Mixing
Comparing two files in R and adding the matched rows from one file to another using for loop in R
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I have two files. I have read both of the files in data frame in R. "ShortListedGenes" contains 841 genes and "EmpPval" contains 6000 genes. I want to match each gene present in "shortListedGenes" to the genes present in "EmpPval" and extract the relevant information present in the 1st, 2nd and 3rd column of each row (each row represents each gene) in the "EmpPval" file and add it to the "shortListedFile" as new columns. As i am new to R so am using the simple approach and using for loop to get the job done!!! but it's giving the error. The file "shortListedGenes" looks like this
gene hsq hsq.se hsq.Pv ZscoreHsq PValueZ FDR
ENSG00000198502.5 0.909563 0.018102 0 50.24654734
0 0
ENSG00000225138.3 0.876861 0.018487 0 47.43122194
0 0
The second file "EmpPval" looks like
X obsExp.perExp obsExp.Pv obsExp.perExp.1000
ENSG00000000460.12 129 0.886162308 0.129
ENSG00000000971.11 268 0.728160071 0.268
So whenever the gene (1st column of both files) matches I want to extract the values for the corresponding gene from the "EmpPval" file and add it to the "shortListedGenes" file.
The code which I have tried till now is:
shortListedGenes <- read.csv("zs7-fdr0.05-Aorta.csv")
EmpPvAl <- read.csv("EmpiricalPvaluesAorta.csv")
for(i in 1:nrow(shortListedGenes))
{
for(j in 1:nrow(EmpPvAl))
{
if(shortListedGenes$gene[i] == EmpPvAl$X[j])
{
shortListedGenes$obsLessExp <- EmpPvAl$obsExp.perExp
shortListedGenes$obsExp <- EmpPvAl$obsExp.Pv
shortListedGenes$obsLessExpDiv <- EmpPvAl$obsExp.perExp.1000
}
}
}
and it is giving the error Error in Ops.factor(EmpPvAl$X, shortListedGenes$gene[i]) : level sets of factors are different
Any suggestion/help would be appreciated!!!. Thanks
r
asked Jan 3 at 6:06
StarStar
116
add a comment |
I have two files. I have read both of the files in data frame in R. "ShortListedGenes" contains 841 genes and "EmpPval" contains 6000 genes. I want to match each gene present in "shortListedGenes" to the genes present in "EmpPval" and extract the relevant information present in the 1st, 2nd and 3rd column of each row (each row represents each gene) in the "EmpPval" file and add it to the "shortListedFile" as new columns. As i am new to R so am using the simple approach and using for loop to get the job done!!! but it's giving the error. The file "shortListedGenes" looks like this
gene hsq hsq.se hsq.Pv ZscoreHsq PValueZ FDR
ENSG00000198502.5 0.909563 0.018102 0 50.24654734
0 0
ENSG00000225138.3 0.876861 0.018487 0 47.43122194
0 0
The second file "EmpPval" looks like
X obsExp.perExp obsExp.Pv obsExp.perExp.1000
ENSG00000000460.12 129 0.886162308 0.129
ENSG00000000971.11 268 0.728160071 0.268
So whenever the gene (1st column of both files) matches I want to extract the values for the corresponding gene from the "EmpPval" file and add it to the "shortListedGenes" file.
The code which I have tried till now is:
shortListedGenes <- read.csv("zs7-fdr0.05-Aorta.csv")
EmpPvAl <- read.csv("EmpiricalPvaluesAorta.csv")
for(i in 1:nrow(shortListedGenes))
{
for(j in 1:nrow(EmpPvAl))
{
if(shortListedGenes$gene[i] == EmpPvAl$X[j])
{
shortListedGenes$obsLessExp <- EmpPvAl$obsExp.perExp
shortListedGenes$obsExp <- EmpPvAl$obsExp.Pv
shortListedGenes$obsLessExpDiv <- EmpPvAl$obsExp.perExp.1000
}
}
}
and it is giving the error Error in Ops.factor(EmpPvAl$X, shortListedGenes$gene[i]) : level sets of factors are different
Any suggestion/help would be appreciated!!!. Thanks
r
asked Jan 3 at 6:06
StarStar
116
add a comment |
I have two files. I have read both of the files in data frame in R. "ShortListedGenes" contains 841 genes and "EmpPval" contains 6000 genes. I want to match each gene present in "shortListedGenes" to the genes present in "EmpPval" and extract the relevant information present in the 1st, 2nd and 3rd column of each row (each row represents each gene) in the "EmpPval" file and add it to the "shortListedFile" as new columns. As i am new to R so am using the simple approach and using for loop to get the job done!!! but it's giving the error. The file "shortListedGenes" looks like this
gene hsq hsq.se hsq.Pv ZscoreHsq PValueZ FDR
ENSG00000198502.5 0.909563 0.018102 0 50.24654734
0 0
ENSG00000225138.3 0.876861 0.018487 0 47.43122194
0 0
The second file "EmpPval" looks like
X obsExp.perExp obsExp.Pv obsExp.perExp.1000
ENSG00000000460.12 129 0.886162308 0.129
ENSG00000000971.11 268 0.728160071 0.268
So whenever the gene (1st column of both files) matches I want to extract the values for the corresponding gene from the "EmpPval" file and add it to the "shortListedGenes" file.
The code which I have tried till now is:
shortListedGenes <- read.csv("zs7-fdr0.05-Aorta.csv")
EmpPvAl <- read.csv("EmpiricalPvaluesAorta.csv")
for(i in 1:nrow(shortListedGenes))
{
for(j in 1:nrow(EmpPvAl))
{
if(shortListedGenes$gene[i] == EmpPvAl$X[j])
{
shortListedGenes$obsLessExp <- EmpPvAl$obsExp.perExp
shortListedGenes$obsExp <- EmpPvAl$obsExp.Pv
shortListedGenes$obsLessExpDiv <- EmpPvAl$obsExp.perExp.1000
}
}
}
and it is giving the error Error in Ops.factor(EmpPvAl$X, shortListedGenes$gene[i]) : level sets of factors are different
Any suggestion/help would be appreciated!!!. Thanks
r
asked Jan 3 at 6:06
StarStar
116
I have two files. I have read both of the files in data frame in R. "ShortListedGenes" contains 841 genes and "EmpPval" contains 6000 genes. I want to match each gene present in "shortListedGenes" to the genes present in "EmpPval" and extract the relevant information present in the 1st, 2nd and 3rd column of each row (each row represents each gene) in the "EmpPval" file and add it to the "shortListedFile" as new columns. As i am new to R so am using the simple approach and using for loop to get the job done!!! but it's giving the error. The file "shortListedGenes" looks like this
gene hsq hsq.se hsq.Pv ZscoreHsq PValueZ FDR
ENSG00000198502.5 0.909563 0.018102 0 50.24654734
0 0
ENSG00000225138.3 0.876861 0.018487 0 47.43122194
0 0
The second file "EmpPval" looks like
X obsExp.perExp obsExp.Pv obsExp.perExp.1000
ENSG00000000460.12 129 0.886162308 0.129
ENSG00000000971.11 268 0.728160071 0.268
So whenever the gene (1st column of both files) matches I want to extract the values for the corresponding gene from the "EmpPval" file and add it to the "shortListedGenes" file.
The code which I have tried till now is:
shortListedGenes <- read.csv("zs7-fdr0.05-Aorta.csv")
EmpPvAl <- read.csv("EmpiricalPvaluesAorta.csv")
for(i in 1:nrow(shortListedGenes))
{
for(j in 1:nrow(EmpPvAl))
{
if(shortListedGenes$gene[i] == EmpPvAl$X[j])
{
shortListedGenes$obsLessExp <- EmpPvAl$obsExp.perExp
shortListedGenes$obsExp <- EmpPvAl$obsExp.Pv
shortListedGenes$obsLessExpDiv <- EmpPvAl$obsExp.perExp.1000
}
}
}
and it is giving the error Error in Ops.factor(EmpPvAl$X, shortListedGenes$gene[i]) : level sets of factors are different
Any suggestion/help would be appreciated!!!. Thanks
r
r
asked Jan 3 at 6:06
StarStar
116
asked Jan 3 at 6:06
StarStar
116
asked Jan 3 at 6:06
StarStar
116
asked Jan 3 at 6:06
StarStar
116
asked Jan 3 at 6:06
StarStar
116
116
add a comment |
add a comment |
1 Answer
1
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You're looking to do a merge (all.x = TRUE, all.y = FALSE):
shortListedGenes <- merge(shortListedGenes, EmpPvAl[, c("X", "obsExp.perExp",
"obsExp.Pv", "obsExp.perExp.1000")], by.x = "gene", by.y = "X", all.x = TRUE, all.y = FALSE)
The issue with your solution is that the variables need to be created first, and you were missing some index notation in the assignment of each of the three new variables:
# Create variables
shortListedGenes$obsLessExp <- NA
shortListedGenes$obsExp <- NA
shortListedGenes$obsLessExpDiv <- NA
for(i in 1:nrow(shortListedGenes)){
for(j in 1:nrow(EmpPvAl)){
if(shortListedGenes$gene[i] == EmpPvAl$X[j]){
# Index notation for i and j added:
shortListedGenes$obsLessExp[i] <- EmpPvAl$obsExp.perExp[j]
shortListedGenes$obsExp[i] <- EmpPvAl$obsExp.Pv[j]
shortListedGenes$obsLessExpDiv[i] <- EmpPvAl$obsExp.perExp.1000[j]
}
}
}
answered Jan 3 at 6:26
KhaynesKhaynes
727721
Yes, it did work. Thankyou. But can you give some insight why my code was not working? I also tried if(as.character(shortListedGenes$gene[i])) == as.character(EmpPvAl$X[j])) :(
– Star
Jan 3 at 7:21
@star No worries. I've updated my answer to show you what you were missing. Hopefully you notice some speed improvements from usingmergethough :-)
– Khaynes
Jan 3 at 7:44
Yes of course merge is the more efficient way of doing this!!! I wanted to know what I was missing!!! Thankyou so much!!! Much appreciated!! :)
– Star
Jan 3 at 8:38
add a comment |
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1 Answer
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1 Answer
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You're looking to do a merge (all.x = TRUE, all.y = FALSE):
shortListedGenes <- merge(shortListedGenes, EmpPvAl[, c("X", "obsExp.perExp",
"obsExp.Pv", "obsExp.perExp.1000")], by.x = "gene", by.y = "X", all.x = TRUE, all.y = FALSE)
The issue with your solution is that the variables need to be created first, and you were missing some index notation in the assignment of each of the three new variables:
# Create variables
shortListedGenes$obsLessExp <- NA
shortListedGenes$obsExp <- NA
shortListedGenes$obsLessExpDiv <- NA
for(i in 1:nrow(shortListedGenes)){
for(j in 1:nrow(EmpPvAl)){
if(shortListedGenes$gene[i] == EmpPvAl$X[j]){
# Index notation for i and j added:
shortListedGenes$obsLessExp[i] <- EmpPvAl$obsExp.perExp[j]
shortListedGenes$obsExp[i] <- EmpPvAl$obsExp.Pv[j]
shortListedGenes$obsLessExpDiv[i] <- EmpPvAl$obsExp.perExp.1000[j]
}
}
}
answered Jan 3 at 6:26
KhaynesKhaynes
727721
Yes, it did work. Thankyou. But can you give some insight why my code was not working? I also tried if(as.character(shortListedGenes$gene[i])) == as.character(EmpPvAl$X[j])) :(
– Star
Jan 3 at 7:21
@star No worries. I've updated my answer to show you what you were missing. Hopefully you notice some speed improvements from usingmergethough :-)
– Khaynes
Jan 3 at 7:44
Yes of course merge is the more efficient way of doing this!!! I wanted to know what I was missing!!! Thankyou so much!!! Much appreciated!! :)
– Star
Jan 3 at 8:38
add a comment |
You're looking to do a merge (all.x = TRUE, all.y = FALSE):
shortListedGenes <- merge(shortListedGenes, EmpPvAl[, c("X", "obsExp.perExp",
"obsExp.Pv", "obsExp.perExp.1000")], by.x = "gene", by.y = "X", all.x = TRUE, all.y = FALSE)
The issue with your solution is that the variables need to be created first, and you were missing some index notation in the assignment of each of the three new variables:
# Create variables
shortListedGenes$obsLessExp <- NA
shortListedGenes$obsExp <- NA
shortListedGenes$obsLessExpDiv <- NA
for(i in 1:nrow(shortListedGenes)){
for(j in 1:nrow(EmpPvAl)){
if(shortListedGenes$gene[i] == EmpPvAl$X[j]){
# Index notation for i and j added:
shortListedGenes$obsLessExp[i] <- EmpPvAl$obsExp.perExp[j]
shortListedGenes$obsExp[i] <- EmpPvAl$obsExp.Pv[j]
shortListedGenes$obsLessExpDiv[i] <- EmpPvAl$obsExp.perExp.1000[j]
}
}
}
answered Jan 3 at 6:26
KhaynesKhaynes
727721
Yes, it did work. Thankyou. But can you give some insight why my code was not working? I also tried if(as.character(shortListedGenes$gene[i])) == as.character(EmpPvAl$X[j])) :(
– Star
Jan 3 at 7:21
@star No worries. I've updated my answer to show you what you were missing. Hopefully you notice some speed improvements from usingmergethough :-)
– Khaynes
Jan 3 at 7:44
Yes of course merge is the more efficient way of doing this!!! I wanted to know what I was missing!!! Thankyou so much!!! Much appreciated!! :)
– Star
Jan 3 at 8:38
add a comment |
You're looking to do a merge (all.x = TRUE, all.y = FALSE):
shortListedGenes <- merge(shortListedGenes, EmpPvAl[, c("X", "obsExp.perExp",
"obsExp.Pv", "obsExp.perExp.1000")], by.x = "gene", by.y = "X", all.x = TRUE, all.y = FALSE)
The issue with your solution is that the variables need to be created first, and you were missing some index notation in the assignment of each of the three new variables:
# Create variables
shortListedGenes$obsLessExp <- NA
shortListedGenes$obsExp <- NA
shortListedGenes$obsLessExpDiv <- NA
for(i in 1:nrow(shortListedGenes)){
for(j in 1:nrow(EmpPvAl)){
if(shortListedGenes$gene[i] == EmpPvAl$X[j]){
# Index notation for i and j added:
shortListedGenes$obsLessExp[i] <- EmpPvAl$obsExp.perExp[j]
shortListedGenes$obsExp[i] <- EmpPvAl$obsExp.Pv[j]
shortListedGenes$obsLessExpDiv[i] <- EmpPvAl$obsExp.perExp.1000[j]
}
}
}
answered Jan 3 at 6:26
KhaynesKhaynes
727721
You're looking to do a merge (all.x = TRUE, all.y = FALSE):
shortListedGenes <- merge(shortListedGenes, EmpPvAl[, c("X", "obsExp.perExp",
"obsExp.Pv", "obsExp.perExp.1000")], by.x = "gene", by.y = "X", all.x = TRUE, all.y = FALSE)
The issue with your solution is that the variables need to be created first, and you were missing some index notation in the assignment of each of the three new variables:
# Create variables
shortListedGenes$obsLessExp <- NA
shortListedGenes$obsExp <- NA
shortListedGenes$obsLessExpDiv <- NA
for(i in 1:nrow(shortListedGenes)){
for(j in 1:nrow(EmpPvAl)){
if(shortListedGenes$gene[i] == EmpPvAl$X[j]){
# Index notation for i and j added:
shortListedGenes$obsLessExp[i] <- EmpPvAl$obsExp.perExp[j]
shortListedGenes$obsExp[i] <- EmpPvAl$obsExp.Pv[j]
shortListedGenes$obsLessExpDiv[i] <- EmpPvAl$obsExp.perExp.1000[j]
}
}
}
answered Jan 3 at 6:26
KhaynesKhaynes
727721
edited Jan 3 at 7:43
answered Jan 3 at 6:26
KhaynesKhaynes
727721
answered Jan 3 at 6:26
KhaynesKhaynes
727721
answered Jan 3 at 6:26
KhaynesKhaynes
727721
727721
Yes, it did work. Thankyou. But can you give some insight why my code was not working? I also tried if(as.character(shortListedGenes$gene[i])) == as.character(EmpPvAl$X[j])) :(
– Star
Jan 3 at 7:21
@star No worries. I've updated my answer to show you what you were missing. Hopefully you notice some speed improvements from usingmergethough :-)
– Khaynes
Jan 3 at 7:44
Yes of course merge is the more efficient way of doing this!!! I wanted to know what I was missing!!! Thankyou so much!!! Much appreciated!! :)
– Star
Jan 3 at 8:38
add a comment |
Yes, it did work. Thankyou. But can you give some insight why my code was not working? I also tried if(as.character(shortListedGenes$gene[i])) == as.character(EmpPvAl$X[j])) :(
– Star
Jan 3 at 7:21
@star No worries. I've updated my answer to show you what you were missing. Hopefully you notice some speed improvements from usingmergethough :-)
– Khaynes
Jan 3 at 7:44
Yes of course merge is the more efficient way of doing this!!! I wanted to know what I was missing!!! Thankyou so much!!! Much appreciated!! :)
– Star
Jan 3 at 8:38
Yes, it did work. Thankyou. But can you give some insight why my code was not working? I also tried if(as.character(shortListedGenes$gene[i])) == as.character(EmpPvAl$X[j])) :(
– Star
Jan 3 at 7:21
Yes, it did work. Thankyou. But can you give some insight why my code was not working? I also tried if(as.character(shortListedGenes$gene[i])) == as.character(EmpPvAl$X[j])) :(
– Star
Jan 3 at 7:21
@star No worries. I've updated my answer to show you what you were missing. Hopefully you notice some speed improvements from using
merge though :-)– Khaynes
Jan 3 at 7:44
@star No worries. I've updated my answer to show you what you were missing. Hopefully you notice some speed improvements from using
merge though :-)– Khaynes
Jan 3 at 7:44
Yes of course merge is the more efficient way of doing this!!! I wanted to know what I was missing!!! Thankyou so much!!! Much appreciated!! :)
– Star
Jan 3 at 8:38
Yes of course merge is the more efficient way of doing this!!! I wanted to know what I was missing!!! Thankyou so much!!! Much appreciated!! :)
– Star
Jan 3 at 8:38
add a comment |
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