Finding strings that are a certain length and contain specific characters
1
Sample data
a<-c("hour","four","ruoh", "six", "high", "our")
I want to find all strings that contain o & u & h & are 4 characters but the order does not matter.
I want to return "hour","four","ruoh"
this is my attempt
grepl("o+u+r", a) nchar(a)==4
r string grepl
asked Nov 14 '18 at 23:22
bvowebvowe
30818
|
show 2 more comments
1
Sample data
a<-c("hour","four","ruoh", "six", "high", "our")
I want to find all strings that contain o & u & h & are 4 characters but the order does not matter.
I want to return "hour","four","ruoh"
this is my attempt
grepl("o+u+r", a) nchar(a)==4
r string grepl
asked Nov 14 '18 at 23:22
bvowebvowe
30818
What about testing each separately. You first test (with grep) which elements of the vector contains "o", those who pass, you test if they has "u" and those who pass you test for "h".
– Cris
Nov 14 '18 at 23:26
@Cris is this the most simple approach to do so?
– bvowe
Nov 14 '18 at 23:28
5
"four" does not contain o & u & h.
– neilfws
Nov 14 '18 at 23:28
@neilfws I have now done a modification
– bvowe
Nov 14 '18 at 23:34
1
See Regular Expressions: Is there an AND operator?;grepl("(?=.*h)(?=.*o)(?=.*u)", a, perl = TRUE)
– Henrik
Nov 14 '18 at 23:43
|
show 2 more comments
1
1
1
Sample data
a<-c("hour","four","ruoh", "six", "high", "our")
I want to find all strings that contain o & u & h & are 4 characters but the order does not matter.
I want to return "hour","four","ruoh"
this is my attempt
grepl("o+u+r", a) nchar(a)==4
r string grepl
asked Nov 14 '18 at 23:22
bvowebvowe
30818
Sample data
a<-c("hour","four","ruoh", "six", "high", "our")
I want to find all strings that contain o & u & h & are 4 characters but the order does not matter.
I want to return "hour","four","ruoh"
this is my attempt
grepl("o+u+r", a) nchar(a)==4
r string grepl
r string grepl
asked Nov 14 '18 at 23:22
bvowebvowe
30818
asked Nov 14 '18 at 23:22
bvowebvowe
30818
edited Nov 14 '18 at 23:34
bvowe
asked Nov 14 '18 at 23:22
bvowebvowe
30818
asked Nov 14 '18 at 23:22
bvowebvowe
30818
asked Nov 14 '18 at 23:22
bvowebvowe
30818
30818
What about testing each separately. You first test (with grep) which elements of the vector contains "o", those who pass, you test if they has "u" and those who pass you test for "h".
– Cris
Nov 14 '18 at 23:26
@Cris is this the most simple approach to do so?
– bvowe
Nov 14 '18 at 23:28
5
"four" does not contain o & u & h.
– neilfws
Nov 14 '18 at 23:28
@neilfws I have now done a modification
– bvowe
Nov 14 '18 at 23:34
1
See Regular Expressions: Is there an AND operator?;grepl("(?=.*h)(?=.*o)(?=.*u)", a, perl = TRUE)
– Henrik
Nov 14 '18 at 23:43
|
show 2 more comments
What about testing each separately. You first test (with grep) which elements of the vector contains "o", those who pass, you test if they has "u" and those who pass you test for "h".
– Cris
Nov 14 '18 at 23:26
@Cris is this the most simple approach to do so?
– bvowe
Nov 14 '18 at 23:28
5
"four" does not contain o & u & h.
– neilfws
Nov 14 '18 at 23:28
@neilfws I have now done a modification
– bvowe
Nov 14 '18 at 23:34
1
See Regular Expressions: Is there an AND operator?;grepl("(?=.*h)(?=.*o)(?=.*u)", a, perl = TRUE)
– Henrik
Nov 14 '18 at 23:43
What about testing each separately. You first test (with grep) which elements of the vector contains "o", those who pass, you test if they has "u" and those who pass you test for "h".
– Cris
Nov 14 '18 at 23:26
What about testing each separately. You first test (with grep) which elements of the vector contains "o", those who pass, you test if they has "u" and those who pass you test for "h".
– Cris
Nov 14 '18 at 23:26
@Cris is this the most simple approach to do so?
– bvowe
Nov 14 '18 at 23:28
@Cris is this the most simple approach to do so?
– bvowe
Nov 14 '18 at 23:28
5
5
"four" does not contain o & u & h.
– neilfws
Nov 14 '18 at 23:28
"four" does not contain o & u & h.
– neilfws
Nov 14 '18 at 23:28
@neilfws I have now done a modification
– bvowe
Nov 14 '18 at 23:34
@neilfws I have now done a modification
– bvowe
Nov 14 '18 at 23:34
1
1
See Regular Expressions: Is there an AND operator?;
grepl("(?=.*h)(?=.*o)(?=.*u)", a, perl = TRUE)– Henrik
Nov 14 '18 at 23:43
See Regular Expressions: Is there an AND operator?;
grepl("(?=.*h)(?=.*o)(?=.*u)", a, perl = TRUE)– Henrik
Nov 14 '18 at 23:43
|
show 2 more comments
3 Answers
3
active
oldest
votes
1
Using grepl with your edited method (r instead of h):
a<-c("hour","four","ruoh", "six", "high", "our")
a[grepl(pattern="o", x=a) & grepl(pattern="u", x=a) & grepl(pattern="r", x=a) & nchar(a)==4]
Returns:
[1] "hour" "four" "ruoh"
answered Nov 14 '18 at 23:42
CrisCris
498311
add a comment |
2
To match strings of length 4 containing the characters h, o, and u use:
grepl("(?=^.{4}$)(?=.*h)(?=.*o)(?=.*u)",
c("hour","four","ruoh", "six", "high", "our"),
perl = TRUE)
[1] TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
(?=^.{4}$): string has length 4.
(?=.*x):xoccurs at any position in string.
answered Nov 15 '18 at 0:36
FlorianFlorian
1,092817
add a comment |
1
You could use strsplit and setdiff, I added an additional edge case to your sample data :
a<-c("hour","four","ruoh", "six", "high", "our","oouh")
a[nchar(a) == 4 &
lengths(lapply(strsplit(a,""),function(x) setdiff(x, c("o","u","h")))) == 1]
# [1] "hour" "ruoh"
or grepl :
a[nchar(a) == 4 & !rowSums(sapply(c("o","u","h"), Negate(grepl), a))]
# [1] "hour" "ruoh" "oouh"
sapply(c("o","u","h"), Negate(grepl), a) gives you a matrix of which word doesn't contain each letter, then the rowSums acts like any applied by row, as it will be coerced to logical.
answered Nov 14 '18 at 23:28
Moody_MudskipperMoody_Mudskipper
23.1k33264
This might have to be tweaked depending on how you want to treat some edge cases (multiple "h" for example)
– Moody_Mudskipper
Nov 14 '18 at 23:31
thanks a bunch @Moody_Mudskipper do you have grepl solution?
– bvowe
Nov 14 '18 at 23:35
see edited answer
– Moody_Mudskipper
Nov 14 '18 at 23:39
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Using grepl with your edited method (r instead of h):
a<-c("hour","four","ruoh", "six", "high", "our")
a[grepl(pattern="o", x=a) & grepl(pattern="u", x=a) & grepl(pattern="r", x=a) & nchar(a)==4]
Returns:
[1] "hour" "four" "ruoh"
answered Nov 14 '18 at 23:42
CrisCris
498311
add a comment |
1
Using grepl with your edited method (r instead of h):
a<-c("hour","four","ruoh", "six", "high", "our")
a[grepl(pattern="o", x=a) & grepl(pattern="u", x=a) & grepl(pattern="r", x=a) & nchar(a)==4]
Returns:
[1] "hour" "four" "ruoh"
answered Nov 14 '18 at 23:42
CrisCris
498311
add a comment |
1
1
1
Using grepl with your edited method (r instead of h):
a<-c("hour","four","ruoh", "six", "high", "our")
a[grepl(pattern="o", x=a) & grepl(pattern="u", x=a) & grepl(pattern="r", x=a) & nchar(a)==4]
Returns:
[1] "hour" "four" "ruoh"
answered Nov 14 '18 at 23:42
CrisCris
498311
Using grepl with your edited method (r instead of h):
a<-c("hour","four","ruoh", "six", "high", "our")
a[grepl(pattern="o", x=a) & grepl(pattern="u", x=a) & grepl(pattern="r", x=a) & nchar(a)==4]
Returns:
[1] "hour" "four" "ruoh"
answered Nov 14 '18 at 23:42
CrisCris
498311
answered Nov 14 '18 at 23:42
CrisCris
498311
answered Nov 14 '18 at 23:42
CrisCris
498311
answered Nov 14 '18 at 23:42
CrisCris
498311
498311
add a comment |
add a comment |
2
To match strings of length 4 containing the characters h, o, and u use:
grepl("(?=^.{4}$)(?=.*h)(?=.*o)(?=.*u)",
c("hour","four","ruoh", "six", "high", "our"),
perl = TRUE)
[1] TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
(?=^.{4}$): string has length 4.
(?=.*x):xoccurs at any position in string.
answered Nov 15 '18 at 0:36
FlorianFlorian
1,092817
add a comment |
2
To match strings of length 4 containing the characters h, o, and u use:
grepl("(?=^.{4}$)(?=.*h)(?=.*o)(?=.*u)",
c("hour","four","ruoh", "six", "high", "our"),
perl = TRUE)
[1] TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
(?=^.{4}$): string has length 4.
(?=.*x):xoccurs at any position in string.
answered Nov 15 '18 at 0:36
FlorianFlorian
1,092817
add a comment |
2
2
2
To match strings of length 4 containing the characters h, o, and u use:
grepl("(?=^.{4}$)(?=.*h)(?=.*o)(?=.*u)",
c("hour","four","ruoh", "six", "high", "our"),
perl = TRUE)
[1] TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
(?=^.{4}$): string has length 4.
(?=.*x):xoccurs at any position in string.
answered Nov 15 '18 at 0:36
FlorianFlorian
1,092817
To match strings of length 4 containing the characters h, o, and u use:
grepl("(?=^.{4}$)(?=.*h)(?=.*o)(?=.*u)",
c("hour","four","ruoh", "six", "high", "our"),
perl = TRUE)
[1] TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
(?=^.{4}$): string has length 4.
(?=.*x):xoccurs at any position in string.
answered Nov 15 '18 at 0:36
FlorianFlorian
1,092817
edited Nov 15 '18 at 1:39
answered Nov 15 '18 at 0:36
FlorianFlorian
1,092817
answered Nov 15 '18 at 0:36
FlorianFlorian
1,092817
answered Nov 15 '18 at 0:36
FlorianFlorian
1,092817
1,092817
add a comment |
add a comment |
1
You could use strsplit and setdiff, I added an additional edge case to your sample data :
a<-c("hour","four","ruoh", "six", "high", "our","oouh")
a[nchar(a) == 4 &
lengths(lapply(strsplit(a,""),function(x) setdiff(x, c("o","u","h")))) == 1]
# [1] "hour" "ruoh"
or grepl :
a[nchar(a) == 4 & !rowSums(sapply(c("o","u","h"), Negate(grepl), a))]
# [1] "hour" "ruoh" "oouh"
sapply(c("o","u","h"), Negate(grepl), a) gives you a matrix of which word doesn't contain each letter, then the rowSums acts like any applied by row, as it will be coerced to logical.
answered Nov 14 '18 at 23:28
Moody_MudskipperMoody_Mudskipper
23.1k33264
This might have to be tweaked depending on how you want to treat some edge cases (multiple "h" for example)
– Moody_Mudskipper
Nov 14 '18 at 23:31
thanks a bunch @Moody_Mudskipper do you have grepl solution?
– bvowe
Nov 14 '18 at 23:35
see edited answer
– Moody_Mudskipper
Nov 14 '18 at 23:39
add a comment |
1
You could use strsplit and setdiff, I added an additional edge case to your sample data :
a<-c("hour","four","ruoh", "six", "high", "our","oouh")
a[nchar(a) == 4 &
lengths(lapply(strsplit(a,""),function(x) setdiff(x, c("o","u","h")))) == 1]
# [1] "hour" "ruoh"
or grepl :
a[nchar(a) == 4 & !rowSums(sapply(c("o","u","h"), Negate(grepl), a))]
# [1] "hour" "ruoh" "oouh"
sapply(c("o","u","h"), Negate(grepl), a) gives you a matrix of which word doesn't contain each letter, then the rowSums acts like any applied by row, as it will be coerced to logical.
answered Nov 14 '18 at 23:28
Moody_MudskipperMoody_Mudskipper
23.1k33264
This might have to be tweaked depending on how you want to treat some edge cases (multiple "h" for example)
– Moody_Mudskipper
Nov 14 '18 at 23:31
thanks a bunch @Moody_Mudskipper do you have grepl solution?
– bvowe
Nov 14 '18 at 23:35
see edited answer
– Moody_Mudskipper
Nov 14 '18 at 23:39
add a comment |
1
1
1
You could use strsplit and setdiff, I added an additional edge case to your sample data :
a<-c("hour","four","ruoh", "six", "high", "our","oouh")
a[nchar(a) == 4 &
lengths(lapply(strsplit(a,""),function(x) setdiff(x, c("o","u","h")))) == 1]
# [1] "hour" "ruoh"
or grepl :
a[nchar(a) == 4 & !rowSums(sapply(c("o","u","h"), Negate(grepl), a))]
# [1] "hour" "ruoh" "oouh"
sapply(c("o","u","h"), Negate(grepl), a) gives you a matrix of which word doesn't contain each letter, then the rowSums acts like any applied by row, as it will be coerced to logical.
answered Nov 14 '18 at 23:28
Moody_MudskipperMoody_Mudskipper
23.1k33264
You could use strsplit and setdiff, I added an additional edge case to your sample data :
a<-c("hour","four","ruoh", "six", "high", "our","oouh")
a[nchar(a) == 4 &
lengths(lapply(strsplit(a,""),function(x) setdiff(x, c("o","u","h")))) == 1]
# [1] "hour" "ruoh"
or grepl :
a[nchar(a) == 4 & !rowSums(sapply(c("o","u","h"), Negate(grepl), a))]
# [1] "hour" "ruoh" "oouh"
sapply(c("o","u","h"), Negate(grepl), a) gives you a matrix of which word doesn't contain each letter, then the rowSums acts like any applied by row, as it will be coerced to logical.
answered Nov 14 '18 at 23:28
Moody_MudskipperMoody_Mudskipper
23.1k33264
edited Nov 14 '18 at 23:39
answered Nov 14 '18 at 23:28
Moody_MudskipperMoody_Mudskipper
23.1k33264
answered Nov 14 '18 at 23:28
Moody_MudskipperMoody_Mudskipper
23.1k33264
answered Nov 14 '18 at 23:28
Moody_MudskipperMoody_Mudskipper
23.1k33264
23.1k33264
This might have to be tweaked depending on how you want to treat some edge cases (multiple "h" for example)
– Moody_Mudskipper
Nov 14 '18 at 23:31
thanks a bunch @Moody_Mudskipper do you have grepl solution?
– bvowe
Nov 14 '18 at 23:35
see edited answer
– Moody_Mudskipper
Nov 14 '18 at 23:39
add a comment |
This might have to be tweaked depending on how you want to treat some edge cases (multiple "h" for example)
– Moody_Mudskipper
Nov 14 '18 at 23:31
thanks a bunch @Moody_Mudskipper do you have grepl solution?
– bvowe
Nov 14 '18 at 23:35
see edited answer
– Moody_Mudskipper
Nov 14 '18 at 23:39
This might have to be tweaked depending on how you want to treat some edge cases (multiple "h" for example)
– Moody_Mudskipper
Nov 14 '18 at 23:31
This might have to be tweaked depending on how you want to treat some edge cases (multiple "h" for example)
– Moody_Mudskipper
Nov 14 '18 at 23:31
thanks a bunch @Moody_Mudskipper do you have grepl solution?
– bvowe
Nov 14 '18 at 23:35
thanks a bunch @Moody_Mudskipper do you have grepl solution?
– bvowe
Nov 14 '18 at 23:35
see edited answer
– Moody_Mudskipper
Nov 14 '18 at 23:39
see edited answer
– Moody_Mudskipper
Nov 14 '18 at 23:39
add a comment |
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What about testing each separately. You first test (with grep) which elements of the vector contains "o", those who pass, you test if they has "u" and those who pass you test for "h".
– Cris
Nov 14 '18 at 23:26
@Cris is this the most simple approach to do so?
– bvowe
Nov 14 '18 at 23:28
5
"four" does not contain o & u & h.
– neilfws
Nov 14 '18 at 23:28
@neilfws I have now done a modification
– bvowe
Nov 14 '18 at 23:34
1
See Regular Expressions: Is there an AND operator?;
grepl("(?=.*h)(?=.*o)(?=.*u)", a, perl = TRUE)– Henrik
Nov 14 '18 at 23:43