Mixing
R - splitting a column of varied string length in a data frame into multiple columns of just one character
I have a data frame like this:
Name S1 S2 S3 Symbol
n_12 2.3 6.1 0 A
n_13 3.4 3.7 0 ACM
n_14 1.3 1.0 0 BN
n_23 2.0 4.1 0 NOPXY
And I am looking to split the last column, Symbol, into multiple columns, each with one character or nothing.
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
n_12 2.3 6.1 0 A
n_13 3.4 3.7 0 A C M
n_14 1.3 1.0 0 B N
n_23 2.0 4.1 0 N O P X Y
Thank you for any and all help with this.
r split
edited Oct 16 '18 at 23:16
divibisan
4,89681833
asked Oct 16 '18 at 22:56
learning5023learning5023
122
add a comment |
I have a data frame like this:
Name S1 S2 S3 Symbol
n_12 2.3 6.1 0 A
n_13 3.4 3.7 0 ACM
n_14 1.3 1.0 0 BN
n_23 2.0 4.1 0 NOPXY
And I am looking to split the last column, Symbol, into multiple columns, each with one character or nothing.
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
n_12 2.3 6.1 0 A
n_13 3.4 3.7 0 A C M
n_14 1.3 1.0 0 B N
n_23 2.0 4.1 0 N O P X Y
Thank you for any and all help with this.
r split
edited Oct 16 '18 at 23:16
divibisan
4,89681833
asked Oct 16 '18 at 22:56
learning5023learning5023
122
add a comment |
I have a data frame like this:
Name S1 S2 S3 Symbol
n_12 2.3 6.1 0 A
n_13 3.4 3.7 0 ACM
n_14 1.3 1.0 0 BN
n_23 2.0 4.1 0 NOPXY
And I am looking to split the last column, Symbol, into multiple columns, each with one character or nothing.
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
n_12 2.3 6.1 0 A
n_13 3.4 3.7 0 A C M
n_14 1.3 1.0 0 B N
n_23 2.0 4.1 0 N O P X Y
Thank you for any and all help with this.
r split
edited Oct 16 '18 at 23:16
divibisan
4,89681833
asked Oct 16 '18 at 22:56
learning5023learning5023
122
I have a data frame like this:
Name S1 S2 S3 Symbol
n_12 2.3 6.1 0 A
n_13 3.4 3.7 0 ACM
n_14 1.3 1.0 0 BN
n_23 2.0 4.1 0 NOPXY
And I am looking to split the last column, Symbol, into multiple columns, each with one character or nothing.
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
n_12 2.3 6.1 0 A
n_13 3.4 3.7 0 A C M
n_14 1.3 1.0 0 B N
n_23 2.0 4.1 0 N O P X Y
Thank you for any and all help with this.
r split
r split
edited Oct 16 '18 at 23:16
divibisan
4,89681833
asked Oct 16 '18 at 22:56
learning5023learning5023
122
edited Oct 16 '18 at 23:16
divibisan
4,89681833
asked Oct 16 '18 at 22:56
learning5023learning5023
122
edited Oct 16 '18 at 23:16
divibisan
4,89681833
edited Oct 16 '18 at 23:16
divibisan
4,89681833
edited Oct 16 '18 at 23:16
divibisan
4,89681833
4,89681833
asked Oct 16 '18 at 22:56
learning5023learning5023
122
asked Oct 16 '18 at 22:56
learning5023learning5023
122
asked Oct 16 '18 at 22:56
learning5023learning5023
122
122
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
One way to do this is with tidyr::separate which splits a single column containing a string into multiple columns containing substrings.
df
Name S1 S2 S3 Symbol
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 ACM
3 n_14 1.3 1.0 0 BN
4 n_23 2.0 4.1 0 NOPXY
The sep= argument for separate accepts either a regex, or a numeric vector listing the positions in the string to split on. Since we want to split after every character, we want to give a numeric sequence from 1 to the length of the longest string (-1, since we don't need to split after the last character). The length of the longest string is calculated with max(nchar(.$Symbol)). Thanks to Rich Scriven for pointing out that nchar is vectorized and so doesn't need to be called with sapply.
We then make a character vector with the names of the columns to split Symbol into. In your case, we can just paste 'Sy' to that same numeric sequence to get c('Sy1', 'Sy2' ...)
df %>%
tidyr::separate(Symbol,
sep = seq_len(max(nchar(.$Symbol)) - 1),
into = paste0('Sy', seq_len(max(nchar(.$Symbol)))))
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 A C M
3 n_14 1.3 1.0 0 B N
4 n_23 2.0 4.1 0 N O P X Y
If you get the following error:
Error in nchar(.$Symbol) : 'nchar()' requires a character vector
then it is likely that df$Symbol is of type factor (the default when creating or loading a data.frame) not character.
You can either provide read.table or data.frame with the argument stringsAsFactor=F to keep the Symbol variable from being converted to factor, or convert it back to character.
Tidyverse option (which can be inserted into the pipe just before the call to tidyr::separate:
df <- df %>%
dplyr::mutate(Symbol = as.character(Symbol))
or with base R:
df$Symbol <- as.character(df$Symbol)
answered Oct 16 '18 at 23:15
divibisandivibisan
4,89681833
That's because becauseSymbolis of typefactornotcharacter. You should use thestringsAsFactor=Fwhen loading/creating the data.frame, or convert it withdplyr::mutate(Symbol = as.character(Symbol))ordf$Symbol <- as.character(df$Symbol)
– divibisan
Oct 17 '18 at 15:58
Ah, I see- that works - thank you so much!
– learning5023
Oct 17 '18 at 16:03
add a comment |
Here's a base R version using strcapture:
ns <- max(nchar(dat$Symbol))
cbind(
dat,
strcapture(
paste(rep("(.)", ns), collapse=""),
format(dat$Symbol, width=ns),
proto=setNames(rep(list(""), ns), paste0("Sy",1:ns))
)
)
A late base R addition using substring, which loops over each of the inputs, including the start and ends of each substring:
dat[paste0("Sy",seq(ns))] <- matrix(substring(rep(dat$Symbol,each=ns),
seq(ns), seq(ns)), ncol=ns, byrow=TRUE)
# Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
#1 n_12 2.3 6.1 0 A A
#2 n_13 3.4 3.7 0 ACM A C M
#3 n_14 1.3 1.0 0 BN B N
#4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:42
thelatemailthelatemail
68k883151
add a comment |
Here's an R base using brute force:
string <- strsplit(df$Symbol, "")
ind <- max(lengths(string))
out <- data.frame(df, do.call(rbind, lapply(string, function(x) {
if(length(x) != ind){
c(x[1:length(x)], x[(length(x)+1):ind] )
}else{
x
}
})))
names(out) <- sub("X(\d)", "Sy\1", names(out))
print(out, na.print = "")
Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A A
2 n_13 3.4 3.7 0 ACM A C M
3 n_14 1.3 1.0 0 BN B N
4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:56
Jilber UrbinaJilber Urbina
43k482114
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f52845046%2fr-splitting-a-column-of-varied-string-length-in-a-data-frame-into-multiple-col%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
One way to do this is with tidyr::separate which splits a single column containing a string into multiple columns containing substrings.
df
Name S1 S2 S3 Symbol
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 ACM
3 n_14 1.3 1.0 0 BN
4 n_23 2.0 4.1 0 NOPXY
The sep= argument for separate accepts either a regex, or a numeric vector listing the positions in the string to split on. Since we want to split after every character, we want to give a numeric sequence from 1 to the length of the longest string (-1, since we don't need to split after the last character). The length of the longest string is calculated with max(nchar(.$Symbol)). Thanks to Rich Scriven for pointing out that nchar is vectorized and so doesn't need to be called with sapply.
We then make a character vector with the names of the columns to split Symbol into. In your case, we can just paste 'Sy' to that same numeric sequence to get c('Sy1', 'Sy2' ...)
df %>%
tidyr::separate(Symbol,
sep = seq_len(max(nchar(.$Symbol)) - 1),
into = paste0('Sy', seq_len(max(nchar(.$Symbol)))))
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 A C M
3 n_14 1.3 1.0 0 B N
4 n_23 2.0 4.1 0 N O P X Y
If you get the following error:
Error in nchar(.$Symbol) : 'nchar()' requires a character vector
then it is likely that df$Symbol is of type factor (the default when creating or loading a data.frame) not character.
You can either provide read.table or data.frame with the argument stringsAsFactor=F to keep the Symbol variable from being converted to factor, or convert it back to character.
Tidyverse option (which can be inserted into the pipe just before the call to tidyr::separate:
df <- df %>%
dplyr::mutate(Symbol = as.character(Symbol))
or with base R:
df$Symbol <- as.character(df$Symbol)
answered Oct 16 '18 at 23:15
divibisandivibisan
4,89681833
That's because becauseSymbolis of typefactornotcharacter. You should use thestringsAsFactor=Fwhen loading/creating the data.frame, or convert it withdplyr::mutate(Symbol = as.character(Symbol))ordf$Symbol <- as.character(df$Symbol)
– divibisan
Oct 17 '18 at 15:58
Ah, I see- that works - thank you so much!
– learning5023
Oct 17 '18 at 16:03
add a comment |
One way to do this is with tidyr::separate which splits a single column containing a string into multiple columns containing substrings.
df
Name S1 S2 S3 Symbol
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 ACM
3 n_14 1.3 1.0 0 BN
4 n_23 2.0 4.1 0 NOPXY
The sep= argument for separate accepts either a regex, or a numeric vector listing the positions in the string to split on. Since we want to split after every character, we want to give a numeric sequence from 1 to the length of the longest string (-1, since we don't need to split after the last character). The length of the longest string is calculated with max(nchar(.$Symbol)). Thanks to Rich Scriven for pointing out that nchar is vectorized and so doesn't need to be called with sapply.
We then make a character vector with the names of the columns to split Symbol into. In your case, we can just paste 'Sy' to that same numeric sequence to get c('Sy1', 'Sy2' ...)
df %>%
tidyr::separate(Symbol,
sep = seq_len(max(nchar(.$Symbol)) - 1),
into = paste0('Sy', seq_len(max(nchar(.$Symbol)))))
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 A C M
3 n_14 1.3 1.0 0 B N
4 n_23 2.0 4.1 0 N O P X Y
If you get the following error:
Error in nchar(.$Symbol) : 'nchar()' requires a character vector
then it is likely that df$Symbol is of type factor (the default when creating or loading a data.frame) not character.
You can either provide read.table or data.frame with the argument stringsAsFactor=F to keep the Symbol variable from being converted to factor, or convert it back to character.
Tidyverse option (which can be inserted into the pipe just before the call to tidyr::separate:
df <- df %>%
dplyr::mutate(Symbol = as.character(Symbol))
or with base R:
df$Symbol <- as.character(df$Symbol)
answered Oct 16 '18 at 23:15
divibisandivibisan
4,89681833
That's because becauseSymbolis of typefactornotcharacter. You should use thestringsAsFactor=Fwhen loading/creating the data.frame, or convert it withdplyr::mutate(Symbol = as.character(Symbol))ordf$Symbol <- as.character(df$Symbol)
– divibisan
Oct 17 '18 at 15:58
Ah, I see- that works - thank you so much!
– learning5023
Oct 17 '18 at 16:03
add a comment |
One way to do this is with tidyr::separate which splits a single column containing a string into multiple columns containing substrings.
df
Name S1 S2 S3 Symbol
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 ACM
3 n_14 1.3 1.0 0 BN
4 n_23 2.0 4.1 0 NOPXY
The sep= argument for separate accepts either a regex, or a numeric vector listing the positions in the string to split on. Since we want to split after every character, we want to give a numeric sequence from 1 to the length of the longest string (-1, since we don't need to split after the last character). The length of the longest string is calculated with max(nchar(.$Symbol)). Thanks to Rich Scriven for pointing out that nchar is vectorized and so doesn't need to be called with sapply.
We then make a character vector with the names of the columns to split Symbol into. In your case, we can just paste 'Sy' to that same numeric sequence to get c('Sy1', 'Sy2' ...)
df %>%
tidyr::separate(Symbol,
sep = seq_len(max(nchar(.$Symbol)) - 1),
into = paste0('Sy', seq_len(max(nchar(.$Symbol)))))
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 A C M
3 n_14 1.3 1.0 0 B N
4 n_23 2.0 4.1 0 N O P X Y
If you get the following error:
Error in nchar(.$Symbol) : 'nchar()' requires a character vector
then it is likely that df$Symbol is of type factor (the default when creating or loading a data.frame) not character.
You can either provide read.table or data.frame with the argument stringsAsFactor=F to keep the Symbol variable from being converted to factor, or convert it back to character.
Tidyverse option (which can be inserted into the pipe just before the call to tidyr::separate:
df <- df %>%
dplyr::mutate(Symbol = as.character(Symbol))
or with base R:
df$Symbol <- as.character(df$Symbol)
answered Oct 16 '18 at 23:15
divibisandivibisan
4,89681833
One way to do this is with tidyr::separate which splits a single column containing a string into multiple columns containing substrings.
df
Name S1 S2 S3 Symbol
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 ACM
3 n_14 1.3 1.0 0 BN
4 n_23 2.0 4.1 0 NOPXY
The sep= argument for separate accepts either a regex, or a numeric vector listing the positions in the string to split on. Since we want to split after every character, we want to give a numeric sequence from 1 to the length of the longest string (-1, since we don't need to split after the last character). The length of the longest string is calculated with max(nchar(.$Symbol)). Thanks to Rich Scriven for pointing out that nchar is vectorized and so doesn't need to be called with sapply.
We then make a character vector with the names of the columns to split Symbol into. In your case, we can just paste 'Sy' to that same numeric sequence to get c('Sy1', 'Sy2' ...)
df %>%
tidyr::separate(Symbol,
sep = seq_len(max(nchar(.$Symbol)) - 1),
into = paste0('Sy', seq_len(max(nchar(.$Symbol)))))
Name S1 S2 S3 Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A
2 n_13 3.4 3.7 0 A C M
3 n_14 1.3 1.0 0 B N
4 n_23 2.0 4.1 0 N O P X Y
If you get the following error:
Error in nchar(.$Symbol) : 'nchar()' requires a character vector
then it is likely that df$Symbol is of type factor (the default when creating or loading a data.frame) not character.
You can either provide read.table or data.frame with the argument stringsAsFactor=F to keep the Symbol variable from being converted to factor, or convert it back to character.
Tidyverse option (which can be inserted into the pipe just before the call to tidyr::separate:
df <- df %>%
dplyr::mutate(Symbol = as.character(Symbol))
or with base R:
df$Symbol <- as.character(df$Symbol)
answered Oct 16 '18 at 23:15
divibisandivibisan
4,89681833
edited Oct 17 '18 at 16:02
answered Oct 16 '18 at 23:15
divibisandivibisan
4,89681833
answered Oct 16 '18 at 23:15
divibisandivibisan
4,89681833
answered Oct 16 '18 at 23:15
divibisandivibisan
4,89681833
4,89681833
That's because becauseSymbolis of typefactornotcharacter. You should use thestringsAsFactor=Fwhen loading/creating the data.frame, or convert it withdplyr::mutate(Symbol = as.character(Symbol))ordf$Symbol <- as.character(df$Symbol)
– divibisan
Oct 17 '18 at 15:58
Ah, I see- that works - thank you so much!
– learning5023
Oct 17 '18 at 16:03
add a comment |
That's because becauseSymbolis of typefactornotcharacter. You should use thestringsAsFactor=Fwhen loading/creating the data.frame, or convert it withdplyr::mutate(Symbol = as.character(Symbol))ordf$Symbol <- as.character(df$Symbol)
– divibisan
Oct 17 '18 at 15:58
Ah, I see- that works - thank you so much!
– learning5023
Oct 17 '18 at 16:03
That's because because
Symbol is of type factor not character. You should use the stringsAsFactor=F when loading/creating the data.frame, or convert it with dplyr::mutate(Symbol = as.character(Symbol)) or df$Symbol <- as.character(df$Symbol)– divibisan
Oct 17 '18 at 15:58
That's because because
Symbol is of type factor not character. You should use the stringsAsFactor=F when loading/creating the data.frame, or convert it with dplyr::mutate(Symbol = as.character(Symbol)) or df$Symbol <- as.character(df$Symbol)– divibisan
Oct 17 '18 at 15:58
Ah, I see- that works - thank you so much!
– learning5023
Oct 17 '18 at 16:03
Ah, I see- that works - thank you so much!
– learning5023
Oct 17 '18 at 16:03
add a comment |
Here's a base R version using strcapture:
ns <- max(nchar(dat$Symbol))
cbind(
dat,
strcapture(
paste(rep("(.)", ns), collapse=""),
format(dat$Symbol, width=ns),
proto=setNames(rep(list(""), ns), paste0("Sy",1:ns))
)
)
A late base R addition using substring, which loops over each of the inputs, including the start and ends of each substring:
dat[paste0("Sy",seq(ns))] <- matrix(substring(rep(dat$Symbol,each=ns),
seq(ns), seq(ns)), ncol=ns, byrow=TRUE)
# Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
#1 n_12 2.3 6.1 0 A A
#2 n_13 3.4 3.7 0 ACM A C M
#3 n_14 1.3 1.0 0 BN B N
#4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:42
thelatemailthelatemail
68k883151
add a comment |
Here's a base R version using strcapture:
ns <- max(nchar(dat$Symbol))
cbind(
dat,
strcapture(
paste(rep("(.)", ns), collapse=""),
format(dat$Symbol, width=ns),
proto=setNames(rep(list(""), ns), paste0("Sy",1:ns))
)
)
A late base R addition using substring, which loops over each of the inputs, including the start and ends of each substring:
dat[paste0("Sy",seq(ns))] <- matrix(substring(rep(dat$Symbol,each=ns),
seq(ns), seq(ns)), ncol=ns, byrow=TRUE)
# Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
#1 n_12 2.3 6.1 0 A A
#2 n_13 3.4 3.7 0 ACM A C M
#3 n_14 1.3 1.0 0 BN B N
#4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:42
thelatemailthelatemail
68k883151
add a comment |
Here's a base R version using strcapture:
ns <- max(nchar(dat$Symbol))
cbind(
dat,
strcapture(
paste(rep("(.)", ns), collapse=""),
format(dat$Symbol, width=ns),
proto=setNames(rep(list(""), ns), paste0("Sy",1:ns))
)
)
A late base R addition using substring, which loops over each of the inputs, including the start and ends of each substring:
dat[paste0("Sy",seq(ns))] <- matrix(substring(rep(dat$Symbol,each=ns),
seq(ns), seq(ns)), ncol=ns, byrow=TRUE)
# Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
#1 n_12 2.3 6.1 0 A A
#2 n_13 3.4 3.7 0 ACM A C M
#3 n_14 1.3 1.0 0 BN B N
#4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:42
thelatemailthelatemail
68k883151
Here's a base R version using strcapture:
ns <- max(nchar(dat$Symbol))
cbind(
dat,
strcapture(
paste(rep("(.)", ns), collapse=""),
format(dat$Symbol, width=ns),
proto=setNames(rep(list(""), ns), paste0("Sy",1:ns))
)
)
A late base R addition using substring, which loops over each of the inputs, including the start and ends of each substring:
dat[paste0("Sy",seq(ns))] <- matrix(substring(rep(dat$Symbol,each=ns),
seq(ns), seq(ns)), ncol=ns, byrow=TRUE)
# Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
#1 n_12 2.3 6.1 0 A A
#2 n_13 3.4 3.7 0 ACM A C M
#3 n_14 1.3 1.0 0 BN B N
#4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:42
thelatemailthelatemail
68k883151
edited Jan 2 at 1:11
answered Oct 16 '18 at 23:42
thelatemailthelatemail
68k883151
answered Oct 16 '18 at 23:42
thelatemailthelatemail
68k883151
answered Oct 16 '18 at 23:42
thelatemailthelatemail
68k883151
68k883151
add a comment |
add a comment |
Here's an R base using brute force:
string <- strsplit(df$Symbol, "")
ind <- max(lengths(string))
out <- data.frame(df, do.call(rbind, lapply(string, function(x) {
if(length(x) != ind){
c(x[1:length(x)], x[(length(x)+1):ind] )
}else{
x
}
})))
names(out) <- sub("X(\d)", "Sy\1", names(out))
print(out, na.print = "")
Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A A
2 n_13 3.4 3.7 0 ACM A C M
3 n_14 1.3 1.0 0 BN B N
4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:56
Jilber UrbinaJilber Urbina
43k482114
add a comment |
Here's an R base using brute force:
string <- strsplit(df$Symbol, "")
ind <- max(lengths(string))
out <- data.frame(df, do.call(rbind, lapply(string, function(x) {
if(length(x) != ind){
c(x[1:length(x)], x[(length(x)+1):ind] )
}else{
x
}
})))
names(out) <- sub("X(\d)", "Sy\1", names(out))
print(out, na.print = "")
Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A A
2 n_13 3.4 3.7 0 ACM A C M
3 n_14 1.3 1.0 0 BN B N
4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:56
Jilber UrbinaJilber Urbina
43k482114
add a comment |
Here's an R base using brute force:
string <- strsplit(df$Symbol, "")
ind <- max(lengths(string))
out <- data.frame(df, do.call(rbind, lapply(string, function(x) {
if(length(x) != ind){
c(x[1:length(x)], x[(length(x)+1):ind] )
}else{
x
}
})))
names(out) <- sub("X(\d)", "Sy\1", names(out))
print(out, na.print = "")
Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A A
2 n_13 3.4 3.7 0 ACM A C M
3 n_14 1.3 1.0 0 BN B N
4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:56
Jilber UrbinaJilber Urbina
43k482114
Here's an R base using brute force:
string <- strsplit(df$Symbol, "")
ind <- max(lengths(string))
out <- data.frame(df, do.call(rbind, lapply(string, function(x) {
if(length(x) != ind){
c(x[1:length(x)], x[(length(x)+1):ind] )
}else{
x
}
})))
names(out) <- sub("X(\d)", "Sy\1", names(out))
print(out, na.print = "")
Name S1 S2 S3 Symbol Sy1 Sy2 Sy3 Sy4 Sy5
1 n_12 2.3 6.1 0 A A
2 n_13 3.4 3.7 0 ACM A C M
3 n_14 1.3 1.0 0 BN B N
4 n_23 2.0 4.1 0 NOPXY N O P X Y
answered Oct 16 '18 at 23:56
Jilber UrbinaJilber Urbina
43k482114
answered Oct 16 '18 at 23:56
Jilber UrbinaJilber Urbina
43k482114
answered Oct 16 '18 at 23:56
Jilber UrbinaJilber Urbina
43k482114
answered Oct 16 '18 at 23:56
Jilber UrbinaJilber Urbina
43k482114
43k482114
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f52845046%2fr-splitting-a-column-of-varied-string-length-in-a-data-frame-into-multiple-col%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
