Filter out all rows with only one period in R
I have this column, Identifier with character values.
structure(list(Identifier = c("RL.K", "RL.K.1", "RL.K.2", "RL.K.3",
"RL.K.4", "RL.K.5", "RL.K.6", "RL.K.7", "RL.K.9", "RL.K.10",
"RI.K", "RI.K.1", "RI.K.2", "RI.K.3", "RI.K.4", "RI.K.5", "RI.K.6",
"RI.K.7", "RI.K.9", "RI.K.10", "RF.K", "RF.K.1")), row.names = c(NA,
-22L), class = c("tbl_df", "tbl", "data.frame"))
How do I filter out the values with only one period? so that I can take out rows 1, 11, and 21
r dplyr
edited Nov 20 '18 at 17:46
Wiktor Stribiżew
310k16131206
asked Nov 20 '18 at 17:43
JasonBaikJasonBaik
17510
add a comment |
I have this column, Identifier with character values.
structure(list(Identifier = c("RL.K", "RL.K.1", "RL.K.2", "RL.K.3",
"RL.K.4", "RL.K.5", "RL.K.6", "RL.K.7", "RL.K.9", "RL.K.10",
"RI.K", "RI.K.1", "RI.K.2", "RI.K.3", "RI.K.4", "RI.K.5", "RI.K.6",
"RI.K.7", "RI.K.9", "RI.K.10", "RF.K", "RF.K.1")), row.names = c(NA,
-22L), class = c("tbl_df", "tbl", "data.frame"))
How do I filter out the values with only one period? so that I can take out rows 1, 11, and 21
r dplyr
edited Nov 20 '18 at 17:46
Wiktor Stribiżew
310k16131206
asked Nov 20 '18 at 17:43
JasonBaikJasonBaik
17510
add a comment |
I have this column, Identifier with character values.
structure(list(Identifier = c("RL.K", "RL.K.1", "RL.K.2", "RL.K.3",
"RL.K.4", "RL.K.5", "RL.K.6", "RL.K.7", "RL.K.9", "RL.K.10",
"RI.K", "RI.K.1", "RI.K.2", "RI.K.3", "RI.K.4", "RI.K.5", "RI.K.6",
"RI.K.7", "RI.K.9", "RI.K.10", "RF.K", "RF.K.1")), row.names = c(NA,
-22L), class = c("tbl_df", "tbl", "data.frame"))
How do I filter out the values with only one period? so that I can take out rows 1, 11, and 21
r dplyr
edited Nov 20 '18 at 17:46
Wiktor Stribiżew
310k16131206
asked Nov 20 '18 at 17:43
JasonBaikJasonBaik
17510
I have this column, Identifier with character values.
structure(list(Identifier = c("RL.K", "RL.K.1", "RL.K.2", "RL.K.3",
"RL.K.4", "RL.K.5", "RL.K.6", "RL.K.7", "RL.K.9", "RL.K.10",
"RI.K", "RI.K.1", "RI.K.2", "RI.K.3", "RI.K.4", "RI.K.5", "RI.K.6",
"RI.K.7", "RI.K.9", "RI.K.10", "RF.K", "RF.K.1")), row.names = c(NA,
-22L), class = c("tbl_df", "tbl", "data.frame"))
How do I filter out the values with only one period? so that I can take out rows 1, 11, and 21
r dplyr
r dplyr
edited Nov 20 '18 at 17:46
Wiktor Stribiżew
310k16131206
asked Nov 20 '18 at 17:43
JasonBaikJasonBaik
17510
edited Nov 20 '18 at 17:46
Wiktor Stribiżew
310k16131206
asked Nov 20 '18 at 17:43
JasonBaikJasonBaik
17510
edited Nov 20 '18 at 17:46
Wiktor Stribiżew
310k16131206
edited Nov 20 '18 at 17:46
Wiktor Stribiżew
310k16131206
edited Nov 20 '18 at 17:46
Wiktor Stribiżew
310k16131206
310k16131206
asked Nov 20 '18 at 17:43
JasonBaikJasonBaik
17510
asked Nov 20 '18 at 17:43
JasonBaikJasonBaik
17510
asked Nov 20 '18 at 17:43
JasonBaikJasonBaik
17510
17510
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
We can count the number of . in the 'Identifier' and create a logical condition for filtering the rows
library(tidyverse)
df1 %>%
filter(str_count(Identifier, "[.]") == 1)
# A tibble: 3 x 1
# Identifier
# <chr>
#1 RL.K
#2 RI.K
#3 RF.K
Or as @WiktorStribizew mentioned, fixed can be wrapped to make it more faster
df1 %>%
filter(str_count(Identifier, fixed(".")) == 1)
Or without using any external libraries,
df1[nchar(gsub("[^.]*", "", df1$Identifier)) == 1,]
Or using gregexpr from base R
df1[lengths(gregexpr(".", df1$Identifier, fixed = TRUE)) == 1,]
answered Nov 20 '18 at 17:44
akrunakrun
400k13190265
Why regex? There is just a dot to find, usestr_count(Identifier, fixed("."))
– Wiktor Stribiżew
Nov 20 '18 at 17:45
1
Wow, that was a quickie!
– JasonBaik
Nov 20 '18 at 17:47
add a comment |
If we're going to use base and grepl, there's a simpler regex code:
df[grepl("\..*\.", df$Identifier),]
(explanation for the regex: \. finds a literal ., .* finds anything, so this code finds cases where there are two literal dots separated by anything)
answered Nov 20 '18 at 17:58
iodiod
3,5792722
add a comment |
A solution using base R. (find all strings with exactly one dot)
grepl("^[^.]*[.][^.]*$", df1$Identifier)
To remove the rows with one dot use:
df1[
!grepl("^[^.]*[.][^.]*$", df1$Identifier),
]
answered Nov 20 '18 at 17:48
Andre ElricoAndre Elrico
5,63311027
1
it nees a!in front ofgreplexpression, since you want to filter out those with only one.for which the regex is searching.
– Gwang-Jin Kim
Nov 20 '18 at 17:59
thanks @Gwang-JinKim. I just realized "filter out" meant "remove".
– Andre Elrico
Nov 22 '18 at 11:09
add a comment |
With as little Regex as possible ;):
has.only.one.dot <- function(str_vec) sapply(strsplit(str_vec, "\."), function(vec) length(vec) == 2)
df[!has.only.one.dot(df$Identifier), ]
However, the list functions sapply and strsplit are slower than regex solution.
has.only.one.dot <- function(str_vec) grepl("\.", str_vec) & ! grepl("\..*\.", str_vec)
df[!has.only.one.dot(df$Identifier), ]
answered Nov 20 '18 at 18:06
Gwang-Jin KimGwang-Jin Kim
2,421116
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
We can count the number of . in the 'Identifier' and create a logical condition for filtering the rows
library(tidyverse)
df1 %>%
filter(str_count(Identifier, "[.]") == 1)
# A tibble: 3 x 1
# Identifier
# <chr>
#1 RL.K
#2 RI.K
#3 RF.K
Or as @WiktorStribizew mentioned, fixed can be wrapped to make it more faster
df1 %>%
filter(str_count(Identifier, fixed(".")) == 1)
Or without using any external libraries,
df1[nchar(gsub("[^.]*", "", df1$Identifier)) == 1,]
Or using gregexpr from base R
df1[lengths(gregexpr(".", df1$Identifier, fixed = TRUE)) == 1,]
answered Nov 20 '18 at 17:44
akrunakrun
400k13190265
Why regex? There is just a dot to find, usestr_count(Identifier, fixed("."))
– Wiktor Stribiżew
Nov 20 '18 at 17:45
1
Wow, that was a quickie!
– JasonBaik
Nov 20 '18 at 17:47
add a comment |
We can count the number of . in the 'Identifier' and create a logical condition for filtering the rows
library(tidyverse)
df1 %>%
filter(str_count(Identifier, "[.]") == 1)
# A tibble: 3 x 1
# Identifier
# <chr>
#1 RL.K
#2 RI.K
#3 RF.K
Or as @WiktorStribizew mentioned, fixed can be wrapped to make it more faster
df1 %>%
filter(str_count(Identifier, fixed(".")) == 1)
Or without using any external libraries,
df1[nchar(gsub("[^.]*", "", df1$Identifier)) == 1,]
Or using gregexpr from base R
df1[lengths(gregexpr(".", df1$Identifier, fixed = TRUE)) == 1,]
answered Nov 20 '18 at 17:44
akrunakrun
400k13190265
Why regex? There is just a dot to find, usestr_count(Identifier, fixed("."))
– Wiktor Stribiżew
Nov 20 '18 at 17:45
1
Wow, that was a quickie!
– JasonBaik
Nov 20 '18 at 17:47
add a comment |
We can count the number of . in the 'Identifier' and create a logical condition for filtering the rows
library(tidyverse)
df1 %>%
filter(str_count(Identifier, "[.]") == 1)
# A tibble: 3 x 1
# Identifier
# <chr>
#1 RL.K
#2 RI.K
#3 RF.K
Or as @WiktorStribizew mentioned, fixed can be wrapped to make it more faster
df1 %>%
filter(str_count(Identifier, fixed(".")) == 1)
Or without using any external libraries,
df1[nchar(gsub("[^.]*", "", df1$Identifier)) == 1,]
Or using gregexpr from base R
df1[lengths(gregexpr(".", df1$Identifier, fixed = TRUE)) == 1,]
answered Nov 20 '18 at 17:44
akrunakrun
400k13190265
We can count the number of . in the 'Identifier' and create a logical condition for filtering the rows
library(tidyverse)
df1 %>%
filter(str_count(Identifier, "[.]") == 1)
# A tibble: 3 x 1
# Identifier
# <chr>
#1 RL.K
#2 RI.K
#3 RF.K
Or as @WiktorStribizew mentioned, fixed can be wrapped to make it more faster
df1 %>%
filter(str_count(Identifier, fixed(".")) == 1)
Or without using any external libraries,
df1[nchar(gsub("[^.]*", "", df1$Identifier)) == 1,]
Or using gregexpr from base R
df1[lengths(gregexpr(".", df1$Identifier, fixed = TRUE)) == 1,]
answered Nov 20 '18 at 17:44
akrunakrun
400k13190265
edited Nov 20 '18 at 18:12
answered Nov 20 '18 at 17:44
akrunakrun
400k13190265
answered Nov 20 '18 at 17:44
akrunakrun
400k13190265
answered Nov 20 '18 at 17:44
akrunakrun
400k13190265
400k13190265
Why regex? There is just a dot to find, usestr_count(Identifier, fixed("."))
– Wiktor Stribiżew
Nov 20 '18 at 17:45
1
Wow, that was a quickie!
– JasonBaik
Nov 20 '18 at 17:47
add a comment |
Why regex? There is just a dot to find, usestr_count(Identifier, fixed("."))
– Wiktor Stribiżew
Nov 20 '18 at 17:45
1
Wow, that was a quickie!
– JasonBaik
Nov 20 '18 at 17:47
Why regex? There is just a dot to find, use
str_count(Identifier, fixed("."))– Wiktor Stribiżew
Nov 20 '18 at 17:45
Why regex? There is just a dot to find, use
str_count(Identifier, fixed("."))– Wiktor Stribiżew
Nov 20 '18 at 17:45
1
1
Wow, that was a quickie!
– JasonBaik
Nov 20 '18 at 17:47
Wow, that was a quickie!
– JasonBaik
Nov 20 '18 at 17:47
add a comment |
If we're going to use base and grepl, there's a simpler regex code:
df[grepl("\..*\.", df$Identifier),]
(explanation for the regex: \. finds a literal ., .* finds anything, so this code finds cases where there are two literal dots separated by anything)
answered Nov 20 '18 at 17:58
iodiod
3,5792722
add a comment |
If we're going to use base and grepl, there's a simpler regex code:
df[grepl("\..*\.", df$Identifier),]
(explanation for the regex: \. finds a literal ., .* finds anything, so this code finds cases where there are two literal dots separated by anything)
answered Nov 20 '18 at 17:58
iodiod
3,5792722
add a comment |
If we're going to use base and grepl, there's a simpler regex code:
df[grepl("\..*\.", df$Identifier),]
(explanation for the regex: \. finds a literal ., .* finds anything, so this code finds cases where there are two literal dots separated by anything)
answered Nov 20 '18 at 17:58
iodiod
3,5792722
If we're going to use base and grepl, there's a simpler regex code:
df[grepl("\..*\.", df$Identifier),]
(explanation for the regex: \. finds a literal ., .* finds anything, so this code finds cases where there are two literal dots separated by anything)
answered Nov 20 '18 at 17:58
iodiod
3,5792722
answered Nov 20 '18 at 17:58
iodiod
3,5792722
answered Nov 20 '18 at 17:58
iodiod
3,5792722
answered Nov 20 '18 at 17:58
iodiod
3,5792722
3,5792722
add a comment |
add a comment |
A solution using base R. (find all strings with exactly one dot)
grepl("^[^.]*[.][^.]*$", df1$Identifier)
To remove the rows with one dot use:
df1[
!grepl("^[^.]*[.][^.]*$", df1$Identifier),
]
answered Nov 20 '18 at 17:48
Andre ElricoAndre Elrico
5,63311027
1
it nees a!in front ofgreplexpression, since you want to filter out those with only one.for which the regex is searching.
– Gwang-Jin Kim
Nov 20 '18 at 17:59
thanks @Gwang-JinKim. I just realized "filter out" meant "remove".
– Andre Elrico
Nov 22 '18 at 11:09
add a comment |
A solution using base R. (find all strings with exactly one dot)
grepl("^[^.]*[.][^.]*$", df1$Identifier)
To remove the rows with one dot use:
df1[
!grepl("^[^.]*[.][^.]*$", df1$Identifier),
]
answered Nov 20 '18 at 17:48
Andre ElricoAndre Elrico
5,63311027
1
it nees a!in front ofgreplexpression, since you want to filter out those with only one.for which the regex is searching.
– Gwang-Jin Kim
Nov 20 '18 at 17:59
thanks @Gwang-JinKim. I just realized "filter out" meant "remove".
– Andre Elrico
Nov 22 '18 at 11:09
add a comment |
A solution using base R. (find all strings with exactly one dot)
grepl("^[^.]*[.][^.]*$", df1$Identifier)
To remove the rows with one dot use:
df1[
!grepl("^[^.]*[.][^.]*$", df1$Identifier),
]
answered Nov 20 '18 at 17:48
Andre ElricoAndre Elrico
5,63311027
A solution using base R. (find all strings with exactly one dot)
grepl("^[^.]*[.][^.]*$", df1$Identifier)
To remove the rows with one dot use:
df1[
!grepl("^[^.]*[.][^.]*$", df1$Identifier),
]
answered Nov 20 '18 at 17:48
Andre ElricoAndre Elrico
5,63311027
edited Nov 22 '18 at 11:11
answered Nov 20 '18 at 17:48
Andre ElricoAndre Elrico
5,63311027
answered Nov 20 '18 at 17:48
Andre ElricoAndre Elrico
5,63311027
answered Nov 20 '18 at 17:48
Andre ElricoAndre Elrico
5,63311027
5,63311027
1
it nees a!in front ofgreplexpression, since you want to filter out those with only one.for which the regex is searching.
– Gwang-Jin Kim
Nov 20 '18 at 17:59
thanks @Gwang-JinKim. I just realized "filter out" meant "remove".
– Andre Elrico
Nov 22 '18 at 11:09
add a comment |
1
it nees a!in front ofgreplexpression, since you want to filter out those with only one.for which the regex is searching.
– Gwang-Jin Kim
Nov 20 '18 at 17:59
thanks @Gwang-JinKim. I just realized "filter out" meant "remove".
– Andre Elrico
Nov 22 '18 at 11:09
1
1
it nees a
! in front of grepl expression, since you want to filter out those with only one . for which the regex is searching.– Gwang-Jin Kim
Nov 20 '18 at 17:59
it nees a
! in front of grepl expression, since you want to filter out those with only one . for which the regex is searching.– Gwang-Jin Kim
Nov 20 '18 at 17:59
thanks @Gwang-JinKim. I just realized "filter out" meant "remove".
– Andre Elrico
Nov 22 '18 at 11:09
thanks @Gwang-JinKim. I just realized "filter out" meant "remove".
– Andre Elrico
Nov 22 '18 at 11:09
add a comment |
With as little Regex as possible ;):
has.only.one.dot <- function(str_vec) sapply(strsplit(str_vec, "\."), function(vec) length(vec) == 2)
df[!has.only.one.dot(df$Identifier), ]
However, the list functions sapply and strsplit are slower than regex solution.
has.only.one.dot <- function(str_vec) grepl("\.", str_vec) & ! grepl("\..*\.", str_vec)
df[!has.only.one.dot(df$Identifier), ]
answered Nov 20 '18 at 18:06
Gwang-Jin KimGwang-Jin Kim
2,421116
add a comment |
With as little Regex as possible ;):
has.only.one.dot <- function(str_vec) sapply(strsplit(str_vec, "\."), function(vec) length(vec) == 2)
df[!has.only.one.dot(df$Identifier), ]
However, the list functions sapply and strsplit are slower than regex solution.
has.only.one.dot <- function(str_vec) grepl("\.", str_vec) & ! grepl("\..*\.", str_vec)
df[!has.only.one.dot(df$Identifier), ]
answered Nov 20 '18 at 18:06
Gwang-Jin KimGwang-Jin Kim
2,421116
add a comment |
With as little Regex as possible ;):
has.only.one.dot <- function(str_vec) sapply(strsplit(str_vec, "\."), function(vec) length(vec) == 2)
df[!has.only.one.dot(df$Identifier), ]
However, the list functions sapply and strsplit are slower than regex solution.
has.only.one.dot <- function(str_vec) grepl("\.", str_vec) & ! grepl("\..*\.", str_vec)
df[!has.only.one.dot(df$Identifier), ]
answered Nov 20 '18 at 18:06
Gwang-Jin KimGwang-Jin Kim
2,421116
With as little Regex as possible ;):
has.only.one.dot <- function(str_vec) sapply(strsplit(str_vec, "\."), function(vec) length(vec) == 2)
df[!has.only.one.dot(df$Identifier), ]
However, the list functions sapply and strsplit are slower than regex solution.
has.only.one.dot <- function(str_vec) grepl("\.", str_vec) & ! grepl("\..*\.", str_vec)
df[!has.only.one.dot(df$Identifier), ]
answered Nov 20 '18 at 18:06
Gwang-Jin KimGwang-Jin Kim
2,421116
edited Nov 20 '18 at 18:11
answered Nov 20 '18 at 18:06
Gwang-Jin KimGwang-Jin Kim
2,421116
answered Nov 20 '18 at 18:06
Gwang-Jin KimGwang-Jin Kim
2,421116
answered Nov 20 '18 at 18:06
Gwang-Jin KimGwang-Jin Kim
2,421116
2,421116
add a comment |
add a comment |
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