PYTHON head_count = [np.equal(x3,i).sum() for i in range(1,18)] head_count The question is: how could I write...

PYTHON

head_count = [np.equal(x3,i).sum() for i in range(1,18)]
head_count

The question is: how could I write code that does the same as np.equal(x3,i).sum() without using Numpy?

Expert Solution

# Python Program illustrating
# numpy.sum() method
import numpy as np

# 1D array
arr = [20, 2, .2, 10, 4]

print("\nSum of arr : ", np.sum(arr))

print("Sum of arr(uint8) : ", np.sum(arr, dtype = np.uint8))
print("Sum of arr(float32) : ", np.sum(arr, dtype = np.float32))

print ("\nIs np.sum(arr).dtype == np.uint : ",
np.sum(arr).dtype == np.uint)

print ("Is np.sum(arr).dtype == np.float : ",
np.sum(arr).dtype == np.float)
------------------------------------------------------

# Python Program illustrating
# numpy.sum() method
import numpy as np

# 2D array
arr = [[14, 17, 12, 33, 44],
   [15, 6, 27, 8, 19],
   [23, 2, 54, 1, 4,]]

print("\nSum of arr : ", np.sum(arr))

print("Sum of arr(uint8) : ", np.sum(arr, dtype = np.uint8))
print("Sum of arr(float32) : ", np.sum(arr, dtype = np.float32))

print ("\nIs np.sum(arr).dtype == np.uint : ",
np.sum(arr).dtype == np.uint)

print ("Is np.sum(arr).dtype == np.uint : ",
np.sum(arr).dtype == np.float)
-------------------------------------------------------------------------------------------------------------------------------------

venereology answered 6 months ago

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