You're not going to have very much fun with the giant arrays that you mentioned. But if you have more reasonably-sized matrices (small enough that the result can fit in memory), the best way to do this is with broadcasting.
import numpy as np
a = np.array(range(5, 10))
b = np.array(range(2, 6))
res = a[None,: ] - b[: , None]
print(res)
#[[3 4 5 6 7] #[2 3 4 5 6] #[1 2 3 4 5] #[0 1 2 3 4]]You can use np.ufunc.outer with np.subtract and then transpose:
a = np.array(range(5, 10)) b = np.array(range(2, 6)) res1 = np.subtract.outer(a, b).T
a = np.array(range(5, 10))
b = np.array(range(2, 6))
res1 = np.subtract.outer(a, b).T
res2 = a[None,: ] - b[: , None] assert np.array_equal(res1, res2)
I have two vectors, v1 and v2. I'd like to anycodings_python subtract each value of v2 from each value of anycodings_python v1 and store the results in another vector. anycodings_python I also would like to work with very large anycodings_python vectors (e.g. 1e6 size), so I think I should anycodings_python be using numpy for performance.,Getting value out of array of object parameter and storing into an empty state in react,Generating a string of values from a set of arrays in Javascript,You can use np.ufunc.outer with anycodings_bigdata np.subtract and then transpose:
Up until now I have:
import numpy
v1 = numpy.array(numpy.random.uniform(-1, 1, size = 1e2))
v2 = numpy.array(numpy.random.uniform(-1, 1, size = 1e2))
vdiff = []
for value in v1:
vdiff.extend([value - v2])You're not going to have very much fun anycodings_bigdata with the giant arrays that you anycodings_bigdata mentioned. But if you have more anycodings_bigdata reasonably-sized matrices (small enough anycodings_bigdata that the result can fit in memory), the anycodings_bigdata best way to do this is with anycodings_bigdata broadcasting.
import numpy as np
a = np.array(range(5, 10))
b = np.array(range(2, 6))
res = a[None,: ] - b[: , None]
print(res)
#[[3 4 5 6 7] #[2 3 4 5 6] #[1 2 3 4 5] #[0 1 2 3 4]]You can use np.ufunc.outer with anycodings_bigdata np.subtract and then transpose:
a = np.array(range(5, 10)) b = np.array(range(2, 6)) res1 = np.subtract.outer(a, b).T
a = np.array(range(5, 10))
b = np.array(range(2, 6))
res1 = np.subtract.outer(a, b).T
res2 = a[None,: ] - b[: , None] assert np.array_equal(res1, res2)
Last Updated : 19 Jul, 2022
Input: arr1[] = {
-1,
-2,
4,
-6,
5,
7
}
arr2[] = {
6,
3,
4,
0
}
x = 8
Output: 4 4
5 3
Input: arr1[] = {
1,
2,
4,
5,
7
}
arr2[] = {
5,
6,
3,
4,
8
}
x = 9
Output: 1 8
4 5
5 41 7 7 1 5 3 4 4
1 7 7 1 5 3 4 4
Found the elements.
6 2 4 4 6 2 7 1
A Naive approach is to simply run two loops and pick elements from both arrays. One by one check that both elements sum is equal to given value x or not. ,Given two unsorted arrays of distinct elements, the task is to find all pairs from both arrays whose sum is equal to x.,An Efficient solution of this problem is to hashing. Hash table is implemented using unordered_set in C++. We store all first array elements in hash table. For elements of second array, we subtract every element from x and check the result in hash table. If result is present, we print the element and key in hash (which is an element of first array). ,Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Examples:
Input: arr1[] = {
-1,
-2,
4,
-6,
5,
7
}
arr2[] = {
6,
3,
4,
0
}
x = 8
Output: 4 4
5 3
Input: arr1[] = {
1,
2,
4,
5,
7
}
arr2[] = {
5,
6,
3,
4,
8
}
x = 9
Output: 1 8
4 5
5 41 7 7 1 5 3 4 4
Tools for reading / writing array data to disk and working with memory-mapped files,As a simple example, suppose we wished to evaluate the function sqrt(x^2 + y^2) across a regular grid of values. The np.meshgrid function takes two 1D arrays and produces two 2D matrices corresponding to all pairs of (x, y) in the two arrays:,np.save and np.load are the two workhorse functions for efficiently saving and loading array data on disk. Arrays are saved by default in an uncompressed raw binary format with file extension .npy.,Standard mathematical functions for fast operations on entire arrays of data without having to write loops
In[83]: names = np.array(['Bob', 'Joe', 'Will', 'Bob', 'Will', 'Joe', 'Joe'])
In[84]: data = np.random.randn(7, 4)
In[85]: names
Out[85]:
array(['Bob', 'Joe', 'Will', 'Bob', 'Will', 'Joe', 'Joe'],
dtype = '|S4')
In[86]: data
Out[86]:
array([
[-0.048, 0.5433, -0.2349, 1.2792],
[-0.268, 0.5465, 0.0939, -2.0445],
[-0.047, -2.026, 0.7719, 0.3103],
[2.1452, 0.8799, -0.0523, 0.0672],
[-1.0023, -0.1698, 1.1503, 1.7289],
[0.1913, 0.4544, 0.4519, 0.5535],
[0.5994, 0.8174, -0.9297, -1.2564]
])“Vectorized” Operations: Optimized Computations on NumPy Arrays,Prescribe the use of NumPy’s vectorized functions for performing optimized numerical computations on arrays.,Compare the performance of a simple non-vectorized computation to a vectorized one.,Consider, for instance, the task of summing the integers 0-9,999 stored in an array. Calling NumPy’s sum function cues optimized C code to iterate over the integers in the array and tally the sum. np.sum is therefore a “vectorized” function. Let’s time how long it takes to compute this sum:
# Demonstrating the application of common
# mathematical operations to a NumPy array
>>>
import numpy as np
>>>
x = np.array([
[0., 1., 2.],
...[3., 4., 5.],
...[6., 7., 8.]
])
# `x ** 2`
squares each entry in the array `x` >>>
x ** 2
array([
[0., 1., 4.],
[9., 16., 25.],
[36., 49., 64.]
])
# `np.sqrt(x)`
takes the square - root
# of each entry in the array `x` >>>
np.sqrt(x)
array([
[0., 1., 1.41421356],
[1.73205081, 2., 2.23606798],
[2.44948974, 2.64575131, 2.82842712]
])
# Slices
return arrays, thus you can operate
# on these too.Add .5 to each entry in row - 0
# of `x` >>>
.5 + x[0,: ]
array([0.5, 1.5, 2.5])# Demonstrating mathematical operations between two arrays
>>>
x = np.array([
[0., 1., 2.],
...[3., 4., 5.],
...[6., 7., 8.]
])
>>>
y = np.array([
[-4., -3.5, -3.],
...[-2.5, -2., -1.5],
...[-1., -0.5, -0.]
])
# `x + y`
will add the corresponding entries of
# the arrays `x`
and `y` >>>
x + y
array([
[-4., -2.5, -1.],
[0.5, 2., 3.5],
[5., 6.5, 8.]
])
>>>
x * y
array([
[-0., -3.5, -6.],
[-7.5, -8., -7.5],
[-6., -3.5, -0.]
])# applying the sequential
function, `np.sum`
# on an array
>>>
x = np.array([
[0., 1., 2.],
...[3., 4., 5.],
...[6., 7., 8.]
])
# summing over all of the array 's entries >>>
np.sum(x)
36.0
# summing over the rows, within each column
# of the array >>>
np.sum(x, axis = 0)
array([9., 12., 15.])>>> import numpy as np # sum an array, using NumPy 's vectorized ' sum ' function >>> np.sum(np.arange(10000)) # takes 11 microseconds on my computer 49995000
# sum an array by explicitly looping over the array in Python # this takes 822 microseconds on my computer >>> total = 0 >>> for i in np.arange(10000): ...total = i + total >>> total 49995000
>>>
import numpy as np
# multiply 2 with each number in the array >>>
2 * np.array([2, 3, 4])
array([4, 6, 8])
# subtract the corresponding entries of the two arrays >>>
np.array([10.2, 3.5, -0.9]) - np.array([8.2, 3.5, 6.5])
array([2., 0., -7.4])
# Take the "dot product" of the two arrays
# "dot product"
means: multiply their corresponding entries and sum the result >>>
np.dot(np.array([1, -3, 4]), np.array([2, 0, 1]))
6