Closed a-torgovitsky closed 4 years ago
a-torgovitsky
commented
4 years ago Oops, I see this is partly because I had `progress = TRUE' on
For dkqs_cone the `progress = TRUE' should just tell us how many bootstraps have been completed and how many are remaining.
a-torgovitsky
commented
4 years ago The progress indicator works for parallel_example but not for a single example with progress = TRUE.
For example
library("linearprog")
func_full_info <- function(df){
beta = NULL
y_list = sort(unique(df[,"Y"]))
n = dim(df)[1]
yn = length(y_list)
for (i in 1:yn){
beta_i = sum((df[,"Y"] == y_list[i]) * (df[,"D"] == 1))/n
beta = c(beta,c(beta_i))
}
beta = as.matrix(beta)
return(beta)
}
func_two_moment <- function(df){
beta = matrix(c(0,0), nrow = 2)
n = dim(df)[1]
beta[1] = sum(df[,"Y"] * df[,"D"])/n
beta[2] = sum(df[,"D"])/n
return(beta)
}
N = dim(sampledata)[1]
J1 = length(unique(sampledata[,"Y"]))
yp = seq(0,1,1/(J1-1))
A_obs_twom = matrix(c(rep(0,J1), yp, rep(0,J1), rep(1, J1)), nrow = 2,
byrow = TRUE)
A_target = matrix(c(yp, yp), nrow = 1)
tau = sqrt(log(N)/N)
r <- dkqs(df = sampledata,
A_obs = A_obs_twom,
A_tgt = A_target,
func_obs = func_two_moment,
beta_tgt = 0.375,
bs_seed = 1,
bs_num = 10000,
p_sig = 2,
tau_input = tau,
solver = "gurobi",
cores = 8,
progress = TRUE)Not sure what the explaintion is...
EDIT Oh, I see, you said in the email that it is not implemented for parallel yet. Makes sense!
a-torgovitsky
commented
4 years ago This output doesn't look so good:
library("linearprog")
func_full_info <- function(df){
beta = NULL
y_list = sort(unique(df[,"Y"]))
n = dim(df)[1]
yn = length(y_list)
for (i in 1:yn){
beta_i = sum((df[,"Y"] == y_list[i]) * (df[,"D"] == 1))/n
beta = c(beta,c(beta_i))
}
beta = as.matrix(beta)
return(beta)
}
func_two_moment <- function(df){
beta = matrix(c(0,0), nrow = 2)
n = dim(df)[1]
beta[1] = sum(df[,"Y"] * df[,"D"])/n
beta[2] = sum(df[,"D"])/n
return(beta)
}
N = dim(sampledata)[1]
J1 = length(unique(sampledata[,"Y"]))
yp = seq(0,1,1/(J1-1))
A_obs_twom = matrix(c(rep(0,J1), yp, rep(0,J1), rep(1, J1)), nrow = 2,
byrow = TRUE)
A_target = matrix(c(yp, yp), nrow = 1)
tau = sqrt(log(N)/N)
dkqs_farg <- list(df = sampledata,
A_obs = A_obs_twom,
A_tgt = A_target,
func_obs = func_two_moment,
bs_seed = 1,
bs_num = 100,
p_sig = 2,
tau_input = tau,
solver = "gurobi",
cores = 1,
progress = FALSE)
invertci_dkqs <- invertci(f = dkqs,
farg = dkqs_farg,
alpha = 0.05,
lb0 = 0,
lb1 = 0.4,
ub0 = 1,
ub1 = 0.6,
tol = 0.001,
df_ci = NULL,
progress = TRUE)Gives this:
< Constructing confidence interval for alpha = 0.05 >
=== Computing upper bound of confidence interval ===
Bootstrap completed!
Iteration Test point Lower bound Upper bound p-value Decision
Left end-pt. 0.60000 0.60000 NA 0.78000 Do not reject
Bootstrap completed!
Right end-pt. 1.00000 NA 1.00000 0.00000 Reject
Bootstrap completed!
1 0.80000 0.60000 1.00000 0.00000 Reject
Bootstrap completed!
2 0.70000 0.60000 0.80000 0.00000 Reject
Bootstrap completed!
3 0.65000 0.60000 0.70000 0.05000 Do not reject
Bootstrap completed!
4 0.67500 0.65000 0.70000 0.00000 Reject
Bootstrap completed!
5 0.66250 0.65000 0.67500 0.00000 Reject
Bootstrap completed!
6 0.65625 0.65000 0.66250 0.02000 Reject
Bootstrap completed!
7 0.65312 0.65000 0.65625 0.03000 Do not reject
Bootstrap completed!
8 0.65469 0.65312 0.65625 0.02000 Reject
Bootstrap completed!
9 0.65391 0.65312 0.65469 0.02000 Reject
Bootstrap completed!
>>> Length of interval is below tolerance level. Bisection method is completed.
=== Computing lower bound of confidence interval ===
Bootstrap completed!
Iteration Test point Lower bound Upper bound p-value Decision
Left end-pt. 0.00000 0.00000 NA 0.00000 Reject
Bootstrap completed!
Right end-pt. 0.40000 NA 0.40000 0.74000 Do not reject
Bootstrap completed!
1 0.20000 0.00000 0.40000 0.00000 Reject
Bootstrap completed!
2 0.30000 0.20000 0.40000 0.00000 Reject
Bootstrap completed!
3 0.35000 0.30000 0.40000 0.00000 Reject
Bootstrap completed!
4 0.37500 0.35000 0.40000 0.26000 Do not reject
Bootstrap completed!
5 0.36250 0.35000 0.37500 0.03000 Do not reject
Bootstrap completed!
6 0.35625 0.35000 0.36250 0.00000 Reject
Bootstrap completed!
7 0.35938 0.35625 0.36250 0.01000 Reject
Bootstrap completed!
8 0.36094 0.35938 0.36250 0.01000 Reject
Bootstrap completed!
9 0.36172 0.36094 0.36250 0.01000 Reject
Bootstrap completed!
>>> Length of interval is below tolerance level. Bisection method is completed.
-----------------------------------
< Significance level: 0.05 >
Tolerance level: 0.001.
Total number of iterations: 20.
Confidence interval: [0.362, 0.654].It's nice to have the Progress x% while it is running, but we don't need Bootstrap completed! to stay there.
conroylau
commented
4 years ago Done for the non-parallel case! Now Progress x% while running and there won't be a completion message staying on the console.
Will continue to work on the parallel case. Thanks!
conroylau
commented
4 years ago Done - updated the code for both parallel and nonparallel case.
The following code for invertci:
func_full_info <- function(df){
beta = NULL
y_list = sort(unique(df[,"Y"]))
n = dim(df)[1]
yn = length(y_list)
for (i in 1:yn){
beta_i = sum((df[,"Y"] == y_list[i]) * (df[,"D"] == 1))/n
beta = c(beta,c(beta_i))
}
beta = as.matrix(beta)
return(beta)
}
func_two_moment <- function(df){
beta = matrix(c(0,0), nrow = 2)
n = dim(df)[1]
beta[1] = sum(df[,"Y"] * df[,"D"])/n
beta[2] = sum(df[,"D"])/n
return(beta)
}
N = dim(sampledata)[1]
J1 = length(unique(sampledata[,"Y"]))
yp = seq(0,1,1/(J1-1))
A_obs_twom = matrix(c(rep(0,J1), yp, rep(0,J1), rep(1, J1)), nrow = 2,
byrow = TRUE)
A_target = matrix(c(yp, yp), nrow = 1)
tau = sqrt(log(N)/N)
dkqs_farg <- list(df = sampledata,
A_obs = A_obs_twom,
A_tgt = A_target,
func_obs = func_two_moment,
bs_seed = 1,
bs_num = 100,
p_sig = 2,
tau_input = tau,
solver = "gurobi",
cores = 1,
progress = FALSE)
invertci_dkqs <- invertci(f = dkqs,
farg = dkqs_farg,
alpha = 0.05,
lb0 = 0,
lb1 = 0.4,
ub0 = 1,
ub1 = 0.6,
tol = 0.001,
df_ci = NULL,
progress = TRUE)gives the following output:
< Constructing confidence interval for alpha = 0.05 >
=== Computing upper bound of confidence interval ===
Iteration Lower bound Upper bound Test point p-value Reject?
Left end pt. 0.60000 NA 0.60000 0.73000 FALSE
Right end pt. NA 1.00000 1.00000 0.00000 TRUE
1 0.60000 1.00000 0.80000 0.00000 TRUE
2 0.60000 0.80000 0.70000 0.00000 TRUE
3 0.60000 0.70000 0.65000 0.05000 FALSE
4 0.65000 0.70000 0.67500 0.00000 TRUE
5 0.65000 0.67500 0.66250 0.00000 TRUE
6 0.65000 0.66250 0.65625 0.02000 TRUE
7 0.65000 0.65625 0.65312 0.03000 FALSE
8 0.65312 0.65625 0.65469 0.02000 TRUE
9 0.65312 0.65469 0.65391 0.02000 TRUE
>>> Length of interval is below tolerance level. Bisection method is completed.
=== Computing lower bound of confidence interval ===
Iteration Lower bound Upper bound Test point p-value Reject?
Left end pt. 0.00000 NA 0.00000 0.00000 TRUE
Right end pt. NA 0.40000 0.40000 0.79000 FALSE
1 0.00000 0.40000 0.20000 0.00000 TRUE
2 0.20000 0.40000 0.30000 0.00000 TRUE
3 0.30000 0.40000 0.35000 0.00000 TRUE
4 0.35000 0.40000 0.37500 0.26000 FALSE
5 0.35000 0.37500 0.36250 0.03000 FALSE
6 0.35000 0.36250 0.35625 0.00000 TRUE
7 0.35625 0.36250 0.35938 0.01000 TRUE
8 0.35938 0.36250 0.36094 0.01000 TRUE
9 0.36094 0.36250 0.36172 0.01000 TRUE
>>> Length of interval is below tolerance level. Bisection method is completed.Applying the print command on invertci_dkqs, i.e.
print(invertci_dkqs)gives
< Significance level: 0.05 >
Total number of iterations: 20.
Confidence interval: [0.36211, 0.65352].The following is a sample when the bootstrap procedure is running with progress in the list dkqs_farg being set to TRUE:
< Constructing confidence interval for alpha = 0.05 >
=== Computing upper bound of confidence interval ===
Iteration Lower bound Upper bound Test point p-value Reject?
Left end pt. 0.60000 NA 0.60000 0.73000 FALSE
Right end pt. NA 1.00000 1.00000 0.00000 TRUE
1 0.60000 1.00000 0.80000 0.00000 TRUE
|======= | 35%Thank you!
It should be analogous to the way
lmworks.For example, in RStudio,
This doesn't print anything
While this
gives the same output as
Our functions should work the same way