Means at a time interval

[Johann Schweighofer]

Good evening, I'm having a datset with age as the independent variable. Age is given as a point number, i.e. 38.82 for a person who is 38 years and one month old. I'm supposed to get averages three times a year with 1/3 of a year between them. What would be a suitable way of doing so? The examples I've seen in the archives here are unfortunately all for cases where months or days are given as variables.

[Clyde Schechter]

How does 38.82 represent 38 years and one month? I don't get the logic. 0.82 years is just under 10 months. What is the coding here?

Averages of what? And what kind of data do you have.

When asking for help with coding, always show example data. And always use the -dataex- command to do that. If you are running version 16 or a fully updated version 15.1 or 14.2, -dataex- is already part of your official Stata installation. If not, run -ssc install dataex- to get it. Either way, run -help dataex- to read the simple instructions for using it. -dataex- will save you time; it is easier and quicker than typing out tables. It includes complete information about aspects of the data that are often critical to answering your question but cannot be seen from tabular displays or screenshots. It also makes it possible for those who want to help you to create a faithful representation of your example to try out their code, which in turn makes it more likely that their answer will actually work in your data.

When asking for help with code, always show example data. When showing example data, always use -dataex-.

[Johann Schweighofer]

Hi,

The 38.82 should say 38.082, as 1 day would be 1/365. The example from the data is:

* Example generated by -dataex-. To install: ssc install dataex
clear
input float age int ned float wg_c
 33.82192    .           .
 38.73425   34    .6387262
  39.5726   15  -.26885077
 37.59726 1847    .3594139
 33.47945  687    1.435173
 48.73973  191     .652995
30.852055   78   -.6590613
 41.23561 1578  -.11746673
37.736988   41    .6570308
 49.07397  275  -.51291037
 46.49041  347    .9565516
  49.5726 1097   -.4174039
33.983562   76  -.02404823
30.260275   87           .
 47.23836  245           .
 48.99178  288   -.5132165
 43.98904  258  -.39206505
 45.90411  214  -.29457772
 34.02192   47  -.05003124
 46.49041  530   .13128398
 39.30411   24   -.1025589
 44.15068  234   .19848554
 41.03014  211   .02514625
35.564384  216  -1.0500528
37.983562  134    .3762084
 39.06849  168  -.19183367
 34.70685   72  -.17194057
 48.97534    .           .
 40.89589 1184   -.4963008
 43.81918   77   .19263116
 32.81096  302  -1.1576884
30.561644   62   .17143446
 39.90137  213  -.04189511
 41.15342  213   -.3270496
 34.68219   27  -.22967665
31.476713  189    .1932102
 39.49041  609   .28018147
 48.72877   22           .
33.813698  103           .
35.750687  512  -.09796067
37.246574  178   -.1961045
 49.70137   33   .16929014
 48.83014   15 -.008842388
34.728767 2106   .09558595
34.767124  127      .31053
 48.28219  144   -.1133074
 30.67671   51    .7986139
 31.89589  275   1.0878267
     33.4  883   -.7202019
 43.54795  101   .10834202
33.471233 2453  .037116393
36.232876   73           .
37.265755   66           .
 38.26301   66           .
 39.26027   67           .
 40.25753   66           .
  41.2548   68           .
 42.25206   66           .
 43.28767   60           .
 44.26575   66           .
 30.19452 4168    1.477306
 44.08493    .           .
 39.40548  719   -.4020715
 31.61918  146    .0449803
32.769863  111   .08347853
 33.60822   80   .06447457
     34.8   99  .031088047
36.052055   96           .
34.942467  230   -.9551919
 45.11507  167  .035923693
 43.40822   60  -.10340602
 38.40274  306  -1.0120952
37.230137   18    .3292662
35.564384  111  -.06246318
30.860273  435   .09615602
 44.31781  215  -.52729386
34.515068   64           .
 49.77808  250  -.09715204
 46.50959  108    .2386982
 41.23288    .           .
38.065754   20   .06112232
 42.82466  238   -.3209757
 48.41096    .           .
 33.36986  116           .
36.854794   54           .
 31.52055  320  .024024496
 44.07123 3811  -.50606906
 30.28493  232    .2231882
 35.10685 2133   .54615825
37.172604  647  -1.3095566
     48.6    .           .
 49.07671  279   -.9648237
 37.89041  219   .21694164
 42.19178 2029    .1415003
 42.73699   68           .
32.315067  215  -.04253931
34.663013    .           .
 32.20548  833   .13294433
 38.36986  210 .0044039316
 30.72877  182   .02291842
end

I'm not sure if this is displayed right, the three variables are - from left to right - age (when getting unemployed), ned (for nonemployment duration) and wg_c (for wage change after fnding a new employment). So far, i found out that Stata requires

(thelibrary is closing, I will finish the post later.

[Johann Schweighofer]

(I can't edit the post above, so I'll continue here.)

So far i found out that Stata requires time series data to be integer values, so any option of using the time series tools doesn't work (and would be questionable given that it's not a time series anyways). I was looking at an solution via egen, but didn't see how i could make cuts every 4 months. Examples i have found have fixed boundaries, i.e. "egen urbcat1 = cut(urb), at(0,34,68,101)". I'm not really sure how to get the cutoff at each 1/3.

The problem is part of a larger task which basically involves recreating these 4 diagramms. I have the regressor for this given and should know how recreate that, but finding the details is part of this exercise. It's part of a course and we have seen Stata before, but everything about that was theoretical.

[Clyde Schechter]

gen  age_group = floor(10*floor(3*age)/3)

This will give you an integer value variable called age_group. age_group 300 will apply to all ages from 30.0 to 30 + 1/3, agegroup 303 will apply to ages from 30+1/3 to 30+2/3. agegroup 306 will apply to ages from 30+2/3 to 31. 310 gives you 31 to 31+1/3, etc. That is, read the age_group number as if it had a decimal point before the last digit and think of .3 as 1/3, .6 as 2/3.

[William Lisowski]

Although Clyde beat me to the posting, here's the approach I took. The variable gage is the age at the center of the interval represented by the group, e.g 30 1/6, 30 3/6, and 30 5/6.

generate group = floor(3*(age-30))+1
generate gage = 30 + (group-1)/3 + 1/6
table group, contents(n age min age max age min gage) format(%9.2f)

. table group, contents(n age min age max age min gage) format(%9.2f)

----------------------------------------------------------
    group |     N(age)    min(age)    max(age)   min(gage)
----------+-----------------------------------------------
        1 |          3       30.19       30.28       30.17
        2 |          1       30.56       30.56       30.50
        3 |          4       30.68       30.86       30.83
        5 |          3       31.48       31.62       31.50
        6 |          1       31.90       31.90       31.83
        7 |          2       32.21       32.32       32.17
        9 |          2       32.77       32.81       32.83
       11 |          5       33.37       33.61       33.50
       12 |          3       33.81       33.98       33.83
       13 |          1       34.02       34.02       34.17
       14 |          2       34.52       34.66       34.50
       15 |          6       34.68       34.94       34.83
       16 |          1       35.11       35.11       35.17
       17 |          2       35.56       35.56       35.50
       18 |          1       35.75       35.75       35.83
       19 |          2       36.05       36.23       36.17
       21 |          1       36.85       36.85       36.83
       22 |          4       37.17       37.27       37.17
       23 |          1       37.60       37.60       37.50
       24 |          3       37.74       37.98       37.83
       25 |          2       38.07       38.26       38.17
       26 |          2       38.37       38.40       38.50
       27 |          1       38.73       38.73       38.83
       28 |          3       39.07       39.30       39.17
       29 |          3       39.41       39.57       39.50
       30 |          1       39.90       39.90       39.83
       31 |          1       40.26       40.26       40.17
       33 |          1       40.90       40.90       40.83
       34 |          5       41.03       41.25       41.17
       37 |          2       42.19       42.25       42.17
       39 |          2       42.74       42.82       42.83
       40 |          1       43.29       43.29       43.17
       41 |          2       43.41       43.55       43.50
       42 |          2       43.82       43.99       43.83
       43 |          5       44.07       44.32       44.17
       46 |          1       45.12       45.12       45.17
       48 |          1       45.90       45.90       45.83
       50 |          3       46.49       46.51       46.50
       52 |          1       47.24       47.24       47.17
       55 |          1       48.28       48.28       48.17
       56 |          2       48.41       48.60       48.50
       57 |          5       48.73       48.99       48.83
       58 |          2       49.07       49.08       49.17
       59 |          1       49.57       49.57       49.50
       60 |          2       49.70       49.78       49.83
----------------------------------------------------------

[Nick Cox]

Stata requires time series data to be integer values

Not so; only time variables which want to include a tsset or xtset.

Even if age is a predictor, I can't think that being integer is essential for anything. It seems that you have a bunch of people with different ages. That's not a time series or panel data so far as I can see.