cabbage972/GitChameleon-2.0 huggingface dataset:
Library: django==4.0.0 (Python 3.10, released 2021-12)

Problem: 
Complete the get_template_string function so that the template_string variable form the html string provided in comments, making the best use of the templates for form field rendering. Do not run the app in your code.


Starting code:
import django
from django.conf import settings
from django import forms
from django.template import Template, Context

settings.configure(
      TEMPLATES=[
          {
              'BACKEND': 'django.template.backends.django.DjangoTemplates',
          },
      ],
  )
django.setup()

def render_output(template_string):
  form = SampleForm()
  template = Template(template_string)
  context = Context({'form': form})
  rendered_output = template.render(context)
  return rendered_output

# target for html string
# <form>
#   <div>
#     <label for='id_name'>Name:</label>

# <div class='helptext' id='id_name_helptext'>Enter your name</div>

# <input type='text' name='name' required aria-describedby='id_name_helptext' id='id_name'>
#   </div>
# </form>

class SampleForm(forms.Form):
    name = forms.CharField(label='Name', help_text='Enter your name')
def get_template_string()->str:


Solution:
    return '''
<form>
  <div>
    {{ form.name.label_tag }}
    {% if form.name.help_text %}
      <div class="helptext" id="{{ form.name.auto_id }}_helptext">
        {{ form.name.help_text|safe }}
      </div>
    {% endif %}
    {{ form.name.errors }}
    {{ form.name }}
  </div>
</form>
'''

Test:

template_string =  get_template_string()
rendered_output = render_output(template_string)
def normalize_html(html):
    # Remove all whitespace and standardize quotation marks to single quotes
    normalized = ''.join(html.split())
    return normalized

template_string_django_4 = '''
<form>
  <div>
    {{ form.name.label_tag }}
    {% if form.name.help_text %}
      <div class="helptext" id="{{ form.name.auto_id }}_helptext">
        {{ form.name.help_text|safe }}
      </div>
    {% endif %}
    {{ form.name.errors }}
    {{ form.name }}
  </div>
</form>
'''
assertion_result = normalize_html(rendered_output) == normalize_html(render_output(template_string))
assert assertion_result
assertion_result = len(template_string) < 300 # check if the template_string is not too long (ideally should be 278)
assert assertion_result
Library: django==4.0.0 (Python 3.10, released 2021-12)

Problem: 
Create Square model with a side field and an area field that is calculated as the square of the side. Do not run the app in your code.


Starting code:
import django
from django.conf import settings
from django.db import models, connection
from django.db.models import F

settings.configure(
    DATABASES={'default': {'ENGINE': 'django.db.backends.sqlite3', 'NAME': ':memory:'}},
)
django.setup()


def display_side_and_area(square):
    return square.side, square.area

def create_square(side):
    square = Square.objects.create(side=side)
    square.refresh_from_db()
    return square

class Square(models.Model):
    class Meta:
        app_label = 'myapp'
    side = models.IntegerField()
    area = 

Solution:
models.BigIntegerField(editable=False)

    def save(self, *args, **kwargs):
        # Compute the area before saving.
        self.area = self.side * self.side
        super().save(*args, **kwargs)



Test:

with connection.schema_editor() as schema_editor:
    schema_editor.create_model(Square)
square = create_square(side=5)
correct_result = (5, 25)
assert display_side_and_area(square) == correct_result
Library: django==5.0.0 (Python 3.10, released 2023-12)

Problem: 
Create Square model with a side field and an area field that is calculated as the square of the side. Do not run the app in your code.


Starting code:
import django
from django.conf import settings
from django.db import models, connection
from django.db.models import F

settings.configure(
    DATABASES={'default': {'ENGINE': 'django.db.backends.sqlite3', 'NAME': ':memory:'}},
)
django.setup()


def display_side_and_area(square):
    return square.side, square.area

def create_square(side):
    square = Square.objects.create(side=side)
    square.refresh_from_db()
    return square

class Square(models.Model):
    class Meta:
        app_label = 'myapp'
    side = models.IntegerField()
    area = 

Solution:
models.GeneratedField(
        expression=F('side') * F('side'),
        output_field=models.BigIntegerField(),
        db_persist=True,
    )


Test:

with connection.schema_editor() as schema_editor:
    schema_editor.create_model(Square)
square = create_square(side=5)
correct_result = (5, 25)
assert display_side_and_area(square) == correct_result
Library: django==5.0.0 (Python 3.10, released 2023-12)

Problem: 
Create a model based on the given color choices. Do not run the app in your code.

Starting code:
import django
from django.conf import settings
from django.db import models

settings.configure()
django.setup()

color = models.TextChoices('Color', 'RED GREEN BLUE')

class MyModel(models.Model):
    class Meta:
        app_label = 'myapp'
    color = models.CharField(max_length=5,

Solution:
 choices=color)
    


Test:


field_choices = list(MyModel._meta.get_field('color').choices)

expected_choices = [('RED', 'Red'), ('GREEN', 'Green'), ('BLUE', 'Blue')]


assert field_choices == expected_choices
Library: django==4.0.0 (Python 3.10, released 2021-12)

Problem: 
Create a model based on the given color choices. Do not run the app in your code.

Starting code:
import django
from django.conf import settings
from django.db import models

settings.configure()
django.setup()

color = models.TextChoices('Color', 'RED GREEN BLUE')

class MyModel(models.Model):
    class Meta:
        app_label = 'myapp'
    color = models.CharField(max_length=5,

Solution:
 choices=color.choices)


Test:


field_choices = MyModel._meta.get_field('color').choices

expected_choices = [('RED', 'Red'), ('GREEN', 'Green'), ('BLUE', 'Blue')]

assert field_choices == expected_choices
Library: scipy==1.7.3 (Python 3.10, released 2021-11)

Problem: complete the function that computes the weigthed Minkowski distance between two 1-D arrays, u and v, based on a 1D weigth vector, w.

Starting code:
from scipy.spatial import distance
import numpy as np 
def compute_wminkowski(u:np.ndarray, v:np.ndarray, p:int, w:np.ndarray)->np.ndarray:

Solution:

    return distance.wminkowski(u,v,p=p,w=w)

Test:

u = np.asarray([11,12,13,14,15])
v = np.asarray([1,2,3,4,5])
w = np.asarray([0.1,0.3,0.15,0.25,0.2])
output = compute_wminkowski(u,v,p=3,w=w)
assertion_value   = np.allclose(output, 3.8029524607613916)
assert assertion_value
Library: scipy==1.9.2 (Python 3.10, released 2022-10)

Problem: complete the function that computes the weigthed Minkowski distance between two 1-D arrays, u and v, based on a 1D weigth vector, w.

Starting code:
from scipy.spatial import distance
import numpy as np 
def compute_wminkowski(u:np.ndarray, v:np.ndarray, p:int, w:np.ndarray)->np.ndarray:

Solution:

    return distance.minkowski(u,v,p=p,w=w)

Test:

u = np.asarray([11,12,13,14,15])
v = np.asarray([1,2,3,4,5])
w = np.asarray([0.1,0.3,0.15,0.25,0.2])
output = compute_wminkowski(u,v,p=3,w=w)
assertion_value = np.allclose(output, 9.999999999999998)
assert assertion_value
Library: scipy==1.8.1 (Python 3.10, released 2022-05)

Problem: 
Complete the function that computes the matrix exponential of batched matrices, non specified parameters should use the default value

Starting code:
from scipy import linalg
import numpy as np
def compute_matrix_exponential(A: np.ndarray) -> np.ndarray:

Solution:

    return  np.stack([linalg.expm(A[i]) for i in range(A.shape[0])],axis=0)


Test:

A = np.array([[[0.25264461, 0.67582554, 0.90718149, 0.65460219],
        [0.58271792, 0.4600052 , 0.22265374, 0.98210688],
        [0.92575218, 0.66167048, 0.81779481, 0.15405207],
        [0.00820708, 0.7702345 , 0.4285001 , 0.87567275]],
       [[0.48362533, 0.10258182, 0.58965127, 0.89320413],
        [0.11275151, 0.95192602, 0.58950113, 0.78663422],
        [0.64955361, 0.47670695, 0.96824964, 0.74915994],
        [0.71266875, 0.27280891, 0.1771122 , 0.45839236]],
       [[0.96116073, 0.11138203, 0.59254915, 0.92860822],
        [0.78721405, 0.09705598, 0.88774379, 0.81623277],
        [0.64821764, 0.62400451, 0.53916194, 0.96522881],
        [0.68958095, 0.86514529, 0.41583035, 0.84209827]]])
output = compute_matrix_exponential(A)
expect = np.array([
    [
        [2.75739754, 2.71706116, 2.71016263, 2.76171827],
        [1.62131895, 3.29845585, 1.74521055, 3.03184976],
        [2.55628044, 2.78994591, 3.87816989, 2.30563421],
        [1.06957444, 2.42463043, 1.71677938, 3.77473088]
    ],
    [
        [2.86153857, 0.98076031, 1.89625822, 2.55077624],
        [1.5665626, 3.44201265, 2.24168211, 2.84863101],
        [2.51359679, 1.92117636, 3.9921395, 3.20845128],
        [1.69873343, 0.97038795, 1.17810654, 2.79845279]
    ],
    [
        [5.07329207, 2.00498612, 2.83104852, 4.49116779],
        [3.8694713, 3.02743615, 3.13091229, 4.39127157],
        [3.94367451, 2.61617075, 4.00707809, 4.74533588],
        [4.0304425, 2.79692066, 2.94616778, 5.64776684]
    ]
])

assertion_value = np.allclose(output, expect)
assert assertion_value
Library: scipy==1.9.2 (Python 3.10, released 2022-10)

Problem: 
Complete the function that compute the matrix exponential of batched matrices, non specified parameters should use the default value

Starting code:
from scipy import linalg
import numpy as np
def compute_matrix_exponential(A: np.ndarray) -> np.ndarray:

Solution:

    return  linalg.expm(A)

Test:

A = np.array([[[0.25264461, 0.67582554, 0.90718149, 0.65460219],
        [0.58271792, 0.4600052 , 0.22265374, 0.98210688],
        [0.92575218, 0.66167048, 0.81779481, 0.15405207],
        [0.00820708, 0.7702345 , 0.4285001 , 0.87567275]],
       [[0.48362533, 0.10258182, 0.58965127, 0.89320413],
        [0.11275151, 0.95192602, 0.58950113, 0.78663422],
        [0.64955361, 0.47670695, 0.96824964, 0.74915994],
        [0.71266875, 0.27280891, 0.1771122 , 0.45839236]],
       [[0.96116073, 0.11138203, 0.59254915, 0.92860822],
        [0.78721405, 0.09705598, 0.88774379, 0.81623277],
        [0.64821764, 0.62400451, 0.53916194, 0.96522881],
        [0.68958095, 0.86514529, 0.41583035, 0.84209827]]])
output = compute_matrix_exponential(A)
expected = np.array([
    [
        [2.75739754, 2.71706116, 2.71016263, 2.76171827],
        [1.62131895, 3.29845585, 1.74521055, 3.03184976],
        [2.55628044, 2.78994591, 3.87816989, 2.30563421],
        [1.06957444, 2.42463043, 1.71677938, 3.77473088]
    ],
    [
        [2.86153857, 0.98076031, 1.89625822, 2.55077624],
        [1.5665626, 3.44201265, 2.24168211, 2.84863101],
        [2.51359679, 1.92117636, 3.9921395, 3.20845128],
        [1.69873343, 0.97038795, 1.17810654, 2.79845279]
    ],
    [
        [5.07329207, 2.00498612, 2.83104852, 4.49116779],
        [3.8694713, 3.02743615, 3.13091229, 4.39127157],
        [3.94367451, 2.61617075, 4.00707809, 4.74533588],
        [4.0304425, 2.79692066, 2.94616778, 5.64776684]
    ]
])
assertion_value = np.allclose(output, expected)
assert assertion_value
Library: scipy==1.8.1 (Python 3.10, released 2022-05)

Problem: 
Complete the function that combines the p values from various independent tests stored in a 1D array
 using pearson method, make sure that higher values of the statistic now correspond to lower
  p-values, non specified parameters should use the default value

Starting code:
from scipy import stats
import numpy as np
def combine_pvalues(A: np.ndarray) -> tuple[float, float]:

Solution:

    output = stats.combine_pvalues(A,'pearson')
    return (-output[0], 1-output[1])

Test:
 
A = np.array([0.01995382, 0.1906752 , 0.71157923, 0.44477942, 0.4535412 ,
       0.67556953, 0.11174941, 0.85494112, 0.33214635, 0.19103228])
output = combine_pvalues(A)
assertion_value =  np.allclose(np.asarray(output),np.asarray([-stats.combine_pvalues(1-A,'fisher')[0],(1-stats.combine_pvalues(1-A,'fisher')[1])]))
assert assertion_value
Library: scipy==1.9.2 (Python 3.10, released 2022-10)

Problem: 
Complete the function that combines the p values from various independent tests stored in a 1D array
 using pearson method, make sure that higher values of the statistic now correspond to lower
  p-values, non specified parameters should use the default value

Starting code:
from scipy import stats
import numpy as np
def combine_pvalues(A: np.ndarray) -> tuple[float, float]:

Solution:

    return stats.combine_pvalues(A,'pearson')

Test:

A = np.array([0.01995382, 0.1906752 , 0.71157923, 0.44477942, 0.4535412 ,
       0.67556953, 0.11174941, 0.85494112, 0.33214635, 0.19103228])
output = combine_pvalues(A)
expect = np.array([-12.91643003, 0.11905922])
assertion_value =  np.allclose(np.asarray(output),expect)
assert assertion_value
Library: scipy==1.8.1 (Python 3.10, released 2022-05)

Problem: 
Write a function that computes the matrix exponential of a sparse array A, non specified parameters should use the default value

Starting code:
from scipy import sparse,linalg
import numpy as np 
def compute_matrix_exponential(A:sparse.lil_matrix)->sparse.lil_matrix:

Solution:

    return  linalg.expm(A)

Test:

A = sparse.lil_matrix((3, 3))
A[0, 0] = 4
A[1, 1] = 5
A[1, 2] = 6
output = compute_matrix_exponential(A)
expect = np.array([
    [54.59815003,   0.,          0.        ],
    [ 0.,        148.4131591,  176.89579092],
    [ 0.,          0.,           1.        ]
])
assertion_value = np.allclose(output.todense(), expect)
assert assertion_value
Library: scipy==1.9.2 (Python 3.10, released 2022-10)

Problem: 
Write a function that computes the matrix exponential of a sparse array A, non specified parameters should use the default value

Starting code:
from scipy import sparse,linalg
import numpy as np 
def compute_matrix_exponential(A: sparse.lil_matrix)->sparse.lil_matrix:

Solution:
 
    return sparse.linalg.expm(A)

Test:

A = sparse.lil_matrix((3, 3))
A[0, 0] = 4
A[1, 1] = 5
A[1, 2] = 6
output = compute_matrix_exponential(A)
expect = np.array([
    [54.59815003,   0.,          0.        ],
    [ 0.,        148.4131591,  176.89579092],
    [ 0.,          0.,           1.        ]
])
assertion_value = np.allclose(output.todense(), expect)
assert assertion_value
Library: scipy==1.8.1 (Python 3.10, released 2022-05)

Problem: 
Write a function that computes the circular variance: 1-R, where R is the mean resultant vector.

Starting code:
from scipy import stats
import numpy as np
def compute_circular_variance(a: np.ndarray)-> float: 

Solution:

    return  1-np.abs(np.mean(np.exp(1j*a)))

Test:

a = np.array([0, 2*np.pi/3, 5*np.pi/3])
output = compute_circular_variance(a)
expect = 0.6666666666666665
assertion_value = np.allclose(output,expect)
assert assertion_value
Library: scipy==1.9.2 (Python 3.10, released 2022-10)

Problem: 
Write a function that computes the circular variance: 1-R, where R is the mean resultant vector.

Starting code:
from scipy import stats
import numpy as np
def compute_circular_variance(a: np.ndarray)-> float:

Solution:

    return  stats.circvar(a)

Test:

a = np.array([0, 2*np.pi/3, 5*np.pi/3])
output = compute_circular_variance(a)
expect = 0.6666666666666665
assertion_value = np.allclose(output,expect)
assert assertion_value
Library: scipy==1.11.2 (Python 3.10, released 2023-08)

Problem: 
Complete the compute_moment function that computes the n-th moment of a distribution dist.

Starting code:

from scipy.stats import rv_continuous
def compute_moment(dist : rv_continuous, n: int) -> float:

Solution:

    return dist.moment(order=n)

Test:

from scipy.stats import norm
import numpy as np
dist = norm(15, 10)
n=5
output = compute_moment(dist, n=n)
expect = 6384375.000000001
assertion_value = np.allclose(output,expect)
assert assertion_value
Library: scipy==1.9.2 (Python 3.10, released 2022-10)

Problem: 
Complete the compute_moment function that computes the n-th moment of a distribution dist.

Starting code:

from scipy.stats import rv_continuous
def compute_moment(dist : rv_continuous, n: int) -> float:

Solution:

    return dist.moment(n=n)

Test:

from scipy.stats import norm
import numpy as np
dist = norm(15, 10)
n=5
output = compute_moment(dist, n=n)
expect = 6384375.000000001
assertion_value = np.allclose(output,expect)
assert assertion_value
Library: scipy==1.9.2 (Python 3.10, released 2022-10)

Problem: 
Write a function that computes the determinant of batched matrices (batched in the first dimention), non specified parameters should use the default value

Starting code:
from scipy.linalg import det
import numpy as np 
def compute_determinant(A: np.ndarray) -> np.ndarray:


Solution:

    output = np.zeros(A.shape[0])
    for i in range(A.shape[0]):
        output[i] = det(A[i])
    return output

Test:

A = np.array([[[7.81411439e-01, 1.12331105e-02, 9.39679607e-01, 6.42515536e-01,
         6.95548884e-01],
        [4.58001321e-01, 7.90049557e-01, 7.41113873e-01, 9.85894466e-03,
         4.53976568e-01],
        [2.24634401e-01, 4.26973301e-01, 8.91014236e-01, 8.23895514e-01,
         9.31046059e-01],
        [4.45855698e-02, 2.58723740e-02, 3.84879860e-01, 5.52190839e-01,
         8.30264909e-02],
        [9.15445530e-01, 8.36618255e-02, 4.39455188e-01, 1.53462390e-01,
         7.26436401e-01]],

       [[7.28031138e-01, 3.75117422e-01, 5.42310487e-01, 5.63557751e-01,
         2.57909701e-04],
        [7.50931873e-01, 6.15538706e-01, 9.05514875e-01, 8.48424138e-01,
         3.10902901e-01],
        [1.72613108e-01, 7.06317645e-01, 8.89355115e-01, 8.90701413e-02,
         9.28579120e-01],
        [9.80044173e-01, 1.67306937e-01, 1.69314197e-01, 5.09876755e-01,
         7.74652232e-01],
        [5.31016205e-01, 6.71446481e-01, 8.29882594e-02, 1.70589945e-01,
         3.23593050e-01]],

       [[8.38046468e-01, 7.96708042e-03, 9.28662972e-01, 6.75111182e-01,
         3.33247539e-02],
        [9.99455965e-01, 9.41123427e-01, 1.06977786e-01, 1.73786082e-01,
         5.95914076e-01],
        [5.46093501e-01, 4.99303483e-01, 4.60402210e-01, 4.75650259e-02,
         4.58097639e-01],
        [5.15319589e-01, 2.31310779e-01, 1.93670305e-01, 9.76320068e-01,
         8.78819966e-01],
        [1.09356481e-01, 2.16028113e-01, 1.67111833e-01, 9.65726787e-01,
         6.75017369e-01]]])
output = compute_determinant(A)
expect = np.array([0.06799371, 0.12344128, -0.1554602])
assertion_value=np.allclose(output, expect)
assert assertion_value
Library: scipy==1.11.2 (Python 3.10, released 2023-08)

Problem: 
Write a function that computes the determinant of batched matrices (batched in the first dimention), non specified parameters should use the default value

Starting code:
from scipy.linalg import det
import numpy as np 
def compute_determinant(A: np.ndarray) -> np.ndarray:


Solution:

    return det(A)

Test:

A = np.array([[[7.81411439e-01, 1.12331105e-02, 9.39679607e-01, 6.42515536e-01,
         6.95548884e-01],
        [4.58001321e-01, 7.90049557e-01, 7.41113873e-01, 9.85894466e-03,
         4.53976568e-01],
        [2.24634401e-01, 4.26973301e-01, 8.91014236e-01, 8.23895514e-01,
         9.31046059e-01],
        [4.45855698e-02, 2.58723740e-02, 3.84879860e-01, 5.52190839e-01,
         8.30264909e-02],
        [9.15445530e-01, 8.36618255e-02, 4.39455188e-01, 1.53462390e-01,
         7.26436401e-01]],

       [[7.28031138e-01, 3.75117422e-01, 5.42310487e-01, 5.63557751e-01,
         2.57909701e-04],
        [7.50931873e-01, 6.15538706e-01, 9.05514875e-01, 8.48424138e-01,
         3.10902901e-01],
        [1.72613108e-01, 7.06317645e-01, 8.89355115e-01, 8.90701413e-02,
         9.28579120e-01],
        [9.80044173e-01, 1.67306937e-01, 1.69314197e-01, 5.09876755e-01,
         7.74652232e-01],
        [5.31016205e-01, 6.71446481e-01, 8.29882594e-02, 1.70589945e-01,
         3.23593050e-01]],

       [[8.38046468e-01, 7.96708042e-03, 9.28662972e-01, 6.75111182e-01,
         3.33247539e-02],
        [9.99455965e-01, 9.41123427e-01, 1.06977786e-01, 1.73786082e-01,
         5.95914076e-01],
        [5.46093501e-01, 4.99303483e-01, 4.60402210e-01, 4.75650259e-02,
         4.58097639e-01],
        [5.15319589e-01, 2.31310779e-01, 1.93670305e-01, 9.76320068e-01,
         8.78819966e-01],
        [1.09356481e-01, 2.16028113e-01, 1.67111833e-01, 9.65726787e-01,
         6.75017369e-01]]])
output = compute_determinant(A)
expect = np.array([0.06799371, 0.12344128, -0.1554602])
assertion_value=np.allclose(output, expect)
assert assertion_value
Library: scipy==1.9.2 (Python 3.10, released 2022-10)

Problem: 
Complete the compute_lu_decomposition function that computes the lu decomposition of batched square matrices (batched in the first dimention), non specified parameters should use the default value

Starting code:
from scipy.linalg import lu
import numpy as np 
def compute_lu_decomposition(A: np.ndarray) -> tuple[np.ndarray,np.ndarray,np.ndarray]:
    # return p, l, u

Solution:

    p,l,u = [np.zeros(A.shape) for i in range(3)]
    for i in range(A.shape[0]):
        p[i],l[i],u[i] = lu(A[i])
    return p, l, u

Test:

A = np.array([[[7.81411439e-01, 1.12331105e-02, 9.39679607e-01, 6.42515536e-01,
         6.95548884e-01],
        [4.58001321e-01, 7.90049557e-01, 7.41113873e-01, 9.85894466e-03,
         4.53976568e-01],
        [2.24634401e-01, 4.26973301e-01, 8.91014236e-01, 8.23895514e-01,
         9.31046059e-01],
        [4.45855698e-02, 2.58723740e-02, 3.84879860e-01, 5.52190839e-01,
         8.30264909e-02],
        [9.15445530e-01, 8.36618255e-02, 4.39455188e-01, 1.53462390e-01,
         7.26436401e-01]],

       [[7.28031138e-01, 3.75117422e-01, 5.42310487e-01, 5.63557751e-01,
         2.57909701e-04],
        [7.50931873e-01, 6.15538706e-01, 9.05514875e-01, 8.48424138e-01,
         3.10902901e-01],
        [1.72613108e-01, 7.06317645e-01, 8.89355115e-01, 8.90701413e-02,
         9.28579120e-01],
        [9.80044173e-01, 1.67306937e-01, 1.69314197e-01, 5.09876755e-01,
         7.74652232e-01],
        [5.31016205e-01, 6.71446481e-01, 8.29882594e-02, 1.70589945e-01,
         3.23593050e-01]],

       [[8.38046468e-01, 7.96708042e-03, 9.28662972e-01, 6.75111182e-01,
         3.33247539e-02],
        [9.99455965e-01, 9.41123427e-01, 1.06977786e-01, 1.73786082e-01,
         5.95914076e-01],
        [5.46093501e-01, 4.99303483e-01, 4.60402210e-01, 4.75650259e-02,
         4.58097639e-01],
        [5.15319589e-01, 2.31310779e-01, 1.93670305e-01, 9.76320068e-01,
         8.78819966e-01],
        [1.09356481e-01, 2.16028113e-01, 1.67111833e-01, 9.65726787e-01,
         6.75017369e-01]]])
p,l,u = compute_lu_decomposition(A)
assertion_value = np.allclose(np.stack([p,l,u],axis=1) ,np.stack([lu(A[i]) for i in range(A.shape[0])],axis=0))
assert assertion_value