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Mean Imputation

Gira Ahmed
Posted 10 years ago

I try to generate Simple Random sample with missing data, I do the following

  • generate random sample for x, y
  • delete some observation
  • use mean Imputation to replace it on deleted values

here the program

<< MultivariateStatistics`;
   y = RandomSample[Range[200], 20];
  x = RandomSample[Range[100], 20] + 0.9*y;
  iter = 1000;
  n = 12;
  r = 9;
  k = n - r;
  c1 = 0; c2 = 0; c3 = 0; c4 = 0; c5 = 0; c6 = 0;
  c7 = 0; c8 = 0; c9 = 0; c10 = 0; c11 = 0; c12 = 0;
  c13 = 0; c14 = 0; c15 = 0; c16 = 0; c17 = 0; c18 = 0;
  c19 = 0; c20 = 0; c21 = 0; c22 = 0; c23 = 0; c24 = 0;
  c25 = 0; c26 = 0;
  Array[s1, n]; Array[s2, n]; Array[h, n];
  data = Table[{y[[i]], x[[i]]}, {i, 1, 20}];
  f1 = Mean[y];; f2 = Mean[x];
  Sx2 = Variance[x];
  Sy2 = Variance[y];
  Cy = Sy2/(f1)^2;
  Cx = Sx2/(f2)^2;
  SSS = Correlation[x, y];
  Sxy = SSS*Sqrt[Sx2*Sy2];
  Do 
    sa = RandomkSubset[data, n];
    s1 = Table[sa[[i, 1]], {i, 1, n}];
    s2 = Table[sa[[i, 2]], {i, 1, n}];
    xn = Mean[s2];  
    h = Table[i, {i, 1, n}];
    dropss = RandomKSubset[h, k];
    comp = Complement[h, dropss];
    Do[list1[i] = sa[[dropss[[i]]]], {i, 1, k}];   here doesn't work
      list2 = Table[list1[[i]], {i, 1, k}];
      a1 = Table[list2[[i, 1]], {i, 1, k}];
      a2 = Table[list2[[i, 2]], {i, 1, k}];
    Do[list3[i] = sa[[comp[[i]]]], {i, 1, n - k}];
      list4 = Table[list3[i], {i, 1, n - k}];
      a3 = Table[list4[[i, 1]], {i, 1, n - k}];
      a4 = Table[list4[[i, 2]], {i, 1, n - k}];
      yr = Mean[a3];
      xr = Mean[a4];
  (* Mean method of imputaion*)
  Do[w1[i] = yr, {i, 1, k}];
  w2 = Table[w1[i], {i, 1, k}];
  meth1 = Join[w2, a3];
  m1[j] = Mean[meth1];
  c1 = c1 + m1[j];
  c2 = c2 + (m1[j] - f1)^2;
  ,
  {j, 1, iter}];

need help.

thanks in advance
POSTED BY: Gira Ahmed
6 Replies
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You have a syntax error, instead of 

list[3]=new;

to replace the third element, you should use

list[[3]]=new;

In general, using list manipulation functions will avoid these sort of syntax errors. Here are a few comments.

  Array[s1, n]; Array[s2, n]; Array[h, n];

This is unnecessary. You don't need to initialize the arrays.

Table[{y[[i]], x[[i]]}, {i, 1, 20}];

There are a number of other ways to do this like

MapThread[List,{y,x},1]

  f1 = Mean[y];;

Should not have double semicolons.

Table[i, {i, 1, n}];

Could be Range[n]
POSTED BY: Todd Rowland
Gira Ahmed
Gira Ahmed
Posted 10 years ago

I want to study the effect of mean imputation (and other methods later on..). My main idea is

1- generate a population N=20 sample for two variables x, y

2- then draw all samples without replacement (n=12). (calculate descriptive statistics)

3- drop a random values (k=3) to be missing.

4- compensate this values by mean of other respondent values.

5- compute bias and MSE new imputed data.
POSTED BY: Gira Ahmed

What are you getting as a result?
What do you want as a result??

POSTED BY: Bruce Miller

http://community.wolfram.com/groups/-/m/t/270507 has instructions for formatting code and lists nicely.

POSTED BY: Bruce Miller

It's not clear what is the question here. Is there a problem with the code you have?
POSTED BY: Daniel Lichtblau
Gira Ahmed
Gira Ahmed
Posted 10 years ago

Yes the problem with code RandomKSubset and Do list1

sa = RandomkSubset[data, n];

s1 = Table[sa[[i, 1]], {i, 1, n}];

s2 = Table[sa[[i, 2]], {i, 1, n}];

xn = Mean[s2]; h = Table[i, {i, 1, n}];

dropss = RandomKSubset[h, k]; ?

comp = Complement[h, dropss]; ?

Do[list1[i] = sa[[dropss[[i]]]], {i, 1, k}]; (* here doesn't work ? *)

POSTED BY: Gira Ahmed
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