extract sublist of dataframes from list of dataframes based on condition
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$treatment == "ZEO")
test <- ListofData[(cond)]
The name of the dataframes also contain the information about the treatment, that's why I tried this,but it returns an empty list
test<-ListofData[grep('^[Zeo]+',ListofData)]
Can you please help me to find a way to extract the data frames I need?
r regex list dataframe sublist
edited Nov 19 '18 at 10:11
zx8754
29.8k76399
asked Nov 19 '18 at 10:03
user163731user163731
61
add a comment |
I have a large list of data frames, and I want to create sub lists containing the data frames that fill a condition. Each data frame of the list has the same column names, and they have a column called treatment, which contains the word ZEO or BLEO. I'd like to be able to write a function or a one-liner that allows me to extract all the dataframes that have "ZEO". (note that one dataframe has only one treatment in it, so all the rows of the ListofData$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 Nov 19 '18 at 10:11
zx8754
29.8k76399
asked Nov 19 '18 at 10:03
user163731user163731
61
Maybe tryListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )orListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]
– zx8754
Nov 19 '18 at 10:08
1
@zx8754ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]doesn't work, howeverListofData_ZEO <- lapply(ListofData, function(x) x[ x$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
Nov 19 '18 at 10:34
add a comment |
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$treatment == "ZEO")
test <- ListofData[(cond)]
The name of the dataframes also contain the information about the treatment, that's why I tried this,but it returns an empty list
test<-ListofData[grep('^[Zeo]+',ListofData)]
Can you please help me to find a way to extract the data frames I need?
r regex list dataframe sublist
edited Nov 19 '18 at 10:11
zx8754
29.8k76399
asked Nov 19 '18 at 10:03
user163731user163731
61
I have a large list of data frames, and I want to create sub lists containing the data frames that fill a condition. Each data frame of the list has the same column names, and they have a column called treatment, which contains the word ZEO or BLEO. I'd like to be able to write a function or a one-liner that allows me to extract all the dataframes that have "ZEO". (note that one dataframe has only one treatment in it, so all the rows of the ListofData$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 Nov 19 '18 at 10:11
zx8754
29.8k76399
asked Nov 19 '18 at 10:03
user163731user163731
61
edited Nov 19 '18 at 10:11
zx8754
29.8k76399
asked Nov 19 '18 at 10:03
user163731user163731
61
edited Nov 19 '18 at 10:11
zx8754
29.8k76399
edited Nov 19 '18 at 10:11
zx8754
29.8k76399
edited Nov 19 '18 at 10:11
zx8754
29.8k76399
29.8k76399
asked Nov 19 '18 at 10:03
user163731user163731
61
asked Nov 19 '18 at 10:03
user163731user163731
61
asked Nov 19 '18 at 10:03
user163731user163731
61
61
Maybe tryListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )orListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]
– zx8754
Nov 19 '18 at 10:08
1
@zx8754ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]doesn't work, howeverListofData_ZEO <- lapply(ListofData, function(x) x[ x$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
Nov 19 '18 at 10:34
add a comment |
Maybe tryListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )orListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]
– zx8754
Nov 19 '18 at 10:08
1
@zx8754ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]doesn't work, howeverListofData_ZEO <- lapply(ListofData, function(x) x[ x$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
Nov 19 '18 at 10:34
Maybe try
ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] ) or ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]– zx8754
Nov 19 '18 at 10:08
Maybe try
ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] ) or ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]– zx8754
Nov 19 '18 at 10:08
1
1
@zx8754
ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ] doesn't work, however ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] ) almost does the job! It gives me a list where all the dataframes I need are good and the others that i don't need are still in the list but are empty. Any idea how to make them disappear? Thank you!– user163731
Nov 19 '18 at 10:34
@zx8754
ListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ] doesn't work, however ListofData_ZEO <- lapply(ListofData, function(x) x[ x$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
Nov 19 '18 at 10:34
add a comment |
1 Answer
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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 Nov 19 '18 at 10:38
zx8754zx8754
29.8k76399
add a comment |
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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 Nov 19 '18 at 10:38
zx8754zx8754
29.8k76399
add a comment |
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 Nov 19 '18 at 10:38
zx8754zx8754
29.8k76399
add a comment |
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 Nov 19 '18 at 10:38
zx8754zx8754
29.8k76399
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
#
# $ZEO2
# xx treatment
# 1 3 ZEO
answered Nov 19 '18 at 10:38
zx8754zx8754
29.8k76399
answered Nov 19 '18 at 10:38
zx8754zx8754
29.8k76399
answered Nov 19 '18 at 10:38
zx8754zx8754
29.8k76399
answered Nov 19 '18 at 10:38
zx8754zx8754
29.8k76399
29.8k76399
add a comment |
add a comment |
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Maybe try
ListofData_ZEO <- lapply(ListofData, function(x) x[ x$treatment == "ZEO", ] )orListofData_ZEO <- ListofData[grepl('^[Zeo]+', names(ListofData)) ]– zx8754
Nov 19 '18 at 10:08
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
Nov 19 '18 at 10:34