I have 26 variables and each of them contain numbers ranging from 1 to 61. I want for each case of 1, each case of 2 etc. the number 1 in a new variable. If there is no 1, the variable should contain 2.
So 26 variables with data like:
1 15 28 39 46 1 12 etc.
And I want 61 variables with:
1 2 1 2 2 1 etc.
I have been reading about creating vectors, loops, do if's etc but I can't find the right way to code it. What I have done is just creating 61 variables and writing
do if V1=1 or V2=1 or (etc until V26).
recode newV1=1.
end if.
exe.
**repeat this for all 61 variables.
recode newV1 to newV61(missing=2).
So this is a lot of code and quite a detour from what I imagine it could be.
Anyone who can help me out with this one? Your help is much appreciated!
noumenal is correct, you could do it with two loops. Another way though is to access the VECTOR using the original value though, writing that as 1, and setting all other values to zero.
To illustrate, first I make some fake data (with 4 original variables instead of 26) named X1 to X4.
*Fake Data.
SET SEED 10.
INPUT PROGRAM.
LOOP Id = 1 TO 20.
END CASE.
END LOOP.
END FILE.
END INPUT PROGRAM.
VECTOR X(4,F2.0).
LOOP #i = 1 TO 4.
COMPUTE X(#i) = TRUNC(RV.UNIFORM(1,62)).
END LOOP.
EXECUTE.
Now what this code does is create four vector sets to go along with each variable, then uses DO REPEAT to actually refer to the VECTOR stub. Then finishes up with RECODE - if it is missing it should be coded a 2.
VECTOR V1_ V2_ V3_ V4_ (61,F1.0).
DO REPEAT orig = X1 TO X4 /V = V1_ V2_ V3_ V4_.
COMPUTE V(orig) = 1.
END REPEAT.
RECODE V1_1 TO V4_61 (SYSMIS = 2).
It is a little painful, as for the original VECTOR command you need to write out all of the stubs, but then you can copy-paste that into the DO REPEAT subcommand (or make a macro to do it for you).
For a more simple illustration, if we have our original variable, say A, that can take on integer values from 1 to 61, and we want to expand to our 61 dummy variables, we would then make a vector and then access the location in that vector.
VECTOR DummyVec(61,F1.0).
COMPUTE DummyVec(A) = 1.
For a record if A = 10, then here DummyVec10 will equal 1, and all the others DummyVec variables will still by system missing by default. No need to use DO IF for 61 values.
The rest of the code is just extra to do it in one swoop for multiple original variables.
This should do it:
do repeat NewV=NewV1 to NewV61/vl=1 to 61.
compute NewV=any(vl,v1 to v26).
end repeat.
EXPLANATION:
This syntax will go through values 1 to 61, for each one checking whether any of the variables v1 to v26 has that value. If any of them do, the right NewV will receive the value of 1. If none of them do, the right NewV will receive the value of 0.
Just make sure v1 to v26 are consecutively ordered in the file. if not, then change to:
compute NewV=any(vl,v1, v2, v3, v4 ..... v26).
You need a nested loop: two loops - one outer and one inner.
Related
Ultimately, I want to change scores of 0 to 1, scores of 1 to 2, and scores of 2 to 3. I thought one way to do that was using +1, but I realize I could also use a more complicated if then series.
Here is what I did so far:
I used the existing variable (x) to create a new variable (y=x+1) using SPSS syntax. I only want to do this for variables with values >=0 (this was my approach to excluding cells with missing data; the range for x is 0-2).
I can create x+1, but it overwrites the existing variables.
DO REPEAT x =var_1 TO var_86.
if (x>=0) x=(x+1).
end repeat.
exe.
I tried this modification, but it doesn't work:
DO REPEAT x = var_1 TO var_86 / y = var_1a TO var_86a.
IF (x >= 0) y=x +1.
END REPEAT.
EXE.
The error message is:
DO REPEAT The form VARX TO VARY to refer to a range of variables has
been used incorrectly. When using VARX TO VARY to create new
variables, X must be an integer less than or equal to the integer Y.
(Can't use A3 TO A1.)
I tried many other configurations including vectors and loops but haven't yet figured out how to do this computation across the range of variables without overwriting the existing ones. Thanks in advance for any recommendations.
The message you are getting is because SPSS doesn't understand the form var_1a TO var_86a.
For the x to y form to work the number has to be at the end of the name, so for example varA_1 to varA_86 should work.
While you're at it, here's a simple way to go about your task:
recode var_1 TO var_86 (0=1)(1=2)(2=3) into varA_1 TO varA_86.
I have the File as following format
Name Number Position
A 1
B 2
C 3
D 4
Now on position A3 , I applied =IF(B2=1,"Goal Keeper",OR(IF(B2=2,"Defender",OR(IF(B2=3,"MidField","Striker"))))) But it giving me an error #value!
Looked up at google, and my formula is correct.
What i basically want it
1- Goalkeeper 2-Defender 3-Midfield 4-Striker
Yes the other way is to to just filter the number and copy paste the text
But I want to do it using formula and want to know where did I go wrong.
Your immediate problem lies with the expression (for example):
OR(IF(B2=3,"MidField","Striker"))
| \__/ \________/ \_______/ |
| bool string string |
\____________________________/
string
The OR function expects a series of boolean values (true or false) and you're giving it a string value from the inner IF.
You don't actually need the or bits in this specific case, the if is a full if-else. So you can just use:
=IF(B1=1,"Goal Keeper",IF(B2=2,"Defender",IF(B2=3,"MidField","Striker")))
This means that B1=1 will result in "Goal Keeper", otherwise it will evaluate IF(B2=2,"Defender",IF(B2=3,"MidField","Striker")).
Then that means that, if B2=2, it will result in "Defender", otherwise it will evaluate IF(B2=3,"MidField","Striker").
Finally, that means the B2=3 will result in "MidField", anything else will give "Striker".
The only situation I can envisage when OR would come in handy here would be when two different numbers were to generate the same string. Let's say both 1 and 4 should give "Goalie", you could use:
=IF(OR(B1=1,B1=4),"Goalie",IF(B2=2,"Defender","MidField"))
Keep in mind that a more general solution would be better implemented with the Excel lookup functions, ones that would search a table (on the spreadsheet somewhere) which mapped the integers to strings. Then, if the mapping needed to change, you would just update the table rather than going back and changing the formula in every single row.
If you are actually tasked with solving the problem by using the IF and OR function within the same equation, this is the only way I can see how:
=IF(OR(B1=1, B1 = 2, B1 = 3, B1 = 4),IF(B1 = 1, "Goal Keeper", IF(B1 = 2,"Defender",IF(B1 = 3,"MidField","Striker")))
If B1 does not equal 1-4, the OR function will return FALSE and completely bypass all of the nested IF statements.
In Stata, I am trying to use a foreach loop where I am looping over numbers from, say, 05-11. The problem is that I wish to keep the 0 as part of the value. I need to do this because the 0 appears in variable names. For example, I may have variables named Y2005, Y2006, Var05, Var06, etc. Here is an example of the code that I tried:
foreach year of numlist 05/09 {
...do stuff with Y20`year` or with Var`year`
}
This gives me an error that e.g. Y205 is not found. (I think that what is happening is that it is treating 05 as 5.)
Also note that I can't add a 0 in at the end of e.g. Y20 to get Y200 because of the 10 and 11 values.
Is there a work-around or an obvious thing I am not doing?
Another work-around is
forval y = 5/11 {
local Y : di %02.0f `y'
<code using local Y, which must be treated as a string>
}
The middle line could be based on
`: di %02.0f `y''
so that using another macro can be avoided, but at the cost of making the code more cryptic.
Here I've exploited the extra fact that foreach over such a simple numlist is replaceable with forvalues.
The main trick here is documented here. This trick avoids the very slight awkwardness of treating 5/9 differently from 10/11.
Note. To understand what is going on, it often helps to use display interactively on very simple examples. The detail here is that Stata is happily indifferent to leading zeros when presented with numbers. Usually this is immaterial to you, or indeed a feature as when you appreciate that Stata does not insist on a leading zero for numbers less than 1.
. di 05
5
. di 0.3
.3
. di .3
.3
Here we really need the leading zero, and the art is to see that the problem is one of string manipulation, the strings such as "08" just happening to contain numeric characters. Agreed that this is obvious only when understood.
There's probably a better solution but here's how this one goes:
clear
set more off
*----- example data -----
input ///
var2008 var2009 var2010 var2011 var2012
0 1 2 3 4
end
*----- what you want -----
numlist "10(1)12"
local nums 08 09 `r(numlist)'
foreach x of local nums {
display var20`x'
}
The 01...09 you can insert manually. The rest you build with numlist. Put all that in a local, and finally use it in the loop.
As you say, the problem with your code is that Stata will read 5 when given 05, if you've told it is a number (which you do using numlist in the loop).
Another solution would be to use an if command to count the number of characters in the looping value, and then if needed you can add a leading zero by reassigning the local.
clear
input var2008 var2009 var2010 var2011 var2012
0 1 2 3 4
end
foreach year of numlist 08/12{
if length("`year'") == 1 local year 0`year'
di var20`year'
}
There is an integer array, for eg.
{3,1,2,7,5,6}
One can move forward through the array either each element at a time or can jump a few elements based on the value at that index. For e.g., one can go from 3 to 1 or 3 to 7, then one can go from 1 to 2 or 1 to 2(no jumping possible here), then one can go 2 to 7 or 2 to 5, then one can go 7 to 5 only coz index of 7 is 3 and adding 7 to 3 = 10 and there is no tenth element.
I have to only count the number of possible paths to reach the end of the array from start index.
I could only do it recursively and naively which runs in exponential time.
Somebody plz help.
My recommendation: use dynamic programming.
If this key word is sufficient and you want the challenge to find a possible solution on your own, dont read any further!
Here a possible DP-algorithm on the example input {3,1,2,7,5,6}. It will be your job to adjust on the general problem.
create array sol length 6 with just zeros in it. the array will hold the number of ways.
sol[5] = 1;
for (i = 4; i>=0;i--) {
sol[i] = sol[i+1];
if (i+input[i] < 6 && input[i] != 1)
sol[i] += sol[i+input[i]];
}
return sol[0];
runtime O(n)
As for the directed graph solution hinted in the comments :
Each cell in the array represents a node. Make an directed edge from each node to the node accessable. Basically you can then count more easily the number of ways by just looking at the outdegrees on the nodes (since there is no directed cycle) however it is a lot of boiler plate to actual program it.
Adjusting the recursive solution
another solution would be to pruning. This is basically equivalent to the DP-algorithm. The exponentiel time comes from the fact, that you calculate values several times. Eg function is recfunc(index). The initial call recFunc(0) calls recFunc(1) and recFunc(3) and so on. However recFunc(3) is bound to be called somewhen again, which leads to a repeated recursive calculation. To prune this you add a Map to hold all already calculated values. If you make a call recFunc(x) you lookup in the map if x was already calculated. If yes, return the stored value. If not, calculate, store and return it. This way you get a O(n) too.
Can anyone tell me if there is a way (in MATLAB) to check whether a certain value is equal to any of the values stored within another array?
The way I intend to use it is to check whether an element index in one matrix is equal to the values stored in another array (where the stored values are the indices of the elements which meet a certain criteria).
So, if the indices of the elements which meet the criteria are stored in the matrix below:
criteriacheck = [3 5 6 8 20];
Going through the main array (called array) and checking if the index matches:
for i = 1:numel(array)
if i == 'Any value stored in criteriacheck'
%# "Do this"
end
end
Does anyone have an idea of how I might go about this?
The excellent answer previously given by #woodchips applies here as well:
Many ways to do this. ismember is the first that comes to mind, since it is a set membership action you wish to take. Thus
X = primes(20);
ismember([15 17],X)
ans =
0 1
Since 15 is not prime, but 17 is, ismember has done its job well here.
Of course, find (or any) will also work. But these are not vectorized in the sense that ismember was. We can test to see if 15 is in the set represented by X, but to test both of those numbers will take a loop, or successive tests.
~isempty(find(X == 15))
~isempty(find(X == 17))
or,
any(X == 15)
any(X == 17)
Finally, I would point out that tests for exact values are dangerous if the numbers may be true floats. Tests against integer values as I have shown are easy. But tests against floating point numbers should usually employ a tolerance.
tol = 10*eps;
any(abs(X - 3.1415926535897932384) <= tol)
you could use the find command
if (~isempty(find(criteriacheck == i)))
% do something
end
Note: Although this answer doesn't address the question in the title, it does address a more fundamental issue with how you are designing your for loop (the solution of which negates having to do what you are asking in the title). ;)
Based on the for loop you've written, your array criteriacheck appears to be a set of indices into array, and for each of these indexed elements you want to do some computation. If this is so, here's an alternative way for you to design your for loop:
for i = criteriacheck
%# Do something with array(i)
end
This will loop over all the values in criteriacheck, setting i to each subsequent value (i.e. 3, 5, 6, 8, and 20 in your example). This is more compact and efficient than looping over each element of array and checking if the index is in criteriacheck.
NOTE: As Jonas points out, you want to make sure criteriacheck is a row vector for the for loop to function properly. You can form any matrix into a row vector by following it with the (:)' syntax, which reshapes it into a column vector and then transposes it into a row vector:
for i = criteriacheck(:)'
...
The original question "Can anyone tell me if there is a way (in MATLAB) to check whether a certain value is equal to any of the values stored within another array?" can be solved without any loop.
Just use the setdiff function.
I think the INTERSECT function is what you are looking for.
C = intersect(A,B) returns the values common to both A and B. The
values of C are in sorted order.
http://www.mathworks.de/de/help/matlab/ref/intersect.html
The question if i == 'Any value stored in criteriacheck can also be answered this way if you consider i a trivial matrix. However, you are proably better off with any(i==criteriacheck)