quant-econ Solutions: Julia Essentials

Solutions for http://quant-econ.net/jl/julia_essentials.html

Exercise 1

Part 1 solution:

Here's one possible solution



In [1]:

    
x_vals = [1, 2, 3]
y_vals = [1, 1, 1]
sum([x * y for (x, y) in zip(x_vals, y_vals)])









    Out[1]:





6

Part 2 solution:

One solution is



In [2]:

    
sum([x % 2 == 0 for x in 0:99])









    Out[2]:





50

This also works



In [3]:

    
sum(map(x -> x % 2 == 0, 0:99))









    Out[3]:





50

Part 3 solution

Here's one possibility



In [4]:

    
pairs = ((2, 5), (4, 2), (9, 8), (12, 10))
sum([(x % 2 == 0) & (y % 2 == 0) for (x, y) in pairs])









    Out[4]:





2

Exercise 2



In [9]:

    
p(x, coeff) = sum([a * x^(i-1) for (i, a) in enumerate(coeff)])









    Out[9]:





p (generic function with 1 method)



In [10]:

    
p(1, (2, 4))









    Out[10]:





6

Exercise 3

Here's one solution:



In [23]:

    
function f(string)
    count = 0
    for letter in string
        if (letter == uppercase(letter)) & isalpha(letter)
            count += 1
        end
    end
    return count
end
    
f("The Rain in Spain")









    Out[23]:





3

Exercise 4

Here's a solution:



In [26]:

    
function f(seq_a, seq_b)
    is_subset = true
    for a in seq_a
        if !(a in seq_b)
            is_subset = false
        end
    end
    return is_subset
end

# == test == #

println(f([1, 2], [1, 2, 3]))
println(f([1, 2, 3], [1, 2]))









    



true
false

If we use the Set data type then the solution is easier



In [29]:

    
f(seq_a, seq_b) = issubset(Set(seq_a), Set(seq_b))

println(f([1, 2], [1, 2, 3]))
println(f([1, 2, 3], [1, 2]))









    



true
false

Exercise 5



In [30]:

    
function linapprox(f, a, b, n, x)
    #=
    Evaluates the piecewise linear interpolant of f at x on the interval 
    [a, b], with n evenly spaced grid points.

    =#
    length_of_interval = b - a
    num_subintervals = n - 1
    step = length_of_interval / num_subintervals  

    # === find first grid point larger than x === #
    point = a
    while point <= x
        point += step
    end
    
    # === x must lie between the gridpoints (point - step) and point === #
    u, v = point - step, point  

    return f(u) + (x - u) * (f(v) - f(u)) / (v - u)
end









    Out[30]:





linapprox (generic function with 1 method)

Let's test it



In [39]:

    
f(x) = x^2
g(x) = linapprox(f, -1, 1, 3, x)









    Out[39]:





g (generic function with 1 method)



In [40]:

    
using PyPlot



In [44]:

    
x_grid = linspace(-1, 1, 100)
y_vals = map(f, x_grid)
y_approx = map(g, x_grid)
plot(x_grid, y_vals, label="true")
plot(x_grid, y_approx, label="approximation")
legend()









    












    Out[44]:





PyObject <matplotlib.legend.Legend object at 0x7f25a0f8d0d0>

Exercise 6



In [2]:

    
f = open("us_cities.txt", "r")
total_pop = 0
for line in eachline(f)
    city, population = split(line, ':')            # Tuple unpacking
    total_pop += int(population)
end
close(f)
println("Total population = $total_pop")









    



Total population = 23831986



In [ ]: