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jrosendo
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README.md
Exercise 0: Environment and libraries
The exercice is validated if all questions of the exercice are validated
Install the virtual environment with requirements.txt
Activate the virtual environment. If you used conda, run conda activate ex00
Does the shell specify the name ex00 of the environment on the left ?
Run python --version
Does it print Python 3.8.x? x could be any number from 0 to 9
Does import jupyter and import numpy run without any error ?
Have you used the followingthe command jupyter notebook --port 8891 ?
Is there a file named Notebook_ex00.ipynb in the working directory ?
Is the following markdown code executed in a markdown cell in the first cell ?
# H1 TITLE
## H2 TITLE
Does the second cell contain print("Buy the dip ?") and return Buy the dip ? in the output section ?
Exercise 1: Your first NumPy array
Add cell and run type(your_numpy_array)
Is the your_numpy_array an NumPy array ? It can be checked with that should be equal to numpy.ndarray.
Run all the cells of the notebook or python main.py
Are the types printed are as follows ?
<class 'int'>
<class 'float'>
<class 'str'>
<class 'dict'>
<class 'list'>
<class 'tuple'>
<class 'set'>
<class 'bool'>
Delete all the cells you added for the audit and restart the notebook
TODO
Exercise 2: Zeros
The exercice is validated is all questions of the exercice are validated
The question 1 is validated is the solution uses np.zeros and if the shape of the array is (300,)
The question 2 is validated if the solution uses reshape and the shape of the array is (3, 100)
Exercise 3: Slicing
The exercice is validated is all questions of the exercice are validated
The question 1 is validated if the solution doesn't involve a for loop or writing all integers from 1 to 100 and if the array is: np.array([1,...,100]). The list from 1 to 100 can be generated with an iterator: range.
The question 2 is validated if the solution is: integers[::2]
The question 3 is validated if the solution is: integers[::-2]
The question 4 is validated if the array is: np.array([0, 1,0,3,4,0,...,0,99,100]). There are at least two ways to get this results without for loop. The first one uses integers[1::3] = 0 and the second involves creating a boolean array that indexes the array:
mask = (integers+1)%3 == 0
integers[mask] = 0
Exercise 4: Random
The exercice is validated is all questions of the exercice are validated
For this exercise, as the results may change depending on the version of the package or the OS, I give the code to correct the exercise. If the code is correct and the output is not the same as mine, it is accepted.
The question 1 is validated if the solution is: np.random.seed(888)
The question 2 is validated if the solution is: np.random.randn(100). The value of the first element is 0.17620087373662233.
The question 3 is validated if the solution is: np.random.randint(1,11,(8,8)).
Given the NumPy version and the seed, you should have this output:
array([[ 7, 4, 8, 10, 2, 1, 1, 10],
[ 4, 1, 7, 4, 3, 5, 2, 8],
[ 3, 9, 7, 4, 9, 6, 10, 5],
[ 7, 10, 3, 10, 2, 1, 3, 7],
[ 3, 2, 3, 2, 10, 9, 5, 4],
[ 4, 1, 9, 7, 1, 4, 3, 5],
[ 3, 2, 10, 8, 6, 3, 9, 4],
[ 4, 4, 9, 2, 8, 5, 9, 5]])
The question 4 is validated if the solution is: np.random.randint(1,18,(4,2,5)).
Given the NumPy version and the seed, you should have this output:
array([[[14, 16, 8, 15, 14],
[17, 13, 1, 4, 17]],
[[ 7, 15, 2, 8, 3],
[ 9, 4, 13, 9, 15]],
[[ 5, 11, 11, 14, 10],
[ 2, 1, 15, 3, 3]],
[[ 3, 10, 5, 16, 13],
[17, 12, 9, 7, 16]]])
Exercise 5: Split, concatenate, reshape arrays
The exercice is validated is all questions of the exercice are validated
The question 1 is validated if the generated array is based on an iterator as range or np.arange. Check that 50 is part of the array.
The question 2 is validated if the generated array is based on an iterator as range or np.arange. Check that 100 is part of the array.
The question 3 is validated if the array is concatenated this way np.concatenate(array1,array2).
The question 4 is validated if the result is:
array([[ 1, ... , 10],
...
[ 91, ... , 100]])
The easiest way is to use array.reshape(10,10).
https://jakevdp.github.io/PythonDataScienceHandbook/ (section: The Basics of NumPy Arrays)
Exercise 6: Broadcasting and Slicing
The exercice is validated is all questions of the exercice are validated
The question 1 is validated if the output is the same as:
np.ones([9,9], dtype=np.int8)
The question 2 is validated if the output is
array([[1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1, 0, 1],
[1, 0, 1, 0, 0, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 0, 0, 1, 0, 1],
[1, 0, 1, 1, 1, 1, 1, 0, 1],
[1, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1]], dtype=int8)
The solution of question 2 is not accepted if the values of the array have been changed one by one manually. The usage of the for loop is not allowed neither.
Here is an example of a possible solution:
x[1:8,1:8] = 0
x[2:7,2:7] = 1
x[3:6,3:6] = 0
x[4,4] = 1
Exercise 7: NaN
The exercice is validated is all questions of the exercice are validated
This question is validated if, without having used a for loop or having filled the array manually, the output is:
[[ 7. 1. 7.]
[nan 2. 2.]
[nan 8. 8.]
[ 9. 3. 9.]
[ 8. 9. 8.]
[nan 2. 2.]
[ 8. 2. 8.]
[nan 6. 6.]
[ 9. 2. 9.]
[ 8. 5. 8.]]
There are two steps in this exercise:
- Create the vector that contains the grade of the first exam if available or the second. This can be done using
np.where:
np.where(np.isnan(grades[:, 0]), grades[:, 1], grades[:, 0])
- Add this vector as third column of the array. Here are two ways:
np.insert(arr = grades, values = new_vector, axis = 1, obj = 2)
np.hstack((grades, new_vector[:, None]))
Exercise 8: Wine
The exercice is validated is all questions of the exercice are validated
This question is validated if the text file has successfully been loaded in a NumPy array with
genfromtxt('winequality-red.csv', delimiter=',') and the reduced arrays weights 76800 bytes
This question is validated if the output is
array([[ 7.4 , 0.7 , 0. , 1.9 , 0.076 , 11. , 34. ,
0.9978, 3.51 , 0.56 , 9.4 , 5. ],
[ 7.4 , 0.66 , 0. , 1.8 , 0.075 , 13. , 40. ,
0.9978, 3.51 , 0.56 , 9.4 , 5. ],
[ 6.7 , 0.58 , 0.08 , 1.8 , 0.097 , 15. , 65. ,
0.9959, 3.28 , 0.54 , 9.2 , 5. ]])
This slicing gives the answer my_data[[1,6,11],:].
This question is validated if the answer if False. There many ways to get the answer: find the maximum or check values greater than 20.
This question is validated if the answer is 10.422983114446529.
This question is validated if the answers is:
pH stats
25 percentile: 3.21
50 percentile: 3.31
75 percentile: 3.4
mean: 3.3111131957473416
min: 2.74
max: 4.01
> *Note: Using `percentile` or `median` may give different results depending on the duplicate values in the column. If you do not have my results please use `percentile`.*
This question is validated if the answer is ~5.2. The first step is to get the percentile 20% of the column sulphates, then create a boolean array that contains True of the value is smaller than the percentile 20%, then select this rows with the column quality and compute the mean.
This question is validated if the output for the best wines is:
array([ 8.56666667, 0.42333333, 0.39111111, 2.57777778, 0.06844444,
13.27777778, 33.44444444, 0.99521222, 3.26722222, 0.76777778,
12.09444444, 8. ])
This question is validated if the output for the bad wines is:
array([ 8.36 , 0.8845 , 0.171 , 2.635 , 0.1225 , 11. ,
24.9 , 0.997464, 3.398 , 0.57 , 9.955 , 3. ])
This can be done in three steps: Get the max, create a boolean mask that indicates rows with max quality, use this mask to subset the rows with the best quality and compute the mean on the axis 0.
Exercise 9: Football tournament
This exercise is validated if the output is:
[[0 3 1 2 4]
[7 6 8 9 5]]