]])
choices = np.array([1, 5, 10])
b = a[:,:,None] - choices
np.absolute(b,b)
i = np.argmin(b, axis = -1)
a = choices[i]
print a
>>>
[[ 1 1 1]
[ 5 10 5]
[10 1 10]]
>>>
The extra dimension was added to a so that each element of choices would be subtracted from each element of a. choices was broadcast against a in the third dimension, This link has a decent graphic. b.shape is (3,3,3). EricsBroadcastingDoc is a pretty good explanation and has a graphic 3-d example at the end.
For the second example:
>>> print b
[[[ 1 5 10]
[ 2 2 7]
[ 1 5 10]]
[[ 3 1 6]
[ 7 3 2]
[ 3 1 6]]
[[ 8 4 1]
[ 0 4 9]
[ 8 4 1]]]
>>> print i
[[0 0 0]
[1 2 1]
[2 0 2]]
>>>
The final assignment uses an Index Array or Integer Array Indexing.
In the second example, notice that there was a tie for element a[0,1] , either one or five could have been substituted.
To explain wwii's excellent answer in a little more detail:
The idea is to create a new dimension which does the job of comparing each element of a to each element in choices using numpy broadcasting. This is easily done for an arbitrary number of dimensions in a using the ellipsis syntax:
>>> b = np.abs(a[..., np.newaxis] - choices)
array([[[ 1, 5, 10],
[ 1, 5, 10],
[ 1, 5, 10]],
[[ 3, 1, 6],
[ 3, 1, 6],
[ 3, 1, 6]],
[[ 8, 4, 1],
[ 8, 4, 1],
[ 8, 4, 1]]])
Taking argmin along the axis you just created (the last axis, with label -1) gives you the desired index in choices that you want to substitute:
>>> np.argmin(b, axis=-1)
array([[0, 0, 0],
[1, 1, 1],
[2, 2, 2]])
Which finally allows you to choose those elements from choices:
>>> d = choices[np.argmin(b, axis=-1)]
>>> d
array([[ 1, 1, 1],
[ 5, 5, 5],
[10, 10, 10]])
For a non-symmetric shape:
Let's say a had shape (2, 5):
>>> a = np.arange(10).reshape((2, 5))
>>> a
array([[0, 1, 2, 3, 4],
[5, 6, 7, 8, 9]])
Then you'd get:
>>> b = np.abs(a[..., np.newaxis] - choices)
>>> b
array([[[ 1, 5, 10],
[ 0, 4, 9],
[ 1, 3, 8],
[ 2, 2, 7],
[ 3, 1, 6]],
[[ 4, 0, 5],
[ 5, 1, 4],
[ 6, 2, 3],
[ 7, 3, 2],
[ 8, 4, 1]]])
This is hard to read, but what it's saying is, b has shape:
>>> b.shape
(2, 5, 3)
The first two dimensions came from the shape of a, which is also (2, 5). The last dimension is the one you just created. To get a better idea:
>>> b[:, :, 0] # = abs(a - 1)
array([[1, 0, 1, 2, 3],
[4, 5, 6, 7, 8]])
>>> b[:, :, 1] # = abs(a - 5)
array([[5, 4, 3, 2, 1],
[0, 1, 2, 3, 4]])
>>> b[:, :, 2] # = abs(a - 10)
array([[10, 9, 8, 7, 6],
[ 5, 4, 3, 2, 1]])
Note how b[:, :, i] is the absolute difference between a and choices[i], for each i = 1, 2, 3.
Hope that helps explain this a little more clearly.
I love broadcasting and would have gone that way myself too. But, with large arrays, I would like to suggest another approach with np.searchsorted that keeps it memory efficient and thus achieves performance benefits, like so -
def searchsorted_app(a, choices):
lidx = np.searchsorted(choices, a, 'left').clip(max=choices.size-1)
ridx = (np.searchsorted(choices, a, 'right')-1).clip(min=0)
cl = np.take(choices,lidx) # Or choices[lidx]
cr = np.take(choices,ridx) # Or choices[ridx]
mask = np.abs(a - cl) > np.abs(a - cr)
cl[mask] = cr[mask]
return cl
Please note that if the elements in choices are not sorted, we need to add in the additional argument sorter with np.searchsorted.
Runtime test -
In [160]: # Setup inputs
...: a = np.random.rand(100,100)
...: choices = np.sort(np.random.rand(100))
...:
In [161]: def broadcasting_app(a, choices): # #wwii's solution
...: return choices[np.argmin(np.abs(a[:,:,None] - choices),-1)]
...:
In [162]: np.allclose(broadcasting_app(a,choices),searchsorted_app(a,choices))
Out[162]: True
In [163]: %timeit broadcasting_app(a, choices)
100 loops, best of 3: 9.3 ms per loop
In [164]: %timeit searchsorted_app(a, choices)
1000 loops, best of 3: 1.78 ms per loop
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