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update layout of all exercises

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Bogumił Kamiński
2022-10-14 13:43:12 +02:00
parent 38398729ce
commit 31d8428f6a
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exercises/exercises04.md
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@@ -12,11 +12,81 @@ Create a matrix of shape 2x3 containing numbers from 1 to 6 (fill the matrix
columnwise with consecutive numbers). Next calculate sum, mean and standard
deviation of each row and each column of this matrix.
<details>
<summary>Solution</summary>
Write:
```
julia> using Statistics
julia> mat = [1 3 5
2 4 6]
2×3 Matrix{Int64}:
1 3 5
2 4 6
julia> sum(mat, dims=1)
1×3 Matrix{Int64}:
3 7 11
julia> sum(mat, dims=2)
2×1 Matrix{Int64}:
9
12
julia> mean(mat, dims=1)
1×3 Matrix{Float64}:
1.5 3.5 5.5
julia> mean(mat, dims=2)
2×1 Matrix{Float64}:
3.0
4.0
julia> std(mat, dims=1)
1×3 Matrix{Float64}:
0.707107 0.707107 0.707107
julia> std(mat, dims=2)
2×1 Matrix{Float64}:
2.0
2.0
```
Observe that the returned statistics are also stored in matrices.
If we compute them for columns (`dims=1`) then the produced matrix has one row.
If we compute them for rows (`dims=2`) then the produced matrix has one column.
</details>
### Exercise 2
For each column of the matrix created in exercise 1 compute its range
(i.e. the difference between maximum and minimum element stored in it).
<details>
<summary>Solution</summary>
Here are some ways you can do it:
```
julia> [maximum(x) - minimum(x) for x in eachcol(mat)]
3-element Vector{Int64}:
1
1
1
julia> map(x -> maximum(x) - minimum(x), eachcol(mat))
3-element Vector{Int64}:
1
1
1
```
Observe that if we used `eachcol` the produced result is a vector (not a matrix
like in exercise 1).
</details>
### Exercise 3
This is data for car speed (mph) and distance taken to stop (ft)
@@ -79,127 +149,8 @@ speed dist
Load this data into Julia (this is part of the exercise) and fit a linear
regression where speed is a feature and distance is target variable.
### Exercise 4
Plot the data loaded in exercise 4. Additionally plot the fitted regression
(you need to check Plots.jl documentation to find a way to do this).
### Exercise 5
A simple code for calculation of Fibonacci numbers for positive
arguments is as follows:
```
fib(n) =n < 3 ? 1 : fib(n-1) + fib(n-2)
```
Using the BenchmarkTools.jl package measure runtime of this function for
`n` ranging from `1` to `20`.
### Exercise 6
Improve the speed of code from exercise 5 by using a dictionary where you
store a mapping of `n` to `fib(n)`. Measure the performance of this function
for the same range of values as in exercise 5.
### Exercise 7
Create a vector containing named tuples representing elements of a 4x4 grid.
So the first element of this vector should be `(x=1, y=1)` and last should be
`(x=4, y=4)`. Store the vector in variable `v`.
### Exercise 8
The `filter` function allows you to select some values of an input collection.
Check its documentation first. Next, use it to keep from the vector `v` from
exercise 7 only elements whose sum is even.
### Exercise 9
Check the documentation of the `filter!` function. Perform the same operation
as asked in exercise 8 but using `filter!`. What is the difference?
### Exercise 10
Write a function that takes a number `n`. Next it generates two independent
random vectors of length `n` and returns their correlation coefficient.
Run this function `10000` times for `n` equal to `10`, `100`, `1000`,
and `10000`.
Create a plot with four histograms of distribution of computed Pearson
correlation coefficient. Check in the Plots.jl package which function can be
used to plot histograms.
# Solutions
<details>
<summary>Show!</summary>
### Exercise 1
Write:
```
julia> using Statistics
julia> mat = [1 3 5
2 4 6]
2×3 Matrix{Int64}:
1 3 5
2 4 6
julia> sum(mat, dims=1)
1×3 Matrix{Int64}:
3 7 11
julia> sum(mat, dims=2)
2×1 Matrix{Int64}:
9
12
julia> mean(mat, dims=1)
1×3 Matrix{Float64}:
1.5 3.5 5.5
julia> mean(mat, dims=2)
2×1 Matrix{Float64}:
3.0
4.0
julia> std(mat, dims=1)
1×3 Matrix{Float64}:
0.707107 0.707107 0.707107
julia> std(mat, dims=2)
2×1 Matrix{Float64}:
2.0
2.0
```
Observe that the returned statistics are also stored in matrices.
If we compute them for columns (`dims=1`) then the produced matrix has one row.
If we compute them for rows (`dims=2`) then the produced matrix has one column.
### Exercise 2
Here are some ways you can do it:
```
julia> [maximum(x) - minimum(x) for x in eachcol(mat)]
3-element Vector{Int64}:
1
1
1
julia> map(x -> maximum(x) - minimum(x), eachcol(mat))
3-element Vector{Int64}:
1
1
1
```
Observe that if we used `eachcol` the produced result is a vector (not a matrix
like in exercise 1).
### Exercise 3
<summary>Solution</summary>
First create a matrix with source data by copy pasting it from the exercise
like this:
@@ -285,8 +236,16 @@ julia> [ones(50) data[:, 1]] \ data[:, 2]
3.9324087591240877
```
</details>
### Exercise 4
Plot the data loaded in exercise 4. Additionally plot the fitted regression
(you need to check Plots.jl documentation to find a way to do this).
<details>
<summary>Solution</summary>
Run the following:
```
using Plots
@@ -296,8 +255,23 @@ scatter(data[:, 1], data[:, 2];
The `smooth=true` keyword argument adds the linear regression line to the plot.
</details>
### Exercise 5
A simple code for calculation of Fibonacci numbers for positive
arguments is as follows:
```
fib(n) =n < 3 ? 1 : fib(n-1) + fib(n-2)
```
Using the BenchmarkTools.jl package measure runtime of this function for
`n` ranging from `1` to `20`.
<details>
<summary>Solution</summary>
Use the following code:
```
julia> using BenchmarkTools
@@ -331,8 +305,17 @@ julia> for i in 1:40
Notice that execution time for number `n` is roughly sum of ececution times
for numbers `n-1` and `n-2`.
</details>
### Exercise 6
Improve the speed of code from exercise 5 by using a dictionary where you
store a mapping of `n` to `fib(n)`. Measure the performance of this function
for the same range of values as in exercise 5.
<details>
<summary>Solution</summary>
Use the following code:
```
@@ -422,8 +405,17 @@ julia> @time fib2(200)
As you can see the code does less allocations and is faster now.
</details>
### Exercise 7
Create a vector containing named tuples representing elements of a 4x4 grid.
So the first element of this vector should be `(x=1, y=1)` and last should be
`(x=4, y=4)`. Store the vector in variable `v`.
<details>
<summary>Solution</summary>
Since we are asked to create a vector we can write:
```
@@ -470,8 +462,17 @@ julia> [(; x, y) for x in 1:4, y in 1:4]
(x = 4, y = 1) (x = 4, y = 2) (x = 4, y = 3) (x = 4, y = 4)
```
</details>
### Exercise 8
The `filter` function allows you to select some values of an input collection.
Check its documentation first. Next, use it to keep from the vector `v` from
exercise 7 only elements whose sum is even.
<details>
<summary>Solution</summary>
To get help on the `filter` function write `?filter`. Next run:
```
@@ -487,8 +488,16 @@ julia> filter(e -> iseven(e.x + e.y), v)
(x = 4, y = 4)
```
</details>
### Exercise 9
Check the documentation of the `filter!` function. Perform the same operation
as asked in exercise 8 but using `filter!`. What is the difference?
<details>
<summary>Solution</summary>
To get help on the `filter!` function write `?filter!`. Next run:
```
@@ -518,8 +527,21 @@ julia> v
Notice that `filter` allocated a new vector, while `filter!` updated the `v`
vector in place.
</details>
### Exercise 10
Write a function that takes a number `n`. Next it generates two independent
random vectors of length `n` and returns their correlation coefficient.
Run this function `10000` times for `n` equal to `10`, `100`, `1000`,
and `10000`.
Create a plot with four histograms of distribution of computed Pearson
correlation coefficient. Check in the Plots.jl package which function can be
used to plot histograms.
<details>
<summary>Solution</summary>
You can use for example the following code:
```