I am trying to generate a variable that is filled using a sequence of values starting at time==1.
The sequence changes everytime the variable rest1w changes from 0 to 1 or vice versa.
Firstly, I think I need to generate x, that is where the sequence restarts (see below example dataset). In my example, this is uniform, but in my full dataset the change varies (i.e. it does not change at every 5th observation).
list time restload trainload rest1w x in 1/15
+-----------------------------------------+
| time restload trainload rest1w x |
|-----------------------------------------|
1. | 1 .1994715 .4780615 0 1 |
2. | 2 .2077734 .471063 0 2 |
3. | 3 .2157595 .4641159 0 3 |
4. | 4 .2234298 .4572202 0 4 |
5. | 5 .2307843 .4503757 0 5 |
|-----------------------------------------|
6. | 6 .2378229 .4435827 1 1 |
7. | 7 .2445457 .436841 1 2 |
8. | 8 .2509527 .4301506 1 3 |
9. | 9 .2570438 .4235116 1 4 |
10. | 10 .2628191 .4169239 1 5 |
|-----------------------------------------|
11. | 11 .2682785 .4103876 0 1 |
12. | 12 .2734221 .4039026 0 2 |
13. | 13 .2782499 .397469 0 3 |
14. | 14 .2827618 .3910867 0 4 |
15. | 15 .2869579 .3847558 0 5 |
+-----------------------------------------+
Secondly, I need to generate a variable load. Which as per below shows how I would like to restart from time==1 everytime the sequence restarts. That is, at the second sequence where rest1w==0, load!=trainload.
The rule is that for each new sequence of 0's the value for load again goes back to the start of time (where time==1). This is demonstrated by the load values in the second sequence of 0's being exactly the same as the first sequence. In other words, where time==1, trainload==.478 then load==.478; BUT where time==11, then load==.478 (the clock essentially restarts for load so time==1) and in sequence where time==15, load==.450 (the same load as for where time==5). This is why I wanted to generate x, as I think I could just use that as my new time variable.
+-----------------------------------------+
| time restload trainload rest1w x load
|-----------------------------------------
1. | 1 .1994715 .4780615 0 1 .4780615
2. | 2 .2077734 .471063 0 2 .471063
3. | 3 .2157595 .4641159 0 3 .4641159
4. | 4 .2234298 .4572202 0 4 .4572202
5. | 5 .2307843 .4503757 0 5 .4503757
|-----------------------------------------
6. | 6 .2378229 .4435827 1 1 .1994715
7. | 7 .2445457 .436841 1 2 .2077734
8. | 8 .2509527 .4301506 1 3 .2157595
9. | 9 .2570438 .4235116 1 4 .2234298
10. | 10 .2628191 .4169239 1 5 .2307843
|-----------------------------------------
11. | 11 .2682785 .4103876 0 1 .4780615
12. | 12 .2734221 .4039026 0 2 .471063
13. | 13 .2782499 .397469 0 3 .4641159
14. | 14 .2827618 .3910867 0 4 .4572202
15. | 15 .2869579 .3847558 0 5 .4503757
+-----------------------------------------+
The below code only gives me an entry for where _n==1:
gen load==.
replace load = restload[_n==1] if rest1w==1
And I like the use of levelsof but haven't been able to get it to work (although it might work once I have generated x, but when using time it doesn't restart the sequence obviously).
gen load=.
levelsof x, local(levels)
foreach l of local levels {
replace load=trainload if rest1w==0
replace load=restload if rest1w==1
}
Thanks for any help!
I ended up cross-posting this on statalist.org and got two workable answers.
http://www.statalist.org/forums/forum/general-stata-discussion/general/1355917-fill-with-values-from-an-earlier-time-point
These were:
gen newtime = 1 if rest1w[_n - 1] != rest1w
replace newtime = newtime[_n - 1] + 1 if newtime == .
gen newload = cond(rest1w == 0, trainload[newtime], restload[newtime])
and...
gen newtime = 1
replace newtime = newtime[_n-1] + 1 if rest1w == rest1w[_n-1]
gen newload = .
replace newload = restload[newtime] if rest1w == 1
replace newload = trainload[newtime] if rest1w == 0
Related
I have a panel dataset in Stata with several countries and each country containing groups. I would like to rank the groups by country, according to the variable var1.
The structure of my dataset is as follows (the rank column is what I would like to achieve). Note that var1 is indeed constant within groups (it is just the within group average of another variable).
--country--|--groupId--|---time----|---var1----|---rank---
1 | 1 | 1 | 50 | 3
1 | 1 | 2 | 50 | 3
1 | 1 | 3 | 50 | 3
1 | 2 | 1 | 90 | 1
1 | 2 | 2 | 90 | 1
1 | 2 | 3 | 90 | 1
1 | 3 | 1 | 60 | 2
1 | 3 | 2 | 60 | 2
1 | 3 | 3 | 60 | 2
2 | 4 | 1 | 15 | 2
2 | 4 | 2 | 15 | 2
2 | 4 | 3 | 15 | 2
2 | 5 | 1 | 10 | 3
2 | 5 | 2 | 10 | 3
2 | 5 | 3 | 10 | 3
2 | 6 | 1 | 80 | 1
2 | 6 | 2 | 80 | 1
2 | 6 | 3 | 80 | 1
Among the options I have tried is:
sort country groupId
by country (groupId): egen rank = rank(var1)
However, I cannot achieve the desired result.
Thanks for the data example. There are two problems with your code. One is that as you want to rank from highest to lowest, you need to negate the argument to rank(). The second is that given the repetitions, you need to rank on one time only and then copy those ranks to other times.
This works with your data example, here edited to be input code. (See also the Stata tag wiki for that principle.)
clear
input country groupId time var1 rank
1 1 1 50 3
1 1 2 50 3
1 1 3 50 3
1 2 1 90 1
1 2 2 90 1
1 2 3 90 1
1 3 1 60 2
1 3 2 60 2
1 3 3 60 2
2 4 1 15 2
2 4 2 15 2
2 4 3 15 2
2 5 1 10 3
2 5 2 10 3
2 5 3 10 3
2 6 1 80 1
2 6 2 80 1
2 6 3 80 1
end
bysort country : egen wanted = rank(-var) if time == 1
bysort country groupId (time) : replace wanted = wanted[1]
assert rank == wanted
I have a dataset with only variable values:
clear
input value new_var
1 1
3 3
5 5
30 1
40 3
50 5
11 1
12 3
13 5
end
How can I generate a new variable new_var containing a repeating sequence of the first three observations in value?
Many ways to do it: here are two:
clear
input value new_var
1 1
3 3
5 5
30 1
40 3
50 5
11 1
12 3
13 5
end
egen index = seq(), to(3)
generate wanted = value[index]
generate direct = cond(mod(_n, 3) == 1, 1, cond(mod(_n, 3) == 2, 3, 5))
list, sep(3)
+-------------------------------------------+
| value new_var index wanted direct |
|-------------------------------------------|
1. | 1 1 1 1 1 |
2. | 3 3 2 3 3 |
3. | 5 5 3 5 5 |
|-------------------------------------------|
4. | 30 1 1 1 1 |
5. | 40 3 2 3 3 |
6. | 50 5 3 5 5 |
|-------------------------------------------|
7. | 11 1 1 1 1 |
8. | 12 3 2 3 3 |
9. | 13 5 3 5 5 |
+-------------------------------------------+
I have two datasets that I have appended together in Stata.
There is one variable, say Age in both data sets. I sorted the data so that the ages are in ascending order. I want to delete the observations in each dataset where the corresponding ages don't match.
Dataset 1:
Obs Age
1 7
2 8
3 10
4 5
Dataset 2:
Obs Age
1 10
2 5
3 9
4 7
Combined and sorted in ascending order:
Obs Age
1 5
2 5
3 7
4 7
5 8
6 9
7 10
8 10
So because the ages when sorted don't match up for observations 5 and 6, I want to delete them. Essentially I want a way to loop through pairs of adjacent numbers and compare their values so that I'm only left with pairs with the same ages.
Looping over observations is inefficient and in the vast majority of cases not necessary.
The following works for me:
clear
input age
5
5
7
7
8
9
10
10
end
generate tag = age != age[_n+1] & age != age[_n-1]
list
+-----------+
| age tag |
|-----------|
1. | 5 0 |
2. | 5 0 |
3. | 7 0 |
4. | 7 0 |
5. | 8 1 |
|-----------|
6. | 9 1 |
7. | 10 0 |
8. | 10 0 |
+-----------+
After getting rid of the relevant observations you get the desired result:
keep if tag == 0
list
+-----------+
| age tag |
|-----------|
1. | 5 0 |
2. | 5 0 |
3. | 7 0 |
4. | 7 0 |
5. | 10 0 |
|-----------|
6. | 10 0 |
+-----------+
I have two columns with data.
One has labels for a group and a second displays values for items in each group. I would like to calculate for each group, the average of only those values that are distinct.
How can I do this in Stata?
EDIT:
See my dataset and desired result below:
Group_label Value
x 12
x 12
x 2
x 1
y 5
y 5
y 5
y 2
y 2
I want to generate the following average:
Group_label Value Average
x 12 5
x 12 5
x 2 5
x 1 5
y 5 3.5
y 5 3.5
y 5 3.5
y 2 3.5
y 2 3.5
So the average for x = (12 + 2 + 1) / 3 and for y = (5 + 2) / 2
I have tried the egen(mean) command but it selects all values for each group label.
I only want to select the distinct values.
This is a two-step solution. You first need to tag distinct values using tag() within egen. Then you use mean() within egen.
The most delicate point is that something like ... if tag will leave missing values in the result for observations not selected. How can you omit duplicated values from the calculation yet also spread the result to their observations? See Section 9 of this paper for the use of cond() together with mean() which is one way to do it, exemplified in the code, and perhaps the most transparent way too. See Section 10 of the same paper for another method, which amuses some people.
For a fairly detailed review of distinct observations, see https://www.stata-journal.com/sjpdf.html?articlenum=dm0042
clear
input str1 Group_label Value
x 12
x 12
x 2
x 1
y 5
y 5
y 5
y 2
y 2
end
egen tag = tag(Group_label Value)
egen mean = mean(cond(tag, Value, .)), by(Group_label)
list, sepby(Group_label)
+-------------------------------+
| Group_~l Value tag mean |
|-------------------------------|
1. | x 12 1 5 |
2. | x 12 0 5 |
3. | x 2 1 5 |
4. | x 1 1 5 |
|-------------------------------|
5. | y 5 1 3.5 |
6. | y 5 0 3.5 |
7. | y 5 0 3.5 |
8. | y 2 1 3.5 |
9. | y 2 0 3.5 |
+-------------------------------+
The following works for me:
clear
input str1 vlab val
"x" 12
"x" 12
"x" 2
"x" 1
"y" 5
"y" 5
"y" 5
"y" 2
"y" 2
end
bysort vlab: generate tag = val != val[_n-1]
bysort vlab: egen mean_val = mean(val) if tag == 1
list
+-----------------------------+
| vlab val tag mean_val |
|-----------------------------|
1. | x 12 1 5 |
2. | x 12 0 . |
3. | x 2 1 5 |
4. | x 1 1 5 |
5. | y 5 1 3.5 |
|-----------------------------|
6. | y 5 0 . |
7. | y 5 0 . |
8. | y 2 1 3.5 |
9. | y 2 0 . |
+-----------------------------+
EDIT:
If you also do:
bysort vlab: replace mean_val = mean_val[_n-1] if mean_val == .
You will get:
list
+-----------------------------+
| vlab val tag mean_val |
|-----------------------------|
1. | x 12 1 5 |
2. | x 12 0 5 |
3. | x 2 1 5 |
4. | x 1 1 5 |
5. | y 5 1 3.5 |
|-----------------------------|
6. | y 5 0 3.5 |
7. | y 5 0 3.5 |
8. | y 2 1 3.5 |
9. | y 2 0 3.5 |
+-----------------------------+
Observations in my dataset are players, and binary variables temp1 up are equal to 1 if the player made a move, and equal to zero otherwise.
I would like to to calculate the maximum number of consecutive moves per player.
+------------+------------+-------+-------+-------+-------+-------+-------+
| simulation | playerlist | temp1 | temp2 | temp3 | temp4 | temp5 | temp6 |
+------------+------------+-------+-------+-------+-------+-------+-------+
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 |
| 1 | 2 | 1 | 0 | 0 | 0 | 1 | 1 |
+------------+------------+-------+-------+-------+-------+-------+-------+
My idea was to generate auxiliary variables in a loop, which would count consecutive duplicates and then apply egen, rowmax():
+------------+------------+------+------+------+------+------+------+------+
| simulation | playerlist | aux1 | aux2 | aux3 | aux4 | aux5 | aux6 | _max |
+------------+------------+------+------+------+------+------+------+------+
| 1 | 1 | 0 | 1 | 2 | 3 | 0 | 0 | 3 |
| 1 | 2 | 1 | 0 | 0 | 0 | 1 | 2 | 2 |
+------------+------------+------+------+------+------+------+------+------+
I am struggling with introducing a local counter variable that would be incrementally increased by 1 if consecutive move is made, and would be reset to zero otherwise (the code below keeps auxiliary variables fixed..):
quietly forval i = 1/42 { /*42 is max number of variables temp*/
local j = 1
gen aux`i'=.
local j = `j'+1
replace aux`i'= `j' if temp`i'!=0
}
Tactical answer
You can concatenate your move* variables into a single string and look for the longest substring of 1s.
egen history = concat(move*)
gen max = 0
quietly forval j = 1/6 {
replace max = `j' if strpos(history, substr("111111", 1, `j'))
}
If the number is much more than 6, use something like
local lookfor : di _dup(42) "1"
quietly forval j = 1/42 {
replace max = `j' if strpos(history, substr("`lookfor'", 1, `j'))
}
Compare also http://www.stata-journal.com/article.html?article=dm0056
Strategic answer
Storing a sequence rowwise is working against the grain so far as Stata is concerned. Much more flexibility is available if you reshape long and tsset your data as panel data. Note that the code here uses tsspell which must be installed from SSC using ssc inst tsspell.
tsspell is dedicated to identifying spells or runs in which some condition remains true. Here the condition is that a variable is 1 and since the only other allowed value is 0 that is equivalent to a variable being positive. tsspell creates three variables, giving spell identifier, sequence within spell and whether the spell is ending. Here the maximum length of spell is just the maximum sequence number for each game.
. input simulation playerlist temp1 temp2 temp3 temp4 temp5 temp6
simulat~n playerl~t temp1 temp2 temp3 temp4 temp5 temp6
1. 1 1 0 1 1 1 0 0
2. 1 2 1 0 0 0 1 1
3. end
. reshape long temp , i(sim playerlist) j(seq)
(note: j = 1 2 3 4 5 6)
Data wide -> long
-----------------------------------------------------------------------------
Number of obs. 2 -> 12
Number of variables 8 -> 4
j variable (6 values) -> seq
xij variables:
temp1 temp2 ... temp6 -> temp
-----------------------------------------------------------------------------
. egen id = group(sim playerlist)
. tsset id seq
panel variable: id (strongly balanced)
time variable: seq, 1 to 6
delta: 1 unit
. tsspell, p(temp)
. egen max = max(_seq), by(id)
. l
+--------------------------------------------------------------------+
| simula~n player~t seq temp id _seq _spell _end max |
|--------------------------------------------------------------------|
1. | 1 1 1 0 1 0 0 0 3 |
2. | 1 1 2 1 1 1 1 0 3 |
3. | 1 1 3 1 1 2 1 0 3 |
4. | 1 1 4 1 1 3 1 1 3 |
5. | 1 1 5 0 1 0 0 0 3 |
|--------------------------------------------------------------------|
6. | 1 1 6 0 1 0 0 0 3 |
7. | 1 2 1 1 2 1 1 1 2 |
8. | 1 2 2 0 2 0 0 0 2 |
9. | 1 2 3 0 2 0 0 0 2 |
10. | 1 2 4 0 2 0 0 0 2 |
|--------------------------------------------------------------------|
11. | 1 2 5 1 2 1 2 0 2 |
12. | 1 2 6 1 2 2 2 1 2 |
+--------------------------------------------------------------------+