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extract sublist of dataframes from list of dataframes based on condition
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I have a large list of data frames, and I want to create sub lists containing the data frames that fill a condition. Each data frame of the list has the same column names, and they have a column called treatment, which contains the word ZEO or BLEO. I'd like to be able to write a function or a one-liner that allows me to extract all the dataframes that have "ZEO". (note that one dataframe has only one treatment in it, so all the rows of the ListofData$dataframe1$treatment are equal to ZEO), The list is large (~300 dataframes) and I have other variables I'd like to be able to extract. So far I tried these methods but they didn't seem to work
cond<- sapply(ListofData, function(x) x$treatment == "ZEO")
test <- ListofData[(cond)]
The name of the dataframes also contain the information about the treatment, that's why I tried this,but it returns an empty list
test<-ListofData[grep('^[Zeo]+',ListofData)]
Can you please help me to find a way to extract the data frames I need?
r regex list dataframe sublist
edited yesterday
zx8754
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asked yesterday
user163731
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up vote
-1
down vote
favorite
I have a large list of data frames, and I want to create sub lists containing the data frames that fill a condition. Each data frame of the list has the same column names, and they have a column called treatment, which contains the word ZEO or BLEO. I'd like to be able to write a function or a one-liner that allows me to extract all the dataframes that have "ZEO". (note that one dataframe has only one treatment in it, so all the rows of the ListofData$dataframe1$treatment are equal to ZEO), The list is large (~300 dataframes) and I have other variables I'd like to be able to extract. So far I tried these methods but they didn't seem to work
cond<- sapply(ListofData, function(x) x$treatment == "ZEO")
test <- ListofData[(cond)]
The name of the dataframes also contain the information about the treatment, that's why I tried this,but it returns an empty list
test<-ListofData[grep('^[Zeo]+',ListofData)]
Can you please help me to find a way to extract the data frames I need?
r regex list dataframe sublist
edited yesterday
zx8754
28.5k76394
asked yesterday
user163731
61
New contributor
user163731 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Maybe tryListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )orListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]
– zx8754
yesterday
1
@zx8754ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]doesn't work, howeverListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )almost does the job! It gives me a list where all the dataframes I need are good and the others that i don't need are still in the list but are empty. Any idea how to make them disappear? Thank you!
– user163731
yesterday
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have a large list of data frames, and I want to create sub lists containing the data frames that fill a condition. Each data frame of the list has the same column names, and they have a column called treatment, which contains the word ZEO or BLEO. I'd like to be able to write a function or a one-liner that allows me to extract all the dataframes that have "ZEO". (note that one dataframe has only one treatment in it, so all the rows of the ListofData$dataframe1$treatment are equal to ZEO), The list is large (~300 dataframes) and I have other variables I'd like to be able to extract. So far I tried these methods but they didn't seem to work
cond<- sapply(ListofData, function(x) x$treatment == "ZEO")
test <- ListofData[(cond)]
The name of the dataframes also contain the information about the treatment, that's why I tried this,but it returns an empty list
test<-ListofData[grep('^[Zeo]+',ListofData)]
Can you please help me to find a way to extract the data frames I need?
r regex list dataframe sublist
edited yesterday
zx8754
28.5k76394
asked yesterday
user163731
61
New contributor
user163731 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a large list of data frames, and I want to create sub lists containing the data frames that fill a condition. Each data frame of the list has the same column names, and they have a column called treatment, which contains the word ZEO or BLEO. I'd like to be able to write a function or a one-liner that allows me to extract all the dataframes that have "ZEO". (note that one dataframe has only one treatment in it, so all the rows of the ListofData$dataframe1$treatment are equal to ZEO), The list is large (~300 dataframes) and I have other variables I'd like to be able to extract. So far I tried these methods but they didn't seem to work
cond<- sapply(ListofData, function(x) x$treatment == "ZEO")
test <- ListofData[(cond)]
The name of the dataframes also contain the information about the treatment, that's why I tried this,but it returns an empty list
test<-ListofData[grep('^[Zeo]+',ListofData)]
Can you please help me to find a way to extract the data frames I need?
r regex list dataframe sublist
r regex list dataframe sublist
edited yesterday
zx8754
28.5k76394
asked yesterday
user163731
61
New contributor
user163731 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
zx8754
28.5k76394
asked yesterday
user163731
61
New contributor
user163731 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
zx8754
28.5k76394
edited yesterday
zx8754
28.5k76394
edited yesterday
zx8754
28.5k76394
28.5k76394
asked yesterday
user163731
61
New contributor
user163731 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
user163731
61
asked yesterday
user163731
61
61
New contributor
user163731 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user163731 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user163731 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Maybe tryListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )orListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]
– zx8754
yesterday
1
@zx8754ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]doesn't work, howeverListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )almost does the job! It gives me a list where all the dataframes I need are good and the others that i don't need are still in the list but are empty. Any idea how to make them disappear? Thank you!
– user163731
yesterday
add a comment |
Maybe tryListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )orListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]
– zx8754
yesterday
1
@zx8754ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]doesn't work, howeverListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )almost does the job! It gives me a list where all the dataframes I need are good and the others that i don't need are still in the list but are empty. Any idea how to make them disappear? Thank you!
– user163731
yesterday
Maybe try
ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] ) or ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]– zx8754
yesterday
Maybe try
ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] ) or ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]– zx8754
yesterday
1
1
@zx8754
ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ] doesn't work, however ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] ) almost does the job! It gives me a list where all the dataframes I need are good and the others that i don't need are still in the list but are empty. Any idea how to make them disappear? Thank you!– user163731
yesterday
@zx8754
ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ] doesn't work, however ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] ) almost does the job! It gives me a list where all the dataframes I need are good and the others that i don't need are still in the list but are empty. Any idea how to make them disappear? Thank you!– user163731
yesterday
add a comment |
1 Answer
1
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0
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Solution using regex should work, see this example:
#example data
ListofData <- list(ZEO1 = data.frame(xx = 1, treatment = "ZEO"),
xx1 = data.frame(xx = 2, treatment = "xx"),
ZEO2= data.frame(xx = 3, treatment = "ZEO"))
#using regex
res <- ListofData[ grepl("^[Zeo]+", names(ListofData)) ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
Here is another solution using column values, this returns empty dataframes, which we exclude using nrow and subset:
# using lapply, then filter
res <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )
res <- res[ sapply(res, nrow) > 0 ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
answered yesterday
zx8754
28.5k76394
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Solution using regex should work, see this example:
#example data
ListofData <- list(ZEO1 = data.frame(xx = 1, treatment = "ZEO"),
xx1 = data.frame(xx = 2, treatment = "xx"),
ZEO2= data.frame(xx = 3, treatment = "ZEO"))
#using regex
res <- ListofData[ grepl("^[Zeo]+", names(ListofData)) ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
Here is another solution using column values, this returns empty dataframes, which we exclude using nrow and subset:
# using lapply, then filter
res <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )
res <- res[ sapply(res, nrow) > 0 ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
answered yesterday
zx8754
28.5k76394
add a comment |
up vote
0
down vote
accepted
Solution using regex should work, see this example:
#example data
ListofData <- list(ZEO1 = data.frame(xx = 1, treatment = "ZEO"),
xx1 = data.frame(xx = 2, treatment = "xx"),
ZEO2= data.frame(xx = 3, treatment = "ZEO"))
#using regex
res <- ListofData[ grepl("^[Zeo]+", names(ListofData)) ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
Here is another solution using column values, this returns empty dataframes, which we exclude using nrow and subset:
# using lapply, then filter
res <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )
res <- res[ sapply(res, nrow) > 0 ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
answered yesterday
zx8754
28.5k76394
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Solution using regex should work, see this example:
#example data
ListofData <- list(ZEO1 = data.frame(xx = 1, treatment = "ZEO"),
xx1 = data.frame(xx = 2, treatment = "xx"),
ZEO2= data.frame(xx = 3, treatment = "ZEO"))
#using regex
res <- ListofData[ grepl("^[Zeo]+", names(ListofData)) ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
Here is another solution using column values, this returns empty dataframes, which we exclude using nrow and subset:
# using lapply, then filter
res <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )
res <- res[ sapply(res, nrow) > 0 ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
answered yesterday
zx8754
28.5k76394
Solution using regex should work, see this example:
#example data
ListofData <- list(ZEO1 = data.frame(xx = 1, treatment = "ZEO"),
xx1 = data.frame(xx = 2, treatment = "xx"),
ZEO2= data.frame(xx = 3, treatment = "ZEO"))
#using regex
res <- ListofData[ grepl("^[Zeo]+", names(ListofData)) ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
Here is another solution using column values, this returns empty dataframes, which we exclude using nrow and subset:
# using lapply, then filter
res <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )
res <- res[ sapply(res, nrow) > 0 ]
res
# $ZEO1
# xx treatment
# 1 1 ZEO
#
# $ZEO2
# xx treatment
# 1 3 ZEO
answered yesterday
zx8754
28.5k76394
answered yesterday
zx8754
28.5k76394
answered yesterday
zx8754
28.5k76394
answered yesterday
zx8754
28.5k76394
28.5k76394
add a comment |
add a comment |
user163731 is a new contributor. Be nice, and check out our Code of Conduct.
user163731 is a new contributor. Be nice, and check out our Code of Conduct.
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Maybe try
ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )orListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]– zx8754
yesterday
1
@zx8754
ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]doesn't work, howeverListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )almost does the job! It gives me a list where all the dataframes I need are good and the others that i don't need are still in the list but are empty. Any idea how to make them disappear? Thank you!– user163731
yesterday