Grep multiple string statements
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In order to remove a specific string from multiple columns we must use this:
df1 <- with(df, df[ grepl( 'word1', df$Col1) | grepl( 'word1', df$Col2) | grepl( 'word1', df$Col3), ])
If we have more than one string like this:
df1 <- with(df, df[ grepl( 'word1', df$Col1) | grepl( 'word1', df$Col2) | grepl( 'word1', df$Col3), ])
df2 <- with(df, df[ grepl( 'word2', df$Col1) | grepl( 'word2', df$Col2) | grepl( 'word2', df$Col3), ])
How is it possible to have one call instead of many for 'word1; and 'word2' be into one ine?
r
add a comment |
In order to remove a specific string from multiple columns we must use this:
df1 <- with(df, df[ grepl( 'word1', df$Col1) | grepl( 'word1', df$Col2) | grepl( 'word1', df$Col3), ])
If we have more than one string like this:
df1 <- with(df, df[ grepl( 'word1', df$Col1) | grepl( 'word1', df$Col2) | grepl( 'word1', df$Col3), ])
df2 <- with(df, df[ grepl( 'word2', df$Col1) | grepl( 'word2', df$Col2) | grepl( 'word2', df$Col3), ])
How is it possible to have one call instead of many for 'word1; and 'word2' be into one ine?
r
3
Usegrepl( 'word1|word2', df$Col1)
– MrFlick
Jan 3 at 21:53
2
If you have more columns to span, you could trydf[rowSums(sapply(df[1:3], grepl, pattern="word1|word2")) > 0,]
where[1:3]
could be any number of columns including[c("Col1","Col2","Col3")]
.
– r2evans
Jan 3 at 22:10
add a comment |
In order to remove a specific string from multiple columns we must use this:
df1 <- with(df, df[ grepl( 'word1', df$Col1) | grepl( 'word1', df$Col2) | grepl( 'word1', df$Col3), ])
If we have more than one string like this:
df1 <- with(df, df[ grepl( 'word1', df$Col1) | grepl( 'word1', df$Col2) | grepl( 'word1', df$Col3), ])
df2 <- with(df, df[ grepl( 'word2', df$Col1) | grepl( 'word2', df$Col2) | grepl( 'word2', df$Col3), ])
How is it possible to have one call instead of many for 'word1; and 'word2' be into one ine?
r
In order to remove a specific string from multiple columns we must use this:
df1 <- with(df, df[ grepl( 'word1', df$Col1) | grepl( 'word1', df$Col2) | grepl( 'word1', df$Col3), ])
If we have more than one string like this:
df1 <- with(df, df[ grepl( 'word1', df$Col1) | grepl( 'word1', df$Col2) | grepl( 'word1', df$Col3), ])
df2 <- with(df, df[ grepl( 'word2', df$Col1) | grepl( 'word2', df$Col2) | grepl( 'word2', df$Col3), ])
How is it possible to have one call instead of many for 'word1; and 'word2' be into one ine?
r
r
asked Jan 3 at 21:49
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KkyrKkyr
337
337
3
Usegrepl( 'word1|word2', df$Col1)
– MrFlick
Jan 3 at 21:53
2
If you have more columns to span, you could trydf[rowSums(sapply(df[1:3], grepl, pattern="word1|word2")) > 0,]
where[1:3]
could be any number of columns including[c("Col1","Col2","Col3")]
.
– r2evans
Jan 3 at 22:10
add a comment |
3
Usegrepl( 'word1|word2', df$Col1)
– MrFlick
Jan 3 at 21:53
2
If you have more columns to span, you could trydf[rowSums(sapply(df[1:3], grepl, pattern="word1|word2")) > 0,]
where[1:3]
could be any number of columns including[c("Col1","Col2","Col3")]
.
– r2evans
Jan 3 at 22:10
3
3
Use
grepl( 'word1|word2', df$Col1)
– MrFlick
Jan 3 at 21:53
Use
grepl( 'word1|word2', df$Col1)
– MrFlick
Jan 3 at 21:53
2
2
If you have more columns to span, you could try
df[rowSums(sapply(df[1:3], grepl, pattern="word1|word2")) > 0,]
where [1:3]
could be any number of columns including [c("Col1","Col2","Col3")]
.– r2evans
Jan 3 at 22:10
If you have more columns to span, you could try
df[rowSums(sapply(df[1:3], grepl, pattern="word1|word2")) > 0,]
where [1:3]
could be any number of columns including [c("Col1","Col2","Col3")]
.– r2evans
Jan 3 at 22:10
add a comment |
1 Answer
1
active
oldest
votes
First you need the combined regex. You can test it at https://regex101.com/ Then you can use apply()
to run it on each column. This will yield a matrix of TRUE
or FALSE
values. 1 row per variable, 1 column per observation. You can apply()
any()
on that matrix to get the selection.
test <- apply(df, 2, grepl, pattern = "word1|word2")
df[apply(test, 2, any), ]
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First you need the combined regex. You can test it at https://regex101.com/ Then you can use apply()
to run it on each column. This will yield a matrix of TRUE
or FALSE
values. 1 row per variable, 1 column per observation. You can apply()
any()
on that matrix to get the selection.
test <- apply(df, 2, grepl, pattern = "word1|word2")
df[apply(test, 2, any), ]
add a comment |
First you need the combined regex. You can test it at https://regex101.com/ Then you can use apply()
to run it on each column. This will yield a matrix of TRUE
or FALSE
values. 1 row per variable, 1 column per observation. You can apply()
any()
on that matrix to get the selection.
test <- apply(df, 2, grepl, pattern = "word1|word2")
df[apply(test, 2, any), ]
add a comment |
First you need the combined regex. You can test it at https://regex101.com/ Then you can use apply()
to run it on each column. This will yield a matrix of TRUE
or FALSE
values. 1 row per variable, 1 column per observation. You can apply()
any()
on that matrix to get the selection.
test <- apply(df, 2, grepl, pattern = "word1|word2")
df[apply(test, 2, any), ]
First you need the combined regex. You can test it at https://regex101.com/ Then you can use apply()
to run it on each column. This will yield a matrix of TRUE
or FALSE
values. 1 row per variable, 1 column per observation. You can apply()
any()
on that matrix to get the selection.
test <- apply(df, 2, grepl, pattern = "word1|word2")
df[apply(test, 2, any), ]
answered Jan 3 at 22:10
ThierryThierry
14.7k43559
14.7k43559
add a comment |
add a comment |
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L,2FFUeOH
3
Use
grepl( 'word1|word2', df$Col1)
– MrFlick
Jan 3 at 21:53
2
If you have more columns to span, you could try
df[rowSums(sapply(df[1:3], grepl, pattern="word1|word2")) > 0,]
where[1:3]
could be any number of columns including[c("Col1","Col2","Col3")]
.– r2evans
Jan 3 at 22:10