NumPy Quiz를 풀어보자!
NumPy Quiz
# Use the numpy library
import numpy as np
def prepare_inputs(inputs):
# TODO: create a 2-dimensional ndarray from the given 1-dimensional list;
# assign it to input_array
input_array = None
# TODO: find the minimum value in input_array and subtract that
# value from all the elements of input_array. Store the
# result in inputs_minus_min
inputs_minus_min = None
# TODO: find the maximum value in inputs_minus_min and divide
# all of the values in inputs_minus_min by the maximum value.
# Store the results in inputs_div_max.
inputs_div_max = None
# return the three arrays we've created
return input_array, inputs_minus_min, inputs_div_max
def multiply_inputs(m1, m2):
# TODO: Check the shapes of the matrices m1 and m2.
# m1 and m2 will be ndarray objects.
#
# Return False if the shapes cannot be used for matrix
# multiplication. You may not use a transpose
pass
# TODO: If you have not returned False, then calculate the matrix product
# of m1 and m2 and return it. Do not use a transpose,
# but you swap their order if necessary
pass
def find_mean(values):
# TODO: Return the average of the values in the given Python list
pass
input_array, inputs_minus_min, inputs_div_max = prepare_inputs([-1,2,7])
print("Input as Array: {}".format(input_array))
print("Input minus min: {}".format(inputs_minus_min))
print("Input Array: {}".format(inputs_div_max))
print("Multiply 1:\n{}".format(multiply_inputs(np.array([[1,2,3],[4,5,6]]), np.array([[1],[2],[3],[4]]))))
print("Multiply 2:\n{}".format(multiply_inputs(np.array([[1,2,3],[4,5,6]]), np.array([[1],[2],[3]]))))
print("Multiply 3:\n{}".format(multiply_inputs(np.array([[1,2,3],[4,5,6]]), np.array([[1,2]]))))
print("Mean == {}".format(find_mean([1,3,4])))NumPy Quiz - My Answer
# Use the numpy library
import numpy as np
def prepare_inputs(inputs):
# TODO: create a 2-dimensional ndarray from the given 1-dimensional list;
# assign it to input_array
input_array = np.array(inputs)[None, :]
# TODO: find the minimum value in input_array and subtract that
# value from all the elements of input_array. Store the
# result in inputs_minus_min
inputs_minus_min = input_array - np.min( input_array )
# TODO: find the maximum value in inputs_minus_min and divide
# all of the values in inputs_minus_min by the maximum value.
# Store the results in inputs_div_max.
inputs_div_max = inputs_minus_min / (np.max( inputs_minus_min) )
# return the three arrays we've created
return input_array, inputs_minus_min, inputs_div_max
def multiply_inputs(m1, m2):
# TODO: Check the shapes of the matrices m1 and m2.
# m1 and m2 will be ndarray objects.
#
# Return False if the shapes cannot be used for matrix
# multiplication. You may not use a transpose
if m1.shape[0] != m2.shape[1] and m1.shape[1] != m2.shape[0]:
return False
# TODO: If you have not returned False, then calculate the matrix product
# of m1 and m2 and return it. Do not use a transpose,
# but you swap their order if necessary
if m1.shape[0] == m2.shape[1]:
return np.matmul( m2, m1 )
else:
return np.matmul( m1, m2 )
def find_mean(values):
# TODO: Return the average of the values in the given Python list
return np.mean(values)
input_array, inputs_minus_min, inputs_div_max = prepare_inputs([-1,2,7])
print("Input as Array: {}".format(input_array))
print("Input minus min: {}".format(inputs_minus_min))
print("Input Array: {}".format(inputs_div_max))
print("Multiply 1:\n{}".format(multiply_inputs(np.array([[1,2,3],[4,5,6]]), np.array([[1],[2],[3],[4]]))))
print("Multiply 2:\n{}".format(multiply_inputs(np.array([[1,2,3],[4,5,6]]), np.array([[1],[2],[3]]))))
print("Multiply 3:\n{}".format(multiply_inputs(np.array([[1,2,3],[4,5,6]]), np.array([[1,2]]))))
print("Mean == {}".format(find_mean([1,3,4])))
Comment 1 :
첫번째 TODO 주어진 1차원의 List로 2차원의 ndarray를 만들라해서...
1. List를 ndarray로 변환한다.
2. 2차원으로 reshape한다.
였는데!
주어진 List의 1차원 Length가 몇인지 알 수 없잖아?
그래서 reshape( ?, ? )가 되버림...
그래서 ndarrray로 변환한 놈을
( [None, :] ) 로 1행 n열 형태 변환함
근데 정답보니...
걍
np.array( [ inputs ] )
중괄호 하나 겉에 쳐주면 되네 ^ㅡ^
Comment 2 :
m1,m2의 multiply 문제에서
나는 if문에 or을 사용하려했다.
하지만 and 임
무슨말이냐...
주어진 두 개의 Matrix가 matrix multiplication이 되지 않는다는 것은
m1의 Row수 != m2의 Col수
m1의 Col수 != m2의 Row수
둘다여야 한다는 것임!
NumPy Quiz - Solution
# Use the numpy library
import numpy as np
######################################################
#
# MESSAGE TO STUDENTS:
#
# This file contains a solution to the coding quiz. Feel free
# to look at it when you are stuck, but try to solve the
# problem on your own first.
#
######################################################
def prepare_inputs(inputs):
# TODO: create a 2-dimensional ndarray from the given 1-dimensional list;
# assign it to input_array
input_array = np.array([inputs])
# TODO: find the minimum value in input_array and subtract that
# value from all the elements of input_array. Store the
# result in inputs_minus_min
# We can use NumPy's min function and element-wise division
inputs_minus_min = input_array - np.min(input_array)
# TODO: find the maximum value in inputs_minus_min and divide
# all of the values in inputs_minus_min by the maximum value.
# Store the results in inputs_div_max.
# We can use NumPy's max function and element-wise division
inputs_div_max = inputs_minus_min / np.max(inputs_minus_min)
return input_array, inputs_minus_min, inputs_div_max
def multiply_inputs(m1, m2):
# Check the shapes of the matrices m1 and m2.
# m1 and m2 will be ndarray objects.
#
# Return False if the shapes cannot be used for matrix
# multiplication. You may not use a transpose
if m1.shape[0] != m2.shape[1] and m1.shape[1] != m2.shape[0]:
return False
# Have not returned False, so calculate the matrix product
# of m1 and m2 and return it. Do not use a transpose,
# but you swap their order if necessary
if m1.shape[1] == m2.shape[0]:
return np.matmul(m1, m2)
else:
return np.matmul(m2, m1)
def find_mean(values):
# Return the average of the values in the given Python list
# NumPy has a lot of helpful methods like this.
return np.mean(values)
NumPy Quiz - Output
Input as Array: [[-1 2 7]]
Input minus min: [[0 3 8]]
Input Array: [[0. 0.375 1. ]]
Multiply 1:
False
Multiply 2:
[[14]
[32]]
Multiply 3:
[[ 9 12 15]]
Mean == 2.6666666666666665
Nice job! That's right!
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