numeconcopenhagen / lectures-2019 Goto Github PK

In VsCode we have established a connection to our Github repository and I can succesfully sync the code created by other members of my group. However, when I try to push my own code to the repository. I get the following error (submitted in the image).

How do I solve this issue?

Hi

We have run into a problem when solving our model. At one point we maximize and end up with an expression containing a function and the derivative of that function (something like: a + f'(x) + b/f(x) = 0). Then we want to substitute in the input of f(x), but keeping f'(x) as it is. Just using "subs" both f(x) and f'(x) gets substituted. We believe this is because python writes the derivative as d/dxf(x) and thus cannot distinguish between the two functions.
We have also tried to use "replace" and then indicating that only the last element should be replaced, but that does not work either.

Does anyone have a suggestion to what else we can try? :)

I get a NameError when assigning b to a

>>> a = b
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'b' is not defined

Hi

In your mail from 27th of March it says:
3. Remember to get GitHub working before the last minute. See how to get your group repository in the bottom of lecture 5 (under "Group sign-up"). Additionally, see step 2 and 3 in the guide on git.
But when pressing the link in Lecture 5 I just get to the assignment and not a guide on how to sign up as a group. We do not have a repostitory on: github.com/NumEconCopenhagen/ so we are rather confused on where to upload our dataproject

Regards, Jacob

Hello!

Is it possible to lock a column to multiple specific variables in your dataframe?
We want to select the variables called "Y" and "Y_L". We can't find a simple way to select these variables. The way we do it now works, but it is a troublesome way to do it. We are right now dropping every variable that we don't need. It works when you don't have that many variables, but it is not that effective when you have a lot.

It would be nice if we could write something like:
"samle1234=samle1234[samle1234.variable==["Y" & "Y_L"]]
But unfortunately this doesn't work.

The code:

output:

We hope that you can help
Best regards

I couldn't raise an issue in the PS1 exercise repository, so I am creating it here instead. Feel free to (re)move the post if necessary.

My issue/question is regarding the solvers in PS1: As the utility is dependent on the initial guess how should we define the initial guess so it is robust to a large variance in the denominator?

To illustrate consider the following:

from scipy import optimize

def utility_function(x,alpha):
    u = 1
    for x_now,alpha_now in zip(x,alpha):
        u *= np.max(x_now,0)**alpha_now
    return u

def expenditures(x,p):
    E = 0
    for x_now,p_now in zip(x,p):
        E += p_now*x_now
    return E

def print_solution(x,alpha,I,p):
    
    # a. x values
    text = 'x = ['
    for x_now in x:
        text += f'{x_now:.2f} '
    text += f']\n'
    
    # b. utility
    u = utility_function(x,alpha)    
    text += f'utility = {u:.3f}\n'
    
    # c. expenditure vs. income
    E =  expenditures(x,p)
    text += f'E = {E:.2f} <= I = {I:.2f}\n'
    
    # d. expenditure shares
    e = p*x/I
    text += 'expenditure shares = ['
    for e_now in e:
        text += f'{e_now:.2f} '
    text += f']'        
        
    print(text)

alpha = np.ones(5)/5
p = np.array([1,2,3,4,5])
I = 10

Solution using a constrained optimizer:

# a. contraint function (negative if violated)
constraints = ({'type': 'ineq', 'fun': lambda x:  I-expenditures(x,p)})
bounds = [(0,I/p_now) for p_now in p]

# b. call optimizer
initial_guess = (I/p)/100# some guess
res = optimize.minimize(
    lambda x: -utility_function(x,alpha),initial_guess,
    method='SLSQP',bounds=bounds,constraints=constraints)

# c. print result
print_solution(res.x,alpha,I,p)

Changing initial_guess = (I/p)/100 to initial_guess = (I/p)/1 changes
the utility from 0.768 to 0.000.

Conversely, for the solution using an unconstrained optimizer:

# a. define objective function
def unconstrained_objective(x,alpha,I,p):
    
    penalty = 0
    E = expenditures(x,p)
    if E >= I:
        ratio = I/E
        x *= ratio # now p*x = I
        penalty = 1000*(E-I)**2
    
    u = utility_function(x,alpha)
    return -u + penalty

# b. call optimizer
initial_guess = (I/p)/10000
res = optimize.minimize(
    unconstrained_objective,initial_guess,
    method='Nelder-Mead',args=(alpha,I,p))

# c. print result
print_solution(res.x,alpha,I,p)  

Changing the code from initial_guess = (I/p)/1 to initial_guess = (I/p)/10000 changes the utility from 0.768 to 0.000

Looking instead at the loops we consider the code snippet using raw loops:

N = 15 # number of points in each dimension
fac = np.linspace(0,1,N) # vector betweein 0 and 1
x_max = I/p # maximum x so E = I

u_best = -np.inf
x_best = np.empty(5)
for x1 in fac:
    for x2 in fac:
        for x3 in fac:
            for x4 in fac:
                for x5 in fac:
                    x = np.array([x1,x2,x3,x4,x5])*x_max
                    E = expenditures(x,p)
                    if E <= I:
                        u_now = utility_function(x,alpha)
                        if u_now > u_best:
                            x_best = x
                            u_best = u_now

print_solution(x_best,alpha,I,p)

If we change the original value N = 15 to N = 6 the utility changes from 0.758 to 0.768 (same as the the solvers above), however changing it to N = 5 yields a utility of 0.000.
How do we determine the "optimal" or correct number of points in each dimension for our linespacing?

Write here if you are searching for other students to form a group with.

Hi Jeppe,
In the first assignment you set rho to 2. This seems to give negative utility of consumtion no matter what. Is this a mistake? It makes more sense if it is 0.2.

Hello

We use data from DST which we have converted into a Dataframe. We want to create a new column called NX which is the net export: NX=X-M.

But how do you do that? We can't figure it out...

code:
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import pydst
dst = pydst.Dst(lang = 'da')

columns_dict = {}
columns_dict['TRANSAKT'] = 'variable'
columns_dict['PRISENHED'] = 'unit'
columns_dict['TID'] = 'year'
columns_dict['INDHOLD'] = 'value'

var_dict = {} # var is for variable
#var_dict['P.1 Output'] = 'Y'
#var_dict['P.3 Final consumption expenditure'] = 'C'
#var_dict['P.3 Government consumption expenditure'] = 'G'
#var_dict['P.5 Gross capital formation'] = 'I'
#var_dict['P.6 Export of goods and services'] = 'X'
#var_dict['P.7 Import of goods and services'] = 'M'

#VARIABLES
var_dict['B.1*g Bruttpmationalprodukt, BNP'] = 'Y'
var_dict['P.31 Privatforbrug'] = 'C'
var_dict['P.31 Forbrugsudgifter i non-profit institutioner rettet mod husholdninger (NPISH)'] = 'NPISH'
var_dict['P.3 Offentlige forbrugsudgifter'] = 'G'
var_dict['P.5g Bruttoinvesteringer'] = 'I'
var_dict['P.6 Eksport af varer og tjenester'] = 'X'
var_dict['P.7 Import af varer og tjenester'] = 'M'

#UNITS
unit_dict = {}
#unit_dict['2010-prices, chained values'] = 'real'
#unit_dict['Current prices'] = 'nominal'
unit_dict['2010-priser, kædede værdier'] = 'realle'
unit_dict['løbende priser'] = 'nominelle'

#CODE
nan1 = dst.get_data(table_id = 'NAN1', variables={'TRANSAKT':[''], 'PRISENHED':[''], 'TID':['*']})
nan1.rename(columns = columns_dict, inplace = True)

for key,value in var_dict.items():
nan1.variable.replace(key,value,inplace=True)

for key,value in unit_dict.items():
nan1.unit.replace(key,value, inplace=True)

I = False
for key,value in var_dict.items():
I = I | (nan1.variable == value)
nan1 = nan1[I]

nan1.groupby(['variable','unit']).describe()

X = nan1["year"]>2000 #laver en ny variabel hvor X inderholder en variabel year som er større end 2000. giver nan1= nan1[x] for at sige at den skal være lig med det.
nan1 = nan1[X]

Y=nan1["unit"]=="Pr. indbygger, løbende priser, (1000 kr.)"
nan1 = nan1[Y]

df=pd.DataFrame(data=nan1)

df.sort_values(by=["year","variable"])

print(df)

Why is the first element of list A equal to zero when I run this code?

A = [1,2,3] 
B = A 
B[0] = 0
print(A[0])