numpy.allclose () in Python

allclose | Arrays | NumPy | Python Methods and Functions

If the following equation is elementwise True, then allclose returns True. 
absolute (arr1 - arr2) & lt; = (atol + rtol * absolute (arr2))
Since the above equation is not symmetric in arr1 and arr2, in some rare cases allclose (arr1, arr2) may differ from allclose (arr2, arr1).

Syntax: numpy.allclose (arr1, arr2, rtol, atol, equal_nan = False)

Parameters:
arr1: [array_like] Input 1st array.
arr2: [array_like] Input 2nd array.
rtol: [float] The relative tolerance parameter.
atol: [float] The absolute tolerance parameter.
equal_nan: [bool] Whether to compare NaN`s as equal. If True, NaN`s in arr1 will be considered equal to NaN`s in arr2 in the output array.

Return: [bool] Returns True if the two arrays are equal within the given tolerance , otherwise it returns False.

Code # 1:

# Python program, explaining
# allclose () function

 

import numpy as geek

  
# input arrays

in_arr1 = geek.array ([ 5e5 , 1e - 7 , 4.000004e6 ])

print ( "1st Input array:" , in_arr1) 

 

in_arr2 = geek.array ([ 5.00001e5 , 1e - 7 , 4e6 ])

print ( "2nd Input array:" , in_arr2) 

 
# setting absolute and relative tolerance

rtol =   1e - 05

atol = 1e - 08

 

res = geek.allclose (in_arr1, in_arr2, rtol, atol)

print ( " Are the two arrays are equal within the tolerance: " , res)

Exit:

 1st Input array: [5.00000000e + 05 1.00000000e-07 4.00000400e + 06] 2nd Input array: [5.00001000e + 05 1.00000000e-07 4.00000000e + 06] Are the two arrays are equal within the tolerance: True 

Code # 2:

# Python program explaining
# allclose () function

 

import numpy as geek

 
# input arrays

in_arr1 = geek.array ([ 5e5 , 1e - 7 , 4.000004e6 ])

print ( "1st Input array:" , in_arr1) 

 

in_arr2 = < code class = "plain"> geek.array ([ 5.00001e5 , 1e - 7 , 4e6 ])

print ( "2nd Input array:" , in_arr2) 

  
# setting absolute and relative tolerance

rtol = 1e - 06

atol = 1e - 09

 

res =   geek.allclose (in_arr1, in_arr2, rtol, atol)

print ( " Are the two arrays are equal within the tolerance: " , res)

Exit:

 1st Input array: [5000000.0, 1e-07, 40000004.0] 2nd Input array: [5000001.0, 1e-07, 40000000.0] Are the two arrays are equal within the tolerance: True 

Code # 3:

# Python program explaining
# allclose () function

 

import numpy as geek

 
# input arrays

i n_arr1 = geek.array ([ 5e5 , 1e - 7 , geek.nan])

print ( "1st Input array:" , in_arr1) 

 

in_arr2 = geek.array ([ 5e5 , 1e - 7 , geek.nan])

print ( " 2nd Input array: " , in_arr2) 

 
# setting absolute and relative tolerance

rtol = 1e - 06

atol = 1e - 09

  

res = geek.allclose (in_arr1, in_arr2, rtol, atol)

print ( "Are the two arrays are equal within the tolerance:" , res)

Exit:

 1st Input array: [500000.0, 1e-07, nan] 2nd Input array: [500000.0, 1e-07, nan] Are the two arrays are equal within the tolerance: False 

Code # 4:

 

# Python program explaining
# allclose () function

 

import numpy as geek

 
# input arrays

in_arr1 = geek.array ([ 5e5 , 1e - 7 , geek.nan])

print ( "1st Input array:" , in_arr1) 

 

in_arr2 = geek.array ([ 5e5 , 1e - 7 , geek.nan])

print ( " 2nd Input array: " , in_arr2) 

  
# setting absolute and relative tolerance

rtol = 1e - 06

atol = 1e - 09

 

res = geek.all close (in_arr1, in_arr2, rtol, atol, 

equal_nan = True )

 

print ( " Are the two arrays are equal within the tolerance: " , res)

Output :

 1st Input array: [500000.0, 1e-07, nan] 2nd Input array: [500000.0, 1e-07, nan] Are the two arrays are equal within the tolerance: True 




numpy.allclose () in Python: StackOverflow Questions

Answer #1

Update: The example below shows the old pandas.rolling_mean function which has been removed in recent versions of pandas. A modern equivalent of the function call below would be

In [8]: pd.Series(x).rolling(window=N).mean().iloc[N-1:].values
Out[8]: 
array([ 0.49815397,  0.49844183,  0.49840518, ...,  0.49488191,
        0.49456679,  0.49427121])

pandas is more suitable for this than NumPy or SciPy. Its function rolling_mean does the job conveniently. It also returns a NumPy array when the input is an array.

It is difficult to beat rolling_mean in performance with any custom pure Python implementation. Here is an example performance against two of the proposed solutions:

In [1]: import numpy as np

In [2]: import pandas as pd

In [3]: def running_mean(x, N):
   ...:     cumsum = np.cumsum(np.insert(x, 0, 0)) 
   ...:     return (cumsum[N:] - cumsum[:-N]) / N
   ...:

In [4]: x = np.random.random(100000)

In [5]: N = 1000

In [6]: %timeit np.convolve(x, np.ones((N,))/N, mode="valid")
10 loops, best of 3: 172 ms per loop

In [7]: %timeit running_mean(x, N)
100 loops, best of 3: 6.72 ms per loop

In [8]: %timeit pd.rolling_mean(x, N)[N-1:]
100 loops, best of 3: 4.74 ms per loop

In [9]: np.allclose(pd.rolling_mean(x, N)[N-1:], running_mean(x, N))
Out[9]: True

There are also nice options as to how to deal with the edge values.

Answer #2

Grasping the idea of numpy.einsum() is very easy if you understand it intuitively. As an example, let"s start with a simple description involving matrix multiplication.


To use numpy.einsum(), all you have to do is to pass the so-called subscripts string as an argument, followed by your input arrays.

Let"s say you have two 2D arrays, A and B, and you want to do matrix multiplication. So, you do:

np.einsum("ij, jk -> ik", A, B)

Here the subscript string ij corresponds to array A while the subscript string jk corresponds to array B. Also, the most important thing to note here is that the number of characters in each subscript string must match the dimensions of the array. (i.e. two chars for 2D arrays, three chars for 3D arrays, and so on.) And if you repeat the chars between subscript strings (j in our case), then that means you want the einsum to happen along those dimensions. Thus, they will be sum-reduced. (i.e. that dimension will be gone)

The subscript string after this ->, will be our resultant array. If you leave it empty, then everything will be summed and a scalar value is returned as result. Else the resultant array will have dimensions according to the subscript string. In our example, it"ll be ik. This is intuitive because we know that for matrix multiplication the number of columns in array A has to match the number of rows in array B which is what is happening here (i.e. we encode this knowledge by repeating the char j in the subscript string)


Here are some more examples illustrating the use/power of np.einsum() in implementing some common tensor or nd-array operations, succinctly.

Inputs

# a vector
In [197]: vec
Out[197]: array([0, 1, 2, 3])

# an array
In [198]: A
Out[198]: 
array([[11, 12, 13, 14],
       [21, 22, 23, 24],
       [31, 32, 33, 34],
       [41, 42, 43, 44]])

# another array
In [199]: B
Out[199]: 
array([[1, 1, 1, 1],
       [2, 2, 2, 2],
       [3, 3, 3, 3],
       [4, 4, 4, 4]])

1) Matrix multiplication (similar to np.matmul(arr1, arr2))

In [200]: np.einsum("ij, jk -> ik", A, B)
Out[200]: 
array([[130, 130, 130, 130],
       [230, 230, 230, 230],
       [330, 330, 330, 330],
       [430, 430, 430, 430]])

2) Extract elements along the main-diagonal (similar to np.diag(arr))

In [202]: np.einsum("ii -> i", A)
Out[202]: array([11, 22, 33, 44])

3) Hadamard product (i.e. element-wise product of two arrays) (similar to arr1 * arr2)

In [203]: np.einsum("ij, ij -> ij", A, B)
Out[203]: 
array([[ 11,  12,  13,  14],
       [ 42,  44,  46,  48],
       [ 93,  96,  99, 102],
       [164, 168, 172, 176]])

4) Element-wise squaring (similar to np.square(arr) or arr ** 2)

In [210]: np.einsum("ij, ij -> ij", B, B)
Out[210]: 
array([[ 1,  1,  1,  1],
       [ 4,  4,  4,  4],
       [ 9,  9,  9,  9],
       [16, 16, 16, 16]])

5) Trace (i.e. sum of main-diagonal elements) (similar to np.trace(arr))

In [217]: np.einsum("ii -> ", A)
Out[217]: 110

6) Matrix transpose (similar to np.transpose(arr))

In [221]: np.einsum("ij -> ji", A)
Out[221]: 
array([[11, 21, 31, 41],
       [12, 22, 32, 42],
       [13, 23, 33, 43],
       [14, 24, 34, 44]])

7) Outer Product (of vectors) (similar to np.outer(vec1, vec2))

In [255]: np.einsum("i, j -> ij", vec, vec)
Out[255]: 
array([[0, 0, 0, 0],
       [0, 1, 2, 3],
       [0, 2, 4, 6],
       [0, 3, 6, 9]])

8) Inner Product (of vectors) (similar to np.inner(vec1, vec2))

In [256]: np.einsum("i, i -> ", vec, vec)
Out[256]: 14

9) Sum along axis 0 (similar to np.sum(arr, axis=0))

In [260]: np.einsum("ij -> j", B)
Out[260]: array([10, 10, 10, 10])

10) Sum along axis 1 (similar to np.sum(arr, axis=1))

In [261]: np.einsum("ij -> i", B)
Out[261]: array([ 4,  8, 12, 16])

11) Batch Matrix Multiplication

In [287]: BM = np.stack((A, B), axis=0)

In [288]: BM
Out[288]: 
array([[[11, 12, 13, 14],
        [21, 22, 23, 24],
        [31, 32, 33, 34],
        [41, 42, 43, 44]],

       [[ 1,  1,  1,  1],
        [ 2,  2,  2,  2],
        [ 3,  3,  3,  3],
        [ 4,  4,  4,  4]]])

In [289]: BM.shape
Out[289]: (2, 4, 4)

# batch matrix multiply using einsum
In [292]: BMM = np.einsum("bij, bjk -> bik", BM, BM)

In [293]: BMM
Out[293]: 
array([[[1350, 1400, 1450, 1500],
        [2390, 2480, 2570, 2660],
        [3430, 3560, 3690, 3820],
        [4470, 4640, 4810, 4980]],

       [[  10,   10,   10,   10],
        [  20,   20,   20,   20],
        [  30,   30,   30,   30],
        [  40,   40,   40,   40]]])

In [294]: BMM.shape
Out[294]: (2, 4, 4)

12) Sum along axis 2 (similar to np.sum(arr, axis=2))

In [330]: np.einsum("ijk -> ij", BM)
Out[330]: 
array([[ 50,  90, 130, 170],
       [  4,   8,  12,  16]])

13) Sum all the elements in array (similar to np.sum(arr))

In [335]: np.einsum("ijk -> ", BM)
Out[335]: 480

14) Sum over multiple axes (i.e. marginalization)
(similar to np.sum(arr, axis=(axis0, axis1, axis2, axis3, axis4, axis6, axis7)))

# 8D array
In [354]: R = np.random.standard_normal((3,5,4,6,8,2,7,9))

# marginalize out axis 5 (i.e. "n" here)
In [363]: esum = np.einsum("ijklmnop -> n", R)

# marginalize out axis 5 (i.e. sum over rest of the axes)
In [364]: nsum = np.sum(R, axis=(0,1,2,3,4,6,7))

In [365]: np.allclose(esum, nsum)
Out[365]: True

15) Double Dot Products (similar to np.sum(hadamard-product) cf. 3)

In [772]: A
Out[772]: 
array([[1, 2, 3],
       [4, 2, 2],
       [2, 3, 4]])

In [773]: B
Out[773]: 
array([[1, 4, 7],
       [2, 5, 8],
       [3, 6, 9]])

In [774]: np.einsum("ij, ij -> ", A, B)
Out[774]: 124

16) 2D and 3D array multiplication

Such a multiplication could be very useful when solving linear system of equations (Ax = b) where you want to verify the result.

# inputs
In [115]: A = np.random.rand(3,3)
In [116]: b = np.random.rand(3, 4, 5)

# solve for x
In [117]: x = np.linalg.solve(A, b.reshape(b.shape[0], -1)).reshape(b.shape)

# 2D and 3D array multiplication :)
In [118]: Ax = np.einsum("ij, jkl", A, x)

# indeed the same!
In [119]: np.allclose(Ax, b)
Out[119]: True

On the contrary, if one has to use np.matmul() for this verification, we have to do couple of reshape operations to achieve the same result like:

# reshape 3D array `x` to 2D, perform matmul
# then reshape the resultant array to 3D
In [123]: Ax_matmul = np.matmul(A, x.reshape(x.shape[0], -1)).reshape(x.shape)

# indeed correct!
In [124]: np.allclose(Ax, Ax_matmul)
Out[124]: True

Bonus: Read more math here : Einstein-Summation and definitely here: Tensor-Notation

Answer #3

(A==B).all()

test if all values of array (A==B) are True.

Note: maybe you also want to test A and B shape, such as A.shape == B.shape

Special cases and alternatives (from dbaupp"s answer and yoavram"s comment)

It should be noted that:

  • this solution can have a strange behavior in a particular case: if either A or B is empty and the other one contains a single element, then it return True. For some reason, the comparison A==B returns an empty array, for which the all operator returns True.
  • Another risk is if A and B don"t have the same shape and aren"t broadcastable, then this approach will raise an error.

In conclusion, if you have a doubt about A and B shape or simply want to be safe: use one of the specialized functions:

np.array_equal(A,B)  # test if same shape, same elements values
np.array_equiv(A,B)  # test if broadcastable shape, same elements values
np.allclose(A,B,...) # test if same shape, elements have close enough values

Answer #4

The way to "start" the array that you want is:

arr = np.empty((0,3), int)

Which is an empty array but it has the proper dimensionality.

>>> arr
array([], shape=(0, 3), dtype=int64)

Then be sure to append along axis 0:

arr = np.append(arr, np.array([[1,2,3]]), axis=0)
arr = np.append(arr, np.array([[4,5,6]]), axis=0)

But, @jonrsharpe is right. In fact, if you"re going to be appending in a loop, it would be much faster to append to a list as in your first example, then convert to a numpy array at the end, since you"re really not using numpy as intended during the loop:

In [210]: %%timeit
   .....: l = []
   .....: for i in xrange(1000):
   .....:     l.append([3*i+1,3*i+2,3*i+3])
   .....: l = np.asarray(l)
   .....: 
1000 loops, best of 3: 1.18 ms per loop

In [211]: %%timeit
   .....: a = np.empty((0,3), int)
   .....: for i in xrange(1000):
   .....:     a = np.append(a, 3*i+np.array([[1,2,3]]), 0)
   .....: 
100 loops, best of 3: 18.5 ms per loop

In [214]: np.allclose(a, l)
Out[214]: True

The numpythonic way to do it depends on your application, but it would be more like:

In [220]: timeit n = np.arange(1,3001).reshape(1000,3)
100000 loops, best of 3: 5.93 µs per loop

In [221]: np.allclose(a, n)
Out[221]: True

Answer #5

Efficient solution

Convolution is much better than straightforward approach, but (I guess) it uses FFT and thus quite slow. However specially for computing the running mean the following approach works fine

def running_mean(x, N):
    cumsum = numpy.cumsum(numpy.insert(x, 0, 0)) 
    return (cumsum[N:] - cumsum[:-N]) / float(N)

The code to check

In[3]: x = numpy.random.random(100000)
In[4]: N = 1000
In[5]: %timeit result1 = numpy.convolve(x, numpy.ones((N,))/N, mode="valid")
10 loops, best of 3: 41.4 ms per loop
In[6]: %timeit result2 = running_mean(x, N)
1000 loops, best of 3: 1.04 ms per loop

Note that numpy.allclose(result1, result2) is True, two methods are equivalent. The greater N, the greater difference in time.

warning: although cumsum is faster there will be increased floating point error that may cause your results to be invalid/incorrect/unacceptable

the comments pointed out this floating point error issue here but i am making it more obvious here in the answer..

# demonstrate loss of precision with only 100,000 points
np.random.seed(42)
x = np.random.randn(100000)+1e6
y1 = running_mean_convolve(x, 10)
y2 = running_mean_cumsum(x, 10)
assert np.allclose(y1, y2, rtol=1e-12, atol=0)
  • the more points you accumulate over the greater the floating point error (so 1e5 points is noticable, 1e6 points is more significant, more than 1e6 and you may want to resetting the accumulators)
  • you can cheat by using np.longdouble but your floating point error still will get significant for relatively large number of points (around >1e5 but depends on your data)
  • you can plot the error and see it increasing relatively fast
  • the convolve solution is slower but does not have this floating point loss of precision
  • the uniform_filter1d solution is faster than this cumsum solution AND does not have this floating point loss of precision

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