update layout of all exercises
This commit is contained in:
Bogumił Kamiński
@@ -13,89 +13,8 @@ independently and uniformly from the [0,1[ interval.
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Create a data frame using data from this matrix using auto-generated
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column names.
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### Exercise 2
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Now, using matrix `mat` create a data frame with randomly generated
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column names. Use the `randstring` function from the `Random` module
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to generate them. Store this data frame in `df` variable.
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### Exercise 3
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Create a new data frame, taking `df` as a source that will have the same
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columns but its column names will be `y1`, `y2`, `y3`, `y4`.
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### Exercise 4
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Create a dictionary holding `column_name => column_vector` pairs
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using data stored in data frame `df`. Save this dictionary in variable `d`.
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### Exercise 5
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Create a data frame back from dictionary `d` from exercise 4. Compare it
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with `df`.
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### Exercise 6
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For data frame `df` compute the dot product between all pairs of its columns.
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Use the `dot` function from the `LinearAlgebra` module.
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### Exercise 7
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Given two data frames:
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```
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julia> df1 = DataFrame(a=1:2, b=11:12)
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2×2 DataFrame
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Row │ a b
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│ Int64 Int64
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─────┼──────────────
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1 │ 1 11
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2 │ 2 12
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julia> df2 = DataFrame(a=1:2, c=101:102)
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2×2 DataFrame
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Row │ a c
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│ Int64 Int64
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─────┼──────────────
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1 │ 1 101
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2 │ 2 102
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```
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vertically concatenate them so that only columns that are present in both
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data frames are kept. Check the documentation of `vcat` to see how to
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do it.
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### Exercise 8
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Now append to `df1` table `df2`, but add only the columns from `df2` that
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are present in `df1`. Check the documentation of `append!` to see how to
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do it.
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### Exercise 9
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Create a `circle` data frame, using the `push!` function that will store
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1000 samples of the following process:
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* draw `x` and `y` uniformly and independently from the [-1,1[ interval;
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* compute a binary variable `inside` that is `true` if `x^2+y^2 < 1`
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and is `false` otherwise.
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Compute summary statistics of this data frame.
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### Exercise 10
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Create a scatterplot of `circle` data frame where its `x` and `y` axis
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will be the plotted points and `inside` variable will determine the color
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of the plotted point.
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# Solutions
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<details>
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<summary>Show!</summary>
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### Exercise 1
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Solution:
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<summary>Solution</summary>
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```
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julia> using DataFrames
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@@ -120,9 +39,16 @@ julia> DataFrame(mat, :auto)
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5 │ 0.714515 0.861872 0.971521 0.176768
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```
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</details>
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### Exercise 2
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Solution:
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Now, using matrix `mat` create a data frame with randomly generated
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column names. Use the `randstring` function from the `Random` module
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to generate them. Store this data frame in `df` variable.
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<details>
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<summary>Solution</summary>
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```
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julia> using Random
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@@ -139,10 +65,16 @@ julia> df = DataFrame(mat, [randstring() for _ in 1:4])
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5 │ 0.714515 0.861872 0.971521 0.176768
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```
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</details>
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### Exercise 3
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Solution:
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Create a new data frame, taking `df` as a source that will have the same
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columns but its column names will be `y1`, `y2`, `y3`, `y4`.
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<details>
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<summary>Solution</summary>
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```
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julia> DataFrame(["y$i" => df[!, i] for i in 1:4])
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5×4 DataFrame
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@@ -170,9 +102,15 @@ julia> rename(df, string.("y", 1:4))
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5 │ 0.714515 0.861872 0.971521 0.176768
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```
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</details>
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### Exercise 4
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Solution:
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Create a dictionary holding `column_name => column_vector` pairs
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using data stored in data frame `df`. Save this dictionary in variable `d`.
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<details>
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<summary>Solution</summary>
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```
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julia> d = Dict([n => df[:, n] for n in names(df)])
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@@ -194,9 +132,15 @@ Dict{Symbol, AbstractVector} with 4 entries:
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Symbol("5Caz55k0") => [0.0353994, 0.0691152, 0.980079, 0.0697535, 0.971521]
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```
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</details>
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### Exercise 5
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Solution:
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Create a data frame back from dictionary `d` from exercise 4. Compare it
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with `df`.
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<details>
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<summary>Solution</summary>
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```
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julia> DataFrame(d)
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@@ -215,9 +159,15 @@ Note that columns of a data frame are now sorted by their names.
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This is done for `Dict` objects because such dictionaries do not have
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a defined order of keys.
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</details>
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### Exercise 6
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Solution:
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For data frame `df` compute the dot product between all pairs of its columns.
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Use the `dot` function from the `LinearAlgebra` module.
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<details>
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<summary>Solution</summary>
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```
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julia> using LinearAlgebra
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@@ -232,9 +182,36 @@ julia> pairwise(dot, eachcol(df))
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1.50558 1.18411 0.909744 1.47431
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```
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</details>
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### Exercise 7
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Solution:
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Given two data frames:
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```
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julia> df1 = DataFrame(a=1:2, b=11:12)
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2×2 DataFrame
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Row │ a b
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│ Int64 Int64
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─────┼──────────────
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1 │ 1 11
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2 │ 2 12
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julia> df2 = DataFrame(a=1:2, c=101:102)
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2×2 DataFrame
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Row │ a c
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│ Int64 Int64
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─────┼──────────────
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1 │ 1 101
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2 │ 2 102
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```
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vertically concatenate them so that only columns that are present in both
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data frames are kept. Check the documentation of `vcat` to see how to
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do it.
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<details>
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<summary>Solution</summary>
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```
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julia> vcat(df1, df2, cols=:intersect)
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@@ -255,9 +232,16 @@ julia> vcat(df1, df2)
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ERROR: ArgumentError: column(s) c are missing from argument(s) 1, and column(s) b are missing from argument(s) 2
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```
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</details>
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### Exercise 8
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Solution:
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Now append to `df1` table `df2`, but add only the columns from `df2` that
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are present in `df1`. Check the documentation of `append!` to see how to
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do it.
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<details>
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<summary>Solution</summary>
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```
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julia> append!(df1, df2, cols=:subset)
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@@ -271,9 +255,20 @@ julia> append!(df1, df2, cols=:subset)
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4 │ 2 missing
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```
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</details>
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### Exercise 9
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Solution
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Create a `circle` data frame, using the `push!` function that will store
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1000 samples of the following process:
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* draw `x` and `y` uniformly and independently from the [-1,1[ interval;
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* compute a binary variable `inside` that is `true` if `x^2+y^2 < 1`
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and is `false` otherwise.
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Compute summary statistics of this data frame.
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<details>
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<summary>Solution</summary>
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```
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circle=DataFrame()
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@@ -287,9 +282,16 @@ describe(circle)
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We note that mean of variable `inside` is approximately π.
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</details>
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### Exercise 10
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Solution:
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Create a scatterplot of `circle` data frame where its `x` and `y` axis
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will be the plotted points and `inside` variable will determine the color
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of the plotted point.
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<details>
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<summary>Solution</summary>
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```
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using Plots
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