I've read in my SPSS file in R and want to recode a new variable if such and such assumptions are made. To be specific:
I want to turn my spssdata_sub$gest variable into a new variable if the following the conditions are met:
spssdata_sub$indusert != 2 & spssdata_sub$ivf != 1 & spssdata_sub$leie != 3 & spssdata_sub$svkompl_II != 7 & spssdata_sub$svkompl_II != 2 & spssdata_sub$svkompl_II != 1
Anyone here who can help me with a code?
Does one of the following codes work for you?
Either this adapted version of Renu's solution
spssdata_sub$gest <- ifelse(spssdata_sub$indusert != 2 & spssdata_sub$ivf != 1 & spssdata_sub$leie != 3 & spssdata_sub$svkompl_II != 7 & spssdata_sub$svkompl_II != 2 & spssdata_sub$svkompl_II != 1, spssdata_sub$gest, NA)
or this code for filtering observations:
library(dplyr)
spssdata_sub_new <- spssdata_sub %>%
filter(indusert != 2 & ivf != 1 & leie != 3 & svkompl_II != 7 & svkompl_II != 2 & ssvkompl_II != 1)
One way is the following, if you really mean either one of the conditions
Mynewdata <- dplyr::filter(spssdata, indusert != 2, ivf != 1, leie != 3,
svkompl_II != 7 & svkompl_II != 2 & svkompl_II != 1)
only keeps entries that are neither, or putting it the other way exludes entries that have either indusert = 2 or ivf = 1 etc... one of the condition is enough to exclude it.
add-on: or something also like that:
Mynewdata <- dplyr::filter(spssdata, indusert != 2, ivf != 1, leie != 3,
!(svkompl_II %in% c(7,2,1))
Related
I have an inner for-loop in R which I have identified as significant bottleneck in my code. The script simulates the effect of a time-varying policy on individuals prior to adulthood. The outer loop runs over a list of cohorts (yob = 1910,...,1930 etc.) that I would like to study. The inner loop counts from ages from a = 5 to a = 17. CSL.details is a data.table that contains the details of each law that I am studying in form of the variables I grab, which vary by year = birthyear + a. To understand the overall effects of the policy by birth cohort, I need to track ca_years1, ca_years2, ca_years3, and ca_years4 for each a.
ages = seq.int(5,17)
state = "Massachusetts"
yob = seq.int(1910, 1930)
for (birthyear in yob){
ca_years1 = 0; ca_years2 = 0; ca_years3 = 0; ca_years4 = 0;
for (a in ages){
thisyear = birthyear + a
# Grab each law for given state and year and implement exemption permit
thislaw <- CSL.details[statename == state & yob == birthyear & thisyear == year]
if (nrow(thislaw) == 0) next
exempt_workpermit = (ca_years2 >= thislaw$workyrs & a >= thislaw$workage & thislaw$workage > 0)
exempt_yearstodropout = (ca_years3 >= thislaw$earlyyrs & a >= thislaw$earlyyrs_condition & thislaw$earlyyrs > 0)
exempt_cont = ((ca_years2 + ca_years4) >= thislaw$contyrs & thislaw$contyrs > 0)
# Increment each law when school is required
if(thislaw$entryage <= a & a < thislaw$exitage){
ca_years1 = ca_years1 + 1
if(!exempt_workpermit){ca_years2 = ca_years2 + 1}
if(!exempt_yearstodropout){ca_years3 = ca_years3 + 1}
}
if(thislaw$contage > a &
a >= thislaw$workage &
!exempt_cont &
thislaw$workage > 0 &
!(thislaw$entryage <= a & a < thislaw$exitage & !exempt_workpermit)
){ca_years4 = ca_years4 + 1}
}
CSL.exposures[statename == state & yob == birthyear]$ca_years1 = ca_years1
CSL.exposures[statename == state & yob == birthyear]$ca_years2 = ca_years2
CSL.exposures[statename == state & yob == birthyear]$ca_years3 = ca_years3
CSL.exposures[statename == state & yob == birthyear]$ca_years4 = ca_years4
}
Is there a data.table solution for replacing the inner-loop? I am an intermediate R coder and it is a bit difficult to think of how to get started. Although I would prefer data.table exclusively, I am open to dplyr-type solutions if they significantly speed up the code.
Edit: here is an example of what CSL.detail looks like, as a copy-pasted data.table.
statename year yob statefip entryage exitage earlyyrs earlyyrs_condition workage workyrs contage contyrs statecompschoolyr
1: Massachusetts 1913 1800 25 7 16 4 14 14 4 16 0 1852
2: Massachusetts 1913 1801 25 7 16 4 14 14 4 16 0 1852
3: Massachusetts 1913 1802 25 7 16 4 14 14 4 16 0 1852
4: Massachusetts 1913 1803 25 7 16 4 14 14 4 16 0 1852
5: Massachusetts 1913 1804 25 7 16 4 14 14 4 16 0 1852
I managed to refactor the code to solve the problem. The key idea is to exploit state and yob as grouping variables (since all calculations happen within a state and yob pair). This completely eliminates the outer loops and requires only a single loop, iterating by age. I am just saving this answer here for reference, but I am not sure that there is a broader lesson for the stackoverflow.com community so feel free to delete. The time savings are on the order of 95%, primarily because it reduces the overhead time to call data.table.
for(a in ages){
# grab running total of years of education compelled by state and year of birth
CSL.details[CSL.exposures, on = .(statename, yob),
`:=` (ca_years1 = i.ca_years1,
ca_years2 = i.ca_years2,
ca_years3 = i.ca_years3,
ca_years4 = i.ca_years4)] %>%
.[year == a + yob,
`:=`(
# create exemptions by age based on number of years of schooling completed
exempt_workpermit = (ca_years2 >= workyrs & a >= workage & workage > 0),
exempt_yearstodropout = (ca_years3 >= earlyyrs & a >= earlyyrs_condition & earlyyrs > 0),
exempt_cont = ((ca_years2 + ca_years4) >= contyrs & contyrs > 0)
), by = .(statename, yob)]
CSL.exposures[
CSL.details[year == a + yob], on = .(yob, statename),
`:=` (exempt_workpermit = i.exempt_workpermit, exempt_yearstodropout = i.exempt_yearstodropout,
exempt_cont = i.exempt_cont, entryage = i.entryage,
exitage = i.exitage, contage = i.contage, workage = i.workage) ] %>%
.[ ,
`:=` (
ca_years1 =
fifelse(entryage <= a & a < exitage,
ca_years1 + 1, ca_years1, na = as.numeric(ca_years1)),
ca_years2 =
fifelse(entryage <= a & a < exitage & !exempt_workpermit,
ca_years2 + 1, ca_years2, na = as.numeric(ca_years2)),
ca_years3 =
fifelse(entryage <= a & a < exitage & !exempt_yearstodropout,
ca_years3 + 1, ca_years3, na = as.numeric(ca_years3)),
ca_years4 =
fifelse(contage > a & a >= workage & !exempt_cont &
workage > 0 &
!(entryage <= a & a < exitage & !exempt_workpermit),
ca_years4 + 1, ca_years4, na = as.numeric(ca_years4))),
by = .(statename, yob)
]
}
I want to add the output of the loop in a new column "Compared_data".
Data set is libraries_four.
for (i in 1:20)
{
if ((Libraries_four[i,"PhyloAlps_iden"] == 1) & (Libraries_four[i,"ArctBorBryo_iden"] == 1 |
Libraries_four[i,"EMBL_143_iden"] == 1 | Libraries_four[i,"PhyloNorway_iden"] == 1 ))
{
print(TRUE)
}
else
{
print(FALSE)
}
}
The code is working fine but I tried the mutate function for the new column but it is not working. Is there any other way to add a new variable/column?
R is vectorised language, you would rarely need an explicit for loop. Try this :
library(dplyr)
Libraries_four <- Libraries_four %>%
mutate(result = PhyloAlps_iden == 1 & ArctBorBryo_iden == 1|
EMBL_143_iden == 1 | PhyloNorway_iden == 1)
This would create a new column called result in Libraries_four dataset.
You can also do this in base R :
Libraries_four <- transform(Libraries_four, result = PhyloAlps_iden == 1 & ArctBorBryo_iden == 1 | EMBL_143_iden == 1 | PhyloNorway_iden == 1)
Hello everyone I am working on a script that I would like to export to a CSV file.
Everything is working well with the exception that I would like to add column names and headers for the below data.
For instance variable A is the summary data of fixed income trades in 2017. I would like Row 1 in the output file to read as such.
Any help would be greatly appreciated. My code is written below. Thanks in advance!!
#SENDS THE RESULTS TO FILE CALLED OUTFILE.TXT WHICH IS OVERWRITTEN EACH TIME SCRIPT IS RUN
sink("outfile.csv")
#SHORT-TERM PRE-REFUNDED TRADE DATA
A = MSRB[which(MSRB$Coupon.Rate >= 2 & MSRB$Year == 2017 & MSRB$Par.Traded >=500 & MSRB$Class == "PRE-REFUNDED"),]
B = MSRB[which(MSRB$Coupon.Rate >= 2 & MSRB$Year == 2017 & MSRB$Par.Traded >=1000 & MSRB$Class == "PRE-REFUNDED"),]
C = MSRB[which(MSRB$Coupon.Rate >= 2 & MSRB$Year == 2018 & MSRB$Par.Traded >=500 & MSRB$Class == "PRE-REFUNDED"),]
D = MSRB[which(MSRB$Coupon.Rate >= 2 & MSRB$Year == 2018 & MSRB$Par.Traded >=1000 & MSRB$Class == "PRE-REFUNDED"),]
E = MSRB[which(MSRB$Coupon.Rate >= 2 & MSRB$Year == 2019 & MSRB$Par.Traded >=500 & MSRB$Class == "PRE-REFUNDED"),]
F = MSRB[which(MSRB$Coupon.Rate >= 2 & MSRB$Year == 2019 & MSRB$Par.Traded >=1000 & MSRB$Class == "PRE-REFUNDED"),]
#SUMMARY OF PRE-REFUNDED DATA
summary(A$Yield)
summary(B$Yield)
summary(C$Yield)
summary(D$Yield)
summary(E$Yield)
summary(F$Yield)
#END OF OUTPUT FILE
sink()
I have a data matrix 1200 (row, sample name)* 20000 (col, gene name), I want to delete row when my interested 5 genes have zero values in all samples
command I used for single gene:
allexp <-preallexp[preallexp$GZMB > 0, ]
but I want to use AND in above command, like this:
allexp <-preallexp[preallexp$GZMB && preallexp$TP53 && preallexp$EGFR && preallexp$BRAF && preallexp$VGEF > 0, ]
but this command doesnt work, please I need your help..How to use AND in above command.
EDIT: in response to OP.
I'm sure there's a much more efficient way to code this, but this is what you're after:
allexp <-preallexp[preallexp$GZMB + preallexp$TP53 + preallexp$EGFR +
preallexp$BRAF + preallexp$VGEF > 0, ]
Unless you have negative expression values I would have thought mkt's should work. But here is mine. It will remove values rows where each of the 5 genes and a value of 0
which(preallexp$GZMB == 0 && preallexp$TP53 &&
preallexp$EGFR == 0 && preallexp$BRAF == 0 && preallexp$VGEF == 0)
This gives so the rows where all 5 genes have a value of zero
So we can remove these rows if from the dataframe like follows
allexp <-preallexp[
-(which(preallexp$GZMB == 0 && preallexp$TP53 &&
preallexp$EGFR == 0 && preallexp$BRAF == 0 && preallexp$VGEF == 0)), ]
For every entry in rows i need to compute two variables as new columns in a data.frame depending conditional on more than 60 other columns. I would like your recommendation on how to realize that elegant (while and for, with, ifelse, foreach, by or ddply?). I don't like to do that manually like i did for the first cases in the example code and i don't care for performance.
Further: Probably i would not need to ask if i would have understood how to use functions like transform (with ddply or by) and what they do. Thus i hope you can recommend good tutorials on that, maybe relating to my case. I found a lot but in different context and was not able to comprehend it entrily or transcribe it for my case.
My case: I have three columns for each of 20 events representing the kind and date of that event. For each row I need to compute (and save to that data.frame) the difference in time between one special event (depending on whether a special kind happened before or after another) and a date fixed for every entry in rows. Furthermore i need to save the date of that event.
This is how i did (it works, but it is running only through the first cases):
#event.2 (1. event month), event.3 (1. event year), event.4 (1. event kind), event.5 (2. event month), event.6 (2. event year), ...
df$dit[(!is.na(df$event.2) & !is.na(df$event.3) & !is.na(df$event.4) & !is.na(df$event.5) & !is.na(df$event.6) & !is.na(df$event.7))
& (
(df$event.4 == 3 & ((1/12*df$event.2)+df$event.3) > df$fixdate) & (df$event.7 == 1 | df$event.7 == 2)
)] = ((1/12*df$event.2)+df$event.3) - df$fixdate
df$date[(!is.na(df$event.2) & !is.na(df$event.3) & !is.na(df$event.4) & !is.na(df$event.5) & !is.na(df$event.6) & !is.na(df$event.7))
& (
(df$event.4 == 3 & ((1/12*df$event.2)+df$event.3) > df$fixdate) & (df$event.7 == 1 | df$event.7 == 2)
)] = ((1/12*df$event.2)+df$event.3)
df$dit[(!is.na(df$event.2) & !is.na(df$event.3) & !is.na(df$event.4) & !is.na(df$event.5) & !is.na(df$event.6) & !is.na(df$event.7))
& (
(df$event.4 == 1 & ((1/12*df$event.2)+df$event.3) > df$fixdate)
| (df$event.4 == 2 & ((1/12*df$event.2)+df$event.3) > df$fixdate)
)] = 0
df$date[(!is.na(df$event.2) & !is.na(df$event.3) & !is.na(df$event.4) & !is.na(df$event.5) & !is.na(df$event.6) & !is.na(df$event.7))
& (
(df$event.4 == 1 & ((1/12*df$event.2)+df$event.3) > df$fixdate)
| (df$event.4 == 2 & ((1/12*df$event.2)+df$event.3) > df$fixdate)
)] = df$fixdate
df$dit[(!is.na(df$event.2) & !is.na(df$event.3) & !is.na(df$event.4) & !is.na(df$event.5) & !is.na(df$event.6) & !is.na(df$event.7))
& (
(
(df$event.4 == 1 & ((1/12*df$event.2)+df$event.3) < df$fixdate)
& (
(df$event.7 == 1 & ((1/12*df$event.5)+df$event.6) > df$fixdate)
| (df$event.7 == 2 & ((1/12*df$event.5)+df$event.6) > df$fixdate)
)
)
|
(
(df$event.4 == 2 & ((1/12*df$event.2)+df$event.3) < df$fixdate)
& (
(df$event.7 == 1 & ((1/12*df$event.5)+df$event.6) > df$fixdate)
| (df$event.7 == 2 & ((1/12*df$event.5)+df$event.6) > df$fixdate)
)
)
)] = ((1/12*df$event.5)+df$event.6) - df$fixdate
df$date[(!is.na(df$event.2) & !is.na(df$event.3) & !is.na(df$event.4) & !is.na(df$event.5) & !is.na(df$event.6) & !is.na(df$event.7))
& (
(
(df$event.4 == 1 & ((1/12*df$event.2)+df$event.3) < df$fixdate)
& (
(df$event.7 == 1 & ((1/12*df$event.5)+df$event.6) > df$fixdate)
| (df$event.7 == 2 & ((1/12*df$event.5)+df$event.6) > df$fixdate)
)
)
|
(
(df$event.4 == 2 & ((1/12*df$event.2)+df$event.3) < df$fixdate)
& (
(df$event.7 == 1 & ((1/12*df$event.5)+df$event.6) > df$fixdate)
| (df$event.7 == 2 & ((1/12*df$event.5)+df$event.6) > df$fixdate)
)
)
)] = ((1/12*df$event.5)+df$event.6)
You can define your conditions as expressions and use them within transform. The idea is to factorize your conditions at most as possible .
COND1 <- expression(!is.na(event.2) & !is.na(event.3) &
!is.na(event.4) & !is.na(event.5) &
!is.na(event.6) & !is.na(event.7))
COND2 <- expression(event.4 == 3 & ((1/12*event.2)+event.3) > fixdate) &
(event.7 == 1 | event.7 == 2))
COND3 <- expression(event.4 == 1 & ((1/12*event.2)+event.3) > fixdate)
COND4 <- expression(event.4 == 2 & ((1/12*event.2)+event.3) > fixdate)
### you continue here with the rest of conditions....
Then using them within transform you can do something like:
transform(df, date = ifelse(eval(COND1) & eval(COND2),((1/12*event.2)+event.3),NA),
transform(df, date = ifelse(eval(COND1) & (eval(COND3)|eval(COND4)),fixdate,NA))
## Note also that the seond "dit" variable is deduced from "date"
transform(df,dit=date-fixdate)