In the dataset, there are two columns "start_year" and "end_year", indicating the year a patient start and end the registration in the GP clinic. I want to know whether each patient was registered in the clinic from 1990 to 2019. Probably compute 20 new variables (1=yes,0=no) for each year.
I used ifelse (R) to compute the variable one by one:
test$pt_1990<-ifelse(test$start_year<=1990 & 1990<=test$end_year,1,0)
Hope loops could have a better solution instead of write 20 lines of same code. Thank u very much
Related
I am currently working on analyzing a within-subject dataset with 8 time-ordered assessment points for each subject.
The variables of interest in this example is ID, time point, and accident.
I want to create two variables: accident_intercept and accident_slope, based on the value on accident at a particular time point.
For the accident_intercept variable, once a participant indicated the occurrence of an accident (e.g., accident = 1) at a specific time point, I want the values for that time point and the remaining time points to be 1.
For the accident_slope variable, once a participant indicated the occurrence of an accident (e.g., accident = 1) at a specific time point, I want the value of that time point to be 0, but count up by 1 for the remaining time points until the end time point, for each subject.
The main challenge here is that the process stated above need to be repeated/looped for each participant that occupies 8 rows of data.
Please see how the newly created variables would look like:
I have looked into the instruction for different SPSS syntax, such as loop, the lag/lead functions. I also tried to break my task into different components and google each one. However, I have not made any progress :)
I would be really grateful of any helps and directions that you provide.
Here is one way to do what you need using aggregate to calculate "accident time":
if accident=1 accidentTime=TimePoint.
aggregate out=* mode=addvariables overwrite=yes /break=ID/accidentTime=max(accidentTime).
if TimePoint>=accidentTime Accident_Intercept=1.
if TimePoint>=accidentTime Accident_Slope=TimePoint-accidentTime.
recode Accident_Slope accidentTime (miss=0).
Here is another approach using the lag function:
compute Accident_Intercept=0.
if accident=1 Accident_Intercept=1.
if $casenum>1 and id=lag(id) and lag(Accident_Intercept)=1 Accident_Intercept=1.
compute Accident_Slope=0.
if $casenum>1 and id=lag(id) and lag(Accident_Intercept)=1 Accident_Slope=lag(Accident_Slope) +1.
exe.
I have a simple daily rainfall data set and would like to calculate the antecedent dry period for each day. Here, I'm defining a dry day to be "<10". I'm fairly unfamiliar with INDEX(), MATCH(), and other fancy array functions but feel like I'll need to use them.
For example, in the image, for 1/17/2020, the values in cells C3:C9=0, C10=1, C11:C13=0. I've tried various versions of COUNTIF(), COUNTIFS(), and IF() functions but I cannot get the step-wise + re-set functionality necessary when extended "dry spells" or brief rain periods occur with gaps. Thanks!
You are right, you need to use Match. Basically you need to search for the next antecedent wet day (of which there are many here in Manchester England at the moment) and subtract 1 (Formula 1):
=MATCH(TRUE,INDEX(B15:B$1000>=10,0),0)-1
where B$1000 may need changing to include all of your data. The use of Index here is just a bit of a hack to avoid having to enter the formula as an array formula.
As you can see there is an issue when you come to the end of the range which I will come to in a minute.
In this case, we want to count the number of antecedent dry days to the end of the range (Formula 2):
=IFERROR(MATCH(TRUE,INDEX(B4:B$1000>=10,0),0)-1,COUNTIF(B4:B$1000,"<10"))
If the range ended with a dry spell, you would get this:
I think this is the hardest to date I have had to crack - so hard I had a hard time finding a good headline.
So we have a site where trucks come and buy say Gravel, or sand or other building materials.
Sometimes they also unload demolition waste first.
I need to find out a couple of things
how many trucks (and from what companys) came empty
if they came empty what did they buy from us.
what companys are sending full trucks and what are sending empty trucks.
a tope 10 of materials they will drive to us from to buy even when coming empty to our facility.
a list of all the order numbers that they drove to us til fill and came with empty trucks. ( I have distances linked to order numbers, so now I can estimate the value of our products)
The data I have available:
I have a full data set of when what customer buys what and / or pay to deliver.
E.G.:
I can see the parts I need to split the data into I think it should be something like this
find all unique licence plates
somehow map if they bought materials within 30 minutes of
offloading demolition waste (most trucks will come between 2 and 10
times per day)
Present all this data (on a normal day we have about 800 trucks = 2000 lines since they weigh in, weigh out, and then some buy something = 2 more weigh lines)
I can easily find unique licence plates per day (either by formula or by Excel function Data/delete doublets,
but after that I have no clue where to start.
I think I need some sheets in between, where I somehow mark if a material was bought from an "empty truck" and I need a counter for that .. somehow...
Any help on how to get started is appreciated.
It seems like the best way to start is with a helper column (in the following exampes, I have chosen "Column M") to flag whether the truck arrived empty.
In the helper column, you can use something similar to the following formula.
{=IF(ISBLANK(B2),0,IF(C2="In",0,IF(B2=$B$2:$B$13,IF($C$2:$C$13="In",IF($A$2:$A$13>(A2-TIME(0,30,0)),0,1),1),1)))}
This is an array formula, which means you have to press ctrl+shift+enter after pasting it in the cell. Then you can copy that cell down the column.
Just to explain, the first if statement knows the truck is not arriving empty if Column C is 'In'. The second if statement creates an array and tests to see if other the same truck appears in other rows. The third if statement checks to see if the same truck checked 'In' in the matching rows, and the fourth if statement verifies if the time they checked in was less than thirty minutes ago. You can adjust the length by editing the TIME(0,30,0) function. The format is TIME(hours,minuites,seconds). Unless the truck matches all three of the second, third and fourth if statements, it is marked as coming empty.
Once you have this helper column, just about all of your tasks are quite simple.
1a: How many trucks came empty? Sum Column M
1b: How many trucks from what company? Create a unique list of companies. Then create a COUNTIFS formula based on Column M = 1 and Column K = Company. For example, if C32 had Company B then the formula =COUNTIFS($M$2:$M$13,1,$K$2:$K$13,C32) would return 2
1c: How many times did a truck come empty? Similar to 1b, create a unique list of License Plates, then use a COUNTIFS based on Column M = 1 and Column B = License Plate.
2: Similar to 1b, just use a unique list of products tested against Column F
3: Similar to 1b, just create a second column, next to the first that uses =COUNTIFS($M$2:$M$13,0,$K$2:$K$13,C53,$C$2:$C$13,"In") Which tests that Column M reports the truck did not come empty, that matches the company in Column K and that the truck came 'In' so you don't double count the same truck when it goes 'out'
4: Just sort list created by number 2. You can highlight the range, right-click and select "Sort" > "Custom Sort", then select the column you want to sort on and largest to smallest.
5: There are a couple of different ways, you could do this. The formula
{=TEXTJOIN(", ",TRUE,IF($M$2:$M$13=1,$J$2:$J$13,""))}
(again, entered as an array formula)
would create a comma separated list of order numbers. An alternative if you want a column of order numbers (but would only work if they are actually numbers), is to paste the formula {=MAX(IF($M$2:$M$13=1,$J$2:$J$13,))} in the first row of the column (in my example, its O2) and then {=MAX(IF($M$2:$M$13=1,IF($J$2:$J$13<O2,$J$2:$J$13,)))} in the row below (change the reference to O2 if you pasted it in a different spot)(again, note that both of these are array formulas). Then copy and paste the second formula down the column. When order numbers of trucks that came in empty are exhausted, the formula will report 0.
I have an array that looks something like this:
[["Sunday", [user1, user2]], ["Sunday", [user1, user4]], ["Monday", [user3, user2]]]
The array essentially has all permutations of a given day with a unique pair of users. I obtained it by running
%w[Su Mo Tu We Th Fr Sa].product(User.all_pairs)
where User.all_pairs is every unique pair of users.
My goal now is to compose this set of nested arrays into schedules, meaning I want to find every permutation of length 7 with unique days. In other words, I want every potential week. I already have every potential day, and I have every potential pair of users, now I just need to compose them.
I have a hunch that the Array.permutation method is what I need, but I'm not sure how I'd use it in this case. Or perhaps I should use Array.product?
If I understand you correctly, you want all possible weeks where there is one pair of users assigned to each day. You can do it like this:
User.all_pairs.combination(7)
This will give you all possible ways of how you can pick 7 pairs and assign them to the days of the week. But if you are asking for every possible week, then it also matters into which day is which pair assigned, and you also have to take every permutation of those 7 pairs:
User.all_pairs.combination(7).map{|week| week.permutation().to_a}.flatten(1)
Now this will give you all possible weeks, where every week is represented as array containing 7 pairs. For example one of the weeks may look like this:
[(user1, user2), (user1, user3), (user2, user3), (user3, user4), (user1, user4), (user2, user4), (user3, user4)]
However the amount of the weeks will be huge! If you have n users, you will have k = n!/2 pairs, there is p = k! / (7! * (k - 7)!) ways of selecting 7 pairs and p * 7! possible weeks. If you have just 5 users, you get 1946482876800 possible weeks! No matter what you are planning to do with it, it won't be possible.
If you are trying to find the best schedule for a week, you can try to make some greedy algorithm.
I have this dataset with 2 variables: week and brand_chosen, where brand chosen designates which product from e.g. a super market was chosen, an it looks like this.
Week brand_chosen
2 19
2 15
2 50
2 12
3 19
3 16
3 50
4 77
4 19
What I am trying to do is for each line, to note the week in which the brand purchase was made, and check if in the week before that the same brand purchase was made. In case it did, a variable dummy would take the value of 1, otherwise 0.
Because week appears multiple times I cannot take just the lag(week,1), so I probably need to loop through the week variables for each case, until it finds the first different value.
This is what i tried to do
loop i=1 to 70.
do if (week<>lag(week,i) and brand_chosen=lag(brand_chosen,i)).
compute dummy=1.
end loop.
else.
compute dummy=0.
end if.
end loop.
execute.
Where 70 is just an arbitrary number so that I am sure that it will check all the previous cases.
I get two problems with that. First the lag function needs to contain a number from what I understand but "i" is not considered a number here.
The second problem is that i would like to close the loop if the condition is satisfied, and move to the next case but I get an error.
I am new to spss syntax and I am struggling with that one, so any help is greatly appreciated.
I assume that every combination of week--brand_chosen is unique. In this case the solution is quite simple. Just reorder your dataset by brand_chosen and then week, and then run a simple lag command.
This should do the trick:
SORT CASES BY brand_chosen week.
COMPUTE dummy=0.
IF (brand_chosen=LAG(brand_chosen) AND week>LAG(week)) dummy = 1.