Mixing
Read comma-separated file which also have comma within a column
I have a file where the separator between columns is comma (,). However, comma may also occur within columns, i.e. the 'Notes' column:
Id,Notes,Other_ID
100,This text looks good,1000
101,This text,have,comma,2000
I tried to read the csv:
r <- read.csv("test.csv", sep = ",")
As a result, I received it as :
Id.Notes.GUID
100,This text is good,1000
102,This text,have,comma,2000
which is incorrect, as I would like to have the output as
Id Notes GUID
100 This text is good 1000
102 This text,have,comma 2000
Goal is to receive the data intact with the columns irrespective of comma present within a column and shouldn't work like a delimiter.
Thanks in Advance
r csv
edited Jan 2 at 17:21
Henrik
42.3k994110
asked Jan 2 at 16:49
AmbujAmbuj
61
add a comment |
I have a file where the separator between columns is comma (,). However, comma may also occur within columns, i.e. the 'Notes' column:
Id,Notes,Other_ID
100,This text looks good,1000
101,This text,have,comma,2000
I tried to read the csv:
r <- read.csv("test.csv", sep = ",")
As a result, I received it as :
Id.Notes.GUID
100,This text is good,1000
102,This text,have,comma,2000
which is incorrect, as I would like to have the output as
Id Notes GUID
100 This text is good 1000
102 This text,have,comma 2000
Goal is to receive the data intact with the columns irrespective of comma present within a column and shouldn't work like a delimiter.
Thanks in Advance
r csv
edited Jan 2 at 17:21
Henrik
42.3k994110
asked Jan 2 at 16:49
AmbujAmbuj
61
add a comment |
I have a file where the separator between columns is comma (,). However, comma may also occur within columns, i.e. the 'Notes' column:
Id,Notes,Other_ID
100,This text looks good,1000
101,This text,have,comma,2000
I tried to read the csv:
r <- read.csv("test.csv", sep = ",")
As a result, I received it as :
Id.Notes.GUID
100,This text is good,1000
102,This text,have,comma,2000
which is incorrect, as I would like to have the output as
Id Notes GUID
100 This text is good 1000
102 This text,have,comma 2000
Goal is to receive the data intact with the columns irrespective of comma present within a column and shouldn't work like a delimiter.
Thanks in Advance
r csv
edited Jan 2 at 17:21
Henrik
42.3k994110
asked Jan 2 at 16:49
AmbujAmbuj
61
I have a file where the separator between columns is comma (,). However, comma may also occur within columns, i.e. the 'Notes' column:
Id,Notes,Other_ID
100,This text looks good,1000
101,This text,have,comma,2000
I tried to read the csv:
r <- read.csv("test.csv", sep = ",")
As a result, I received it as :
Id.Notes.GUID
100,This text is good,1000
102,This text,have,comma,2000
which is incorrect, as I would like to have the output as
Id Notes GUID
100 This text is good 1000
102 This text,have,comma 2000
Goal is to receive the data intact with the columns irrespective of comma present within a column and shouldn't work like a delimiter.
Thanks in Advance
r csv
r csv
edited Jan 2 at 17:21
Henrik
42.3k994110
asked Jan 2 at 16:49
AmbujAmbuj
61
edited Jan 2 at 17:21
Henrik
42.3k994110
asked Jan 2 at 16:49
AmbujAmbuj
61
edited Jan 2 at 17:21
Henrik
42.3k994110
edited Jan 2 at 17:21
Henrik
42.3k994110
edited Jan 2 at 17:21
Henrik
42.3k994110
42.3k994110
asked Jan 2 at 16:49
AmbujAmbuj
61
asked Jan 2 at 16:49
AmbujAmbuj
61
asked Jan 2 at 16:49
AmbujAmbuj
61
61
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1) read.pattern read.pattern will read the fields according to the provided regular expression. For reproducibility we have used Lines in the Note below but if the data is in a file replace text=Lines with something like "myfile.csv" .
library(gsubfn)
read.pattern(text = Lines, pattern = "^(.*?),(.*),(.*)$", header = TRUE, as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
2) Base R Read the data into a character vector and replace the first and last comma on each line with some character that does not otherwise occur such as semicolon. Then read that.
L.raw <- readLines(textConnection(Lines))
L.semi <- sub(",(.*),", ";\1;", L.raw)
read.table(text = L.semi, header = TRUE, sep = ";", as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
3) gawk If you have a very large input file then it would likely be faster to do as much as possible outside of R. Using gawk we have the following. (On Windows install Rtools if you don't already have gawk and also make sure it is on your path or else refer to it using the entire pathname.) In the BEGIN block first is the number of commas to replace before the field with commas and last is the number of commas to replace after the field with commas. In this case the field with commas is the second of 3 fields so first = last = 1.
# generate test input
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
cat(Lines, file = "ambuj.dat")
# gawk program to replace commas
ambuj.awk <- '
BEGIN { first = 1; last = 1 }
{
nc = gsub(/,/, ",") # number of commas
for(i = nc; i > nc-last; i--) $0 = gensub(/,/, ";", i) # replace last last commas
for(i = 0; i < first; i++) sub(/,/, ";") # replace first first commas
print
}'
cat(ambuj.awk, file = "ambuj.awk")
read.csv(pipe("gawk -f ambuj.awk ambuj.dat"), sep = ";", quote = "",
comment.char = "")
Also you could set colClasses= to speed it up a bit more.
Note
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
answered Jan 2 at 16:56
G. GrothendieckG. Grothendieck
153k10136244
Awesome, worked like a charm, quick question: do we have to define (.*) as many times as number of columns present in data set?
– Ambuj
Jan 2 at 18:10
The pattern inread.patternmust correspond to the data so if the data has a different number of columns or the columns are in a different order then the pattern will have to be modified appropriately.
– G. Grothendieck
Jan 7 at 13:58
One more query I have, the number of comma(,) in that text column are not fixed, at some rows there are 2 and at some other there are 7. How can we determine the maximum number of commas and prevent them to split the text into several parts.
– Ambuj
Jan 7 at 14:06
The number of commas shouldn't matter. The^(.*?),captures everything up to the first comma, the first(.*),captures everything after the first comma until the last comma no matter how many commas that field contains and the(.*)$at the end captures the last field.
– G. Grothendieck
Jan 7 at 14:12
So do I have to enter (.*) as many times as number of text columns present in the database?Or does defining (.*) once will do the task? Because I have 8 columns with 3 numeric columns in beginning, 4th column as text with comma, 5th column blank,6th column number,7th column text and 8th column text.The only column have commas in text is 5th one for which I wrote the code as: ^(.?),(.?),(.?),(.),(.*),(.?),(.),(.*)$. Please correct me if I am wrong.
– Ambuj
Jan 7 at 15:19
|
show 4 more comments
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1 Answer
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active
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
1) read.pattern read.pattern will read the fields according to the provided regular expression. For reproducibility we have used Lines in the Note below but if the data is in a file replace text=Lines with something like "myfile.csv" .
library(gsubfn)
read.pattern(text = Lines, pattern = "^(.*?),(.*),(.*)$", header = TRUE, as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
2) Base R Read the data into a character vector and replace the first and last comma on each line with some character that does not otherwise occur such as semicolon. Then read that.
L.raw <- readLines(textConnection(Lines))
L.semi <- sub(",(.*),", ";\1;", L.raw)
read.table(text = L.semi, header = TRUE, sep = ";", as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
3) gawk If you have a very large input file then it would likely be faster to do as much as possible outside of R. Using gawk we have the following. (On Windows install Rtools if you don't already have gawk and also make sure it is on your path or else refer to it using the entire pathname.) In the BEGIN block first is the number of commas to replace before the field with commas and last is the number of commas to replace after the field with commas. In this case the field with commas is the second of 3 fields so first = last = 1.
# generate test input
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
cat(Lines, file = "ambuj.dat")
# gawk program to replace commas
ambuj.awk <- '
BEGIN { first = 1; last = 1 }
{
nc = gsub(/,/, ",") # number of commas
for(i = nc; i > nc-last; i--) $0 = gensub(/,/, ";", i) # replace last last commas
for(i = 0; i < first; i++) sub(/,/, ";") # replace first first commas
print
}'
cat(ambuj.awk, file = "ambuj.awk")
read.csv(pipe("gawk -f ambuj.awk ambuj.dat"), sep = ";", quote = "",
comment.char = "")
Also you could set colClasses= to speed it up a bit more.
Note
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
answered Jan 2 at 16:56
G. GrothendieckG. Grothendieck
153k10136244
Awesome, worked like a charm, quick question: do we have to define (.*) as many times as number of columns present in data set?
– Ambuj
Jan 2 at 18:10
The pattern inread.patternmust correspond to the data so if the data has a different number of columns or the columns are in a different order then the pattern will have to be modified appropriately.
– G. Grothendieck
Jan 7 at 13:58
One more query I have, the number of comma(,) in that text column are not fixed, at some rows there are 2 and at some other there are 7. How can we determine the maximum number of commas and prevent them to split the text into several parts.
– Ambuj
Jan 7 at 14:06
The number of commas shouldn't matter. The^(.*?),captures everything up to the first comma, the first(.*),captures everything after the first comma until the last comma no matter how many commas that field contains and the(.*)$at the end captures the last field.
– G. Grothendieck
Jan 7 at 14:12
So do I have to enter (.*) as many times as number of text columns present in the database?Or does defining (.*) once will do the task? Because I have 8 columns with 3 numeric columns in beginning, 4th column as text with comma, 5th column blank,6th column number,7th column text and 8th column text.The only column have commas in text is 5th one for which I wrote the code as: ^(.?),(.?),(.?),(.),(.*),(.?),(.),(.*)$. Please correct me if I am wrong.
– Ambuj
Jan 7 at 15:19
|
show 4 more comments
1) read.pattern read.pattern will read the fields according to the provided regular expression. For reproducibility we have used Lines in the Note below but if the data is in a file replace text=Lines with something like "myfile.csv" .
library(gsubfn)
read.pattern(text = Lines, pattern = "^(.*?),(.*),(.*)$", header = TRUE, as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
2) Base R Read the data into a character vector and replace the first and last comma on each line with some character that does not otherwise occur such as semicolon. Then read that.
L.raw <- readLines(textConnection(Lines))
L.semi <- sub(",(.*),", ";\1;", L.raw)
read.table(text = L.semi, header = TRUE, sep = ";", as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
3) gawk If you have a very large input file then it would likely be faster to do as much as possible outside of R. Using gawk we have the following. (On Windows install Rtools if you don't already have gawk and also make sure it is on your path or else refer to it using the entire pathname.) In the BEGIN block first is the number of commas to replace before the field with commas and last is the number of commas to replace after the field with commas. In this case the field with commas is the second of 3 fields so first = last = 1.
# generate test input
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
cat(Lines, file = "ambuj.dat")
# gawk program to replace commas
ambuj.awk <- '
BEGIN { first = 1; last = 1 }
{
nc = gsub(/,/, ",") # number of commas
for(i = nc; i > nc-last; i--) $0 = gensub(/,/, ";", i) # replace last last commas
for(i = 0; i < first; i++) sub(/,/, ";") # replace first first commas
print
}'
cat(ambuj.awk, file = "ambuj.awk")
read.csv(pipe("gawk -f ambuj.awk ambuj.dat"), sep = ";", quote = "",
comment.char = "")
Also you could set colClasses= to speed it up a bit more.
Note
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
answered Jan 2 at 16:56
G. GrothendieckG. Grothendieck
153k10136244
Awesome, worked like a charm, quick question: do we have to define (.*) as many times as number of columns present in data set?
– Ambuj
Jan 2 at 18:10
The pattern inread.patternmust correspond to the data so if the data has a different number of columns or the columns are in a different order then the pattern will have to be modified appropriately.
– G. Grothendieck
Jan 7 at 13:58
One more query I have, the number of comma(,) in that text column are not fixed, at some rows there are 2 and at some other there are 7. How can we determine the maximum number of commas and prevent them to split the text into several parts.
– Ambuj
Jan 7 at 14:06
The number of commas shouldn't matter. The^(.*?),captures everything up to the first comma, the first(.*),captures everything after the first comma until the last comma no matter how many commas that field contains and the(.*)$at the end captures the last field.
– G. Grothendieck
Jan 7 at 14:12
So do I have to enter (.*) as many times as number of text columns present in the database?Or does defining (.*) once will do the task? Because I have 8 columns with 3 numeric columns in beginning, 4th column as text with comma, 5th column blank,6th column number,7th column text and 8th column text.The only column have commas in text is 5th one for which I wrote the code as: ^(.?),(.?),(.?),(.),(.*),(.?),(.),(.*)$. Please correct me if I am wrong.
– Ambuj
Jan 7 at 15:19
|
show 4 more comments
1) read.pattern read.pattern will read the fields according to the provided regular expression. For reproducibility we have used Lines in the Note below but if the data is in a file replace text=Lines with something like "myfile.csv" .
library(gsubfn)
read.pattern(text = Lines, pattern = "^(.*?),(.*),(.*)$", header = TRUE, as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
2) Base R Read the data into a character vector and replace the first and last comma on each line with some character that does not otherwise occur such as semicolon. Then read that.
L.raw <- readLines(textConnection(Lines))
L.semi <- sub(",(.*),", ";\1;", L.raw)
read.table(text = L.semi, header = TRUE, sep = ";", as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
3) gawk If you have a very large input file then it would likely be faster to do as much as possible outside of R. Using gawk we have the following. (On Windows install Rtools if you don't already have gawk and also make sure it is on your path or else refer to it using the entire pathname.) In the BEGIN block first is the number of commas to replace before the field with commas and last is the number of commas to replace after the field with commas. In this case the field with commas is the second of 3 fields so first = last = 1.
# generate test input
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
cat(Lines, file = "ambuj.dat")
# gawk program to replace commas
ambuj.awk <- '
BEGIN { first = 1; last = 1 }
{
nc = gsub(/,/, ",") # number of commas
for(i = nc; i > nc-last; i--) $0 = gensub(/,/, ";", i) # replace last last commas
for(i = 0; i < first; i++) sub(/,/, ";") # replace first first commas
print
}'
cat(ambuj.awk, file = "ambuj.awk")
read.csv(pipe("gawk -f ambuj.awk ambuj.dat"), sep = ";", quote = "",
comment.char = "")
Also you could set colClasses= to speed it up a bit more.
Note
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
answered Jan 2 at 16:56
G. GrothendieckG. Grothendieck
153k10136244
1) read.pattern read.pattern will read the fields according to the provided regular expression. For reproducibility we have used Lines in the Note below but if the data is in a file replace text=Lines with something like "myfile.csv" .
library(gsubfn)
read.pattern(text = Lines, pattern = "^(.*?),(.*),(.*)$", header = TRUE, as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
2) Base R Read the data into a character vector and replace the first and last comma on each line with some character that does not otherwise occur such as semicolon. Then read that.
L.raw <- readLines(textConnection(Lines))
L.semi <- sub(",(.*),", ";\1;", L.raw)
read.table(text = L.semi, header = TRUE, sep = ";", as.is = TRUE)
giving:
Id Notes Other_ID
1 100 This text looks good 1000
2 101 This text,have,comma 2000
3) gawk If you have a very large input file then it would likely be faster to do as much as possible outside of R. Using gawk we have the following. (On Windows install Rtools if you don't already have gawk and also make sure it is on your path or else refer to it using the entire pathname.) In the BEGIN block first is the number of commas to replace before the field with commas and last is the number of commas to replace after the field with commas. In this case the field with commas is the second of 3 fields so first = last = 1.
# generate test input
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
cat(Lines, file = "ambuj.dat")
# gawk program to replace commas
ambuj.awk <- '
BEGIN { first = 1; last = 1 }
{
nc = gsub(/,/, ",") # number of commas
for(i = nc; i > nc-last; i--) $0 = gensub(/,/, ";", i) # replace last last commas
for(i = 0; i < first; i++) sub(/,/, ";") # replace first first commas
print
}'
cat(ambuj.awk, file = "ambuj.awk")
read.csv(pipe("gawk -f ambuj.awk ambuj.dat"), sep = ";", quote = "",
comment.char = "")
Also you could set colClasses= to speed it up a bit more.
Note
Lines <- "Id,Notes,Other_ID
100,This text looks good, 1000
101,This text,have,comma,2000"
answered Jan 2 at 16:56
G. GrothendieckG. Grothendieck
153k10136244
edited Jan 8 at 16:21
answered Jan 2 at 16:56
G. GrothendieckG. Grothendieck
153k10136244
answered Jan 2 at 16:56
G. GrothendieckG. Grothendieck
153k10136244
answered Jan 2 at 16:56
G. GrothendieckG. Grothendieck
153k10136244
153k10136244
Awesome, worked like a charm, quick question: do we have to define (.*) as many times as number of columns present in data set?
– Ambuj
Jan 2 at 18:10
The pattern inread.patternmust correspond to the data so if the data has a different number of columns or the columns are in a different order then the pattern will have to be modified appropriately.
– G. Grothendieck
Jan 7 at 13:58
One more query I have, the number of comma(,) in that text column are not fixed, at some rows there are 2 and at some other there are 7. How can we determine the maximum number of commas and prevent them to split the text into several parts.
– Ambuj
Jan 7 at 14:06
The number of commas shouldn't matter. The^(.*?),captures everything up to the first comma, the first(.*),captures everything after the first comma until the last comma no matter how many commas that field contains and the(.*)$at the end captures the last field.
– G. Grothendieck
Jan 7 at 14:12
So do I have to enter (.*) as many times as number of text columns present in the database?Or does defining (.*) once will do the task? Because I have 8 columns with 3 numeric columns in beginning, 4th column as text with comma, 5th column blank,6th column number,7th column text and 8th column text.The only column have commas in text is 5th one for which I wrote the code as: ^(.?),(.?),(.?),(.),(.*),(.?),(.),(.*)$. Please correct me if I am wrong.
– Ambuj
Jan 7 at 15:19
|
show 4 more comments
Awesome, worked like a charm, quick question: do we have to define (.*) as many times as number of columns present in data set?
– Ambuj
Jan 2 at 18:10
The pattern inread.patternmust correspond to the data so if the data has a different number of columns or the columns are in a different order then the pattern will have to be modified appropriately.
– G. Grothendieck
Jan 7 at 13:58
One more query I have, the number of comma(,) in that text column are not fixed, at some rows there are 2 and at some other there are 7. How can we determine the maximum number of commas and prevent them to split the text into several parts.
– Ambuj
Jan 7 at 14:06
The number of commas shouldn't matter. The^(.*?),captures everything up to the first comma, the first(.*),captures everything after the first comma until the last comma no matter how many commas that field contains and the(.*)$at the end captures the last field.
– G. Grothendieck
Jan 7 at 14:12
So do I have to enter (.*) as many times as number of text columns present in the database?Or does defining (.*) once will do the task? Because I have 8 columns with 3 numeric columns in beginning, 4th column as text with comma, 5th column blank,6th column number,7th column text and 8th column text.The only column have commas in text is 5th one for which I wrote the code as: ^(.?),(.?),(.?),(.),(.*),(.?),(.),(.*)$. Please correct me if I am wrong.
– Ambuj
Jan 7 at 15:19
Awesome, worked like a charm, quick question: do we have to define (.*) as many times as number of columns present in data set?
– Ambuj
Jan 2 at 18:10
Awesome, worked like a charm, quick question: do we have to define (.*) as many times as number of columns present in data set?
– Ambuj
Jan 2 at 18:10
The pattern in
read.pattern must correspond to the data so if the data has a different number of columns or the columns are in a different order then the pattern will have to be modified appropriately.– G. Grothendieck
Jan 7 at 13:58
The pattern in
read.pattern must correspond to the data so if the data has a different number of columns or the columns are in a different order then the pattern will have to be modified appropriately.– G. Grothendieck
Jan 7 at 13:58
One more query I have, the number of comma(,) in that text column are not fixed, at some rows there are 2 and at some other there are 7. How can we determine the maximum number of commas and prevent them to split the text into several parts.
– Ambuj
Jan 7 at 14:06
One more query I have, the number of comma(,) in that text column are not fixed, at some rows there are 2 and at some other there are 7. How can we determine the maximum number of commas and prevent them to split the text into several parts.
– Ambuj
Jan 7 at 14:06
The number of commas shouldn't matter. The
^(.*?), captures everything up to the first comma, the first (.*), captures everything after the first comma until the last comma no matter how many commas that field contains and the (.*)$ at the end captures the last field.– G. Grothendieck
Jan 7 at 14:12
The number of commas shouldn't matter. The
^(.*?), captures everything up to the first comma, the first (.*), captures everything after the first comma until the last comma no matter how many commas that field contains and the (.*)$ at the end captures the last field.– G. Grothendieck
Jan 7 at 14:12
So do I have to enter (.*) as many times as number of text columns present in the database?Or does defining (.*) once will do the task? Because I have 8 columns with 3 numeric columns in beginning, 4th column as text with comma, 5th column blank,6th column number,7th column text and 8th column text.The only column have commas in text is 5th one for which I wrote the code as: ^(.?),(.?),(.?),(.),(.*),(.?),(.),(.*)$. Please correct me if I am wrong.
– Ambuj
Jan 7 at 15:19
So do I have to enter (.*) as many times as number of text columns present in the database?Or does defining (.*) once will do the task? Because I have 8 columns with 3 numeric columns in beginning, 4th column as text with comma, 5th column blank,6th column number,7th column text and 8th column text.The only column have commas in text is 5th one for which I wrote the code as: ^(.?),(.?),(.?),(.),(.*),(.?),(.),(.*)$. Please correct me if I am wrong.
– Ambuj
Jan 7 at 15:19
|
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