How to add Double quotes and comma to a column in R dataframe?
0
I have a dataframe with a single column. I want all values in the column to be wrapped in a double quote and comma like below.
df <- data.frame("SN" = 1:4, "Name" = c("John", "Dora","Peter","Lilly"), stringsAsFactors = FALSE)
df$Name
When I extract the Name column, I want it df$Name as
"John",
"Dora",
"Peter",
"Lilly"
I tried paste, gsub and shQuote but it doesn't give me the result that I want. Any help is appreciated.
r dataframe paste gsub double-quotes
edited Nov 14 '18 at 2:33
hrbrmstr
60.8k688150
asked Nov 14 '18 at 2:28
DatamaniacDatamaniac
32
add a comment |
0
I have a dataframe with a single column. I want all values in the column to be wrapped in a double quote and comma like below.
df <- data.frame("SN" = 1:4, "Name" = c("John", "Dora","Peter","Lilly"), stringsAsFactors = FALSE)
df$Name
When I extract the Name column, I want it df$Name as
"John",
"Dora",
"Peter",
"Lilly"
I tried paste, gsub and shQuote but it doesn't give me the result that I want. Any help is appreciated.
r dataframe paste gsub double-quotes
edited Nov 14 '18 at 2:33
hrbrmstr
60.8k688150
asked Nov 14 '18 at 2:28
DatamaniacDatamaniac
32
add a comment |
0
0
0
1
I have a dataframe with a single column. I want all values in the column to be wrapped in a double quote and comma like below.
df <- data.frame("SN" = 1:4, "Name" = c("John", "Dora","Peter","Lilly"), stringsAsFactors = FALSE)
df$Name
When I extract the Name column, I want it df$Name as
"John",
"Dora",
"Peter",
"Lilly"
I tried paste, gsub and shQuote but it doesn't give me the result that I want. Any help is appreciated.
r dataframe paste gsub double-quotes
edited Nov 14 '18 at 2:33
hrbrmstr
60.8k688150
asked Nov 14 '18 at 2:28
DatamaniacDatamaniac
32
I have a dataframe with a single column. I want all values in the column to be wrapped in a double quote and comma like below.
df <- data.frame("SN" = 1:4, "Name" = c("John", "Dora","Peter","Lilly"), stringsAsFactors = FALSE)
df$Name
When I extract the Name column, I want it df$Name as
"John",
"Dora",
"Peter",
"Lilly"
I tried paste, gsub and shQuote but it doesn't give me the result that I want. Any help is appreciated.
r dataframe paste gsub double-quotes
r dataframe paste gsub double-quotes
edited Nov 14 '18 at 2:33
hrbrmstr
60.8k688150
asked Nov 14 '18 at 2:28
DatamaniacDatamaniac
32
edited Nov 14 '18 at 2:33
hrbrmstr
60.8k688150
asked Nov 14 '18 at 2:28
DatamaniacDatamaniac
32
edited Nov 14 '18 at 2:33
hrbrmstr
60.8k688150
edited Nov 14 '18 at 2:33
hrbrmstr
60.8k688150
edited Nov 14 '18 at 2:33
hrbrmstr
60.8k688150
60.8k688150
asked Nov 14 '18 at 2:28
DatamaniacDatamaniac
32
asked Nov 14 '18 at 2:28
DatamaniacDatamaniac
32
asked Nov 14 '18 at 2:28
DatamaniacDatamaniac
32
32
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
data.frame(
SN = 1:4,
Name = c("John", "Dora", "Peter", "Lilly"),
stringsAsFactors = FALSE
) -> xdf
sprintf():
sprintf('"%s",', xdf$Name)
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sprintf('"%s",', xdf$Name), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
paste():
paste('"', xdf$Name, '",', sep = "")
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(paste('"', xdf$Name, '",', sep = ""), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
sub():
sub("$", '",', sub("^", '"', xdf$Name))
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sub("$", '",', sub("^", '"', xdf$Name)), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
FWIW sprintf() wins:
microbenchmark::microbenchmark(
sprintf = sprintf('"%s",', xdf$Name),
paste = paste('"', xdf$Name, '",', sep = ""),
sub = sub("$", '",', sub("^", '"', xdf$Name))
)
## Unit: microseconds
## expr min lq mean median uq max neval
## sprintf 11.705 13.4955 25.91594 19.6095 32.1780 170.989 100
## paste 13.133 15.3150 34.31982 19.0455 31.9615 340.482 100
## sub 31.271 34.8555 64.41124 39.8840 78.0180 533.065 100
UPDATE
I kinda figured you were aiming for:
paste0(sprintf('"%s"', xdf$Name), collapse = ", ")
## [1] ""John", "Dora", "Peter", "Lilly""
cat(paste0(sprintf('"%s"', xdf$Name), collapse = ", "))
## "John", "Dora", "Peter", "Lilly"
answered Nov 14 '18 at 2:36
hrbrmstrhrbrmstr
60.8k688150
Thanks much. If I have to remove the comma from last value of Name column, how do I do that ?
– Datamaniac
Nov 14 '18 at 2:46
I figured you had an actual task at hand. Please add all your requirements and final desired output to the actual question vs get to it piecemeal, but check the update.
– hrbrmstr
Nov 14 '18 at 2:47
The update you posted gives me values instead of a dataframe with a single column. When I use as.data.frame, its an empty list
– Datamaniac
Nov 14 '18 at 3:04
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
data.frame(
SN = 1:4,
Name = c("John", "Dora", "Peter", "Lilly"),
stringsAsFactors = FALSE
) -> xdf
sprintf():
sprintf('"%s",', xdf$Name)
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sprintf('"%s",', xdf$Name), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
paste():
paste('"', xdf$Name, '",', sep = "")
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(paste('"', xdf$Name, '",', sep = ""), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
sub():
sub("$", '",', sub("^", '"', xdf$Name))
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sub("$", '",', sub("^", '"', xdf$Name)), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
FWIW sprintf() wins:
microbenchmark::microbenchmark(
sprintf = sprintf('"%s",', xdf$Name),
paste = paste('"', xdf$Name, '",', sep = ""),
sub = sub("$", '",', sub("^", '"', xdf$Name))
)
## Unit: microseconds
## expr min lq mean median uq max neval
## sprintf 11.705 13.4955 25.91594 19.6095 32.1780 170.989 100
## paste 13.133 15.3150 34.31982 19.0455 31.9615 340.482 100
## sub 31.271 34.8555 64.41124 39.8840 78.0180 533.065 100
UPDATE
I kinda figured you were aiming for:
paste0(sprintf('"%s"', xdf$Name), collapse = ", ")
## [1] ""John", "Dora", "Peter", "Lilly""
cat(paste0(sprintf('"%s"', xdf$Name), collapse = ", "))
## "John", "Dora", "Peter", "Lilly"
answered Nov 14 '18 at 2:36
hrbrmstrhrbrmstr
60.8k688150
Thanks much. If I have to remove the comma from last value of Name column, how do I do that ?
– Datamaniac
Nov 14 '18 at 2:46
I figured you had an actual task at hand. Please add all your requirements and final desired output to the actual question vs get to it piecemeal, but check the update.
– hrbrmstr
Nov 14 '18 at 2:47
The update you posted gives me values instead of a dataframe with a single column. When I use as.data.frame, its an empty list
– Datamaniac
Nov 14 '18 at 3:04
add a comment |
2
data.frame(
SN = 1:4,
Name = c("John", "Dora", "Peter", "Lilly"),
stringsAsFactors = FALSE
) -> xdf
sprintf():
sprintf('"%s",', xdf$Name)
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sprintf('"%s",', xdf$Name), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
paste():
paste('"', xdf$Name, '",', sep = "")
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(paste('"', xdf$Name, '",', sep = ""), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
sub():
sub("$", '",', sub("^", '"', xdf$Name))
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sub("$", '",', sub("^", '"', xdf$Name)), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
FWIW sprintf() wins:
microbenchmark::microbenchmark(
sprintf = sprintf('"%s",', xdf$Name),
paste = paste('"', xdf$Name, '",', sep = ""),
sub = sub("$", '",', sub("^", '"', xdf$Name))
)
## Unit: microseconds
## expr min lq mean median uq max neval
## sprintf 11.705 13.4955 25.91594 19.6095 32.1780 170.989 100
## paste 13.133 15.3150 34.31982 19.0455 31.9615 340.482 100
## sub 31.271 34.8555 64.41124 39.8840 78.0180 533.065 100
UPDATE
I kinda figured you were aiming for:
paste0(sprintf('"%s"', xdf$Name), collapse = ", ")
## [1] ""John", "Dora", "Peter", "Lilly""
cat(paste0(sprintf('"%s"', xdf$Name), collapse = ", "))
## "John", "Dora", "Peter", "Lilly"
answered Nov 14 '18 at 2:36
hrbrmstrhrbrmstr
60.8k688150
Thanks much. If I have to remove the comma from last value of Name column, how do I do that ?
– Datamaniac
Nov 14 '18 at 2:46
I figured you had an actual task at hand. Please add all your requirements and final desired output to the actual question vs get to it piecemeal, but check the update.
– hrbrmstr
Nov 14 '18 at 2:47
The update you posted gives me values instead of a dataframe with a single column. When I use as.data.frame, its an empty list
– Datamaniac
Nov 14 '18 at 3:04
add a comment |
2
2
2
data.frame(
SN = 1:4,
Name = c("John", "Dora", "Peter", "Lilly"),
stringsAsFactors = FALSE
) -> xdf
sprintf():
sprintf('"%s",', xdf$Name)
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sprintf('"%s",', xdf$Name), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
paste():
paste('"', xdf$Name, '",', sep = "")
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(paste('"', xdf$Name, '",', sep = ""), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
sub():
sub("$", '",', sub("^", '"', xdf$Name))
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sub("$", '",', sub("^", '"', xdf$Name)), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
FWIW sprintf() wins:
microbenchmark::microbenchmark(
sprintf = sprintf('"%s",', xdf$Name),
paste = paste('"', xdf$Name, '",', sep = ""),
sub = sub("$", '",', sub("^", '"', xdf$Name))
)
## Unit: microseconds
## expr min lq mean median uq max neval
## sprintf 11.705 13.4955 25.91594 19.6095 32.1780 170.989 100
## paste 13.133 15.3150 34.31982 19.0455 31.9615 340.482 100
## sub 31.271 34.8555 64.41124 39.8840 78.0180 533.065 100
UPDATE
I kinda figured you were aiming for:
paste0(sprintf('"%s"', xdf$Name), collapse = ", ")
## [1] ""John", "Dora", "Peter", "Lilly""
cat(paste0(sprintf('"%s"', xdf$Name), collapse = ", "))
## "John", "Dora", "Peter", "Lilly"
answered Nov 14 '18 at 2:36
hrbrmstrhrbrmstr
60.8k688150
data.frame(
SN = 1:4,
Name = c("John", "Dora", "Peter", "Lilly"),
stringsAsFactors = FALSE
) -> xdf
sprintf():
sprintf('"%s",', xdf$Name)
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sprintf('"%s",', xdf$Name), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
paste():
paste('"', xdf$Name, '",', sep = "")
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(paste('"', xdf$Name, '",', sep = ""), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
sub():
sub("$", '",', sub("^", '"', xdf$Name))
## [1] ""John"," ""Dora"," ""Peter"," ""Lilly","
cat(sub("$", '",', sub("^", '"', xdf$Name)), sep="n")
## "John",
## "Dora",
## "Peter",
## "Lilly",
FWIW sprintf() wins:
microbenchmark::microbenchmark(
sprintf = sprintf('"%s",', xdf$Name),
paste = paste('"', xdf$Name, '",', sep = ""),
sub = sub("$", '",', sub("^", '"', xdf$Name))
)
## Unit: microseconds
## expr min lq mean median uq max neval
## sprintf 11.705 13.4955 25.91594 19.6095 32.1780 170.989 100
## paste 13.133 15.3150 34.31982 19.0455 31.9615 340.482 100
## sub 31.271 34.8555 64.41124 39.8840 78.0180 533.065 100
UPDATE
I kinda figured you were aiming for:
paste0(sprintf('"%s"', xdf$Name), collapse = ", ")
## [1] ""John", "Dora", "Peter", "Lilly""
cat(paste0(sprintf('"%s"', xdf$Name), collapse = ", "))
## "John", "Dora", "Peter", "Lilly"
answered Nov 14 '18 at 2:36
hrbrmstrhrbrmstr
60.8k688150
edited Nov 14 '18 at 2:48
answered Nov 14 '18 at 2:36
hrbrmstrhrbrmstr
60.8k688150
answered Nov 14 '18 at 2:36
hrbrmstrhrbrmstr
60.8k688150
answered Nov 14 '18 at 2:36
hrbrmstrhrbrmstr
60.8k688150
60.8k688150
Thanks much. If I have to remove the comma from last value of Name column, how do I do that ?
– Datamaniac
Nov 14 '18 at 2:46
I figured you had an actual task at hand. Please add all your requirements and final desired output to the actual question vs get to it piecemeal, but check the update.
– hrbrmstr
Nov 14 '18 at 2:47
The update you posted gives me values instead of a dataframe with a single column. When I use as.data.frame, its an empty list
– Datamaniac
Nov 14 '18 at 3:04
add a comment |
Thanks much. If I have to remove the comma from last value of Name column, how do I do that ?
– Datamaniac
Nov 14 '18 at 2:46
I figured you had an actual task at hand. Please add all your requirements and final desired output to the actual question vs get to it piecemeal, but check the update.
– hrbrmstr
Nov 14 '18 at 2:47
The update you posted gives me values instead of a dataframe with a single column. When I use as.data.frame, its an empty list
– Datamaniac
Nov 14 '18 at 3:04
Thanks much. If I have to remove the comma from last value of Name column, how do I do that ?
– Datamaniac
Nov 14 '18 at 2:46
Thanks much. If I have to remove the comma from last value of Name column, how do I do that ?
– Datamaniac
Nov 14 '18 at 2:46
I figured you had an actual task at hand. Please add all your requirements and final desired output to the actual question vs get to it piecemeal, but check the update.
– hrbrmstr
Nov 14 '18 at 2:47
I figured you had an actual task at hand. Please add all your requirements and final desired output to the actual question vs get to it piecemeal, but check the update.
– hrbrmstr
Nov 14 '18 at 2:47
The update you posted gives me values instead of a dataframe with a single column. When I use as.data.frame, its an empty list
– Datamaniac
Nov 14 '18 at 3:04
The update you posted gives me values instead of a dataframe with a single column. When I use as.data.frame, its an empty list
– Datamaniac
Nov 14 '18 at 3:04
add a comment |
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