How to change the values of table with column name

Following code gave me data table

library(RMySQL)

library(reshape)

library(philentropy)

library(distances)

mydb = dbConnect(MySQL(), user='root', password='root', dbname='test_db', host='127.0.0.1')

rs = dbSendQuery(mydb,'select   cv.entity_id,cv.attribute_id, cv.value/1000 as value from   test_1 cv limit 100')

data = fetch(rs,n=-1)

pivotedData = cast(data,entity_id ~ attribute_id) 

distCalcNew = distances(pivotedData,id_variable='entity_id') 

nns <- nearest_neighbor_search(distCalcNew,k=3)

nnsdt <- data.table(nns)

I have a data table in R as follows, data points indicates column indices

    8456 8720 5780

1:    1    2    3

2:    3    3    2

3:    2    1    1

Is it possible to get the following?

    8456 8720 5780

1:  8456 8720 5780

2:  5780 5780 8720

3:  8720 8456 8456

Sorry, I'm new to R

edited Nov 14 '18 at 7:20

asked Nov 14 '18 at 7:00

Nausif

13210

add a comment |

Following code gave me data table

library(RMySQL)

library(reshape)

library(philentropy)

library(distances)

mydb = dbConnect(MySQL(), user='root', password='root', dbname='test_db', host='127.0.0.1')

rs = dbSendQuery(mydb,'select   cv.entity_id,cv.attribute_id, cv.value/1000 as value from   test_1 cv limit 100')

data = fetch(rs,n=-1)

pivotedData = cast(data,entity_id ~ attribute_id) 

distCalcNew = distances(pivotedData,id_variable='entity_id') 

nns <- nearest_neighbor_search(distCalcNew,k=3)

nnsdt <- data.table(nns)

I have a data table in R as follows, data points indicates column indices

    8456 8720 5780

1:    1    2    3

2:    3    3    2

3:    2    1    1

Is it possible to get the following?

    8456 8720 5780

1:  8456 8720 5780

2:  5780 5780 8720

3:  8720 8456 8456

Sorry, I'm new to R

edited Nov 14 '18 at 7:20

asked Nov 14 '18 at 7:00

Nausif

13210

add a comment |

Following code gave me data table

library(RMySQL)

library(reshape)

library(philentropy)

library(distances)

mydb = dbConnect(MySQL(), user='root', password='root', dbname='test_db', host='127.0.0.1')

rs = dbSendQuery(mydb,'select   cv.entity_id,cv.attribute_id, cv.value/1000 as value from   test_1 cv limit 100')

data = fetch(rs,n=-1)

pivotedData = cast(data,entity_id ~ attribute_id) 

distCalcNew = distances(pivotedData,id_variable='entity_id') 

nns <- nearest_neighbor_search(distCalcNew,k=3)

nnsdt <- data.table(nns)

I have a data table in R as follows, data points indicates column indices

    8456 8720 5780

1:    1    2    3

2:    3    3    2

3:    2    1    1

Is it possible to get the following?

    8456 8720 5780

1:  8456 8720 5780

2:  5780 5780 8720

3:  8720 8456 8456

Sorry, I'm new to R

edited Nov 14 '18 at 7:20

asked Nov 14 '18 at 7:00

Nausif

13210

Following code gave me data table

library(RMySQL)

library(reshape)

library(philentropy)

library(distances)

mydb = dbConnect(MySQL(), user='root', password='root', dbname='test_db', host='127.0.0.1')

rs = dbSendQuery(mydb,'select   cv.entity_id,cv.attribute_id, cv.value/1000 as value from   test_1 cv limit 100')

data = fetch(rs,n=-1)

pivotedData = cast(data,entity_id ~ attribute_id) 

distCalcNew = distances(pivotedData,id_variable='entity_id') 

nns <- nearest_neighbor_search(distCalcNew,k=3)

nnsdt <- data.table(nns)

I have a data table in R as follows, data points indicates column indices

    8456 8720 5780

1:    1    2    3

2:    3    3    2

3:    2    1    1

Is it possible to get the following?

    8456 8720 5780

1:  8456 8720 5780

2:  5780 5780 8720

3:  8720 8456 8456

Sorry, I'm new to R

r data.table

edited Nov 14 '18 at 7:20

asked Nov 14 '18 at 7:00

Nausif

13210

edited Nov 14 '18 at 7:20

asked Nov 14 '18 at 7:00

Nausif

13210

edited Nov 14 '18 at 7:20

asked Nov 14 '18 at 7:00

Nausif

13210

asked Nov 14 '18 at 7:00

Nausif

13210

asked Nov 14 '18 at 7:00

Nausif

13210

add a comment |

3 Answers
3

active

oldest

votes

Here is another idea,

m1 <- matrix(names(x)[unlist(x)], ncol = ncol(x))



#tidy up

setNames(data.frame(m1), names(x))

#  8456 8720 5780

#1 8456 8720 5780

#2 5780 5780 8720

#3 8720 8456 8456

answered Nov 14 '18 at 7:22

Sotos

28k51640

add a comment |

You can use lapply to the DataFrame df, so that based on the value, it will retrive the column name in that index

df <-  lapply(df, function(x) {

    return(colnames(df)[x])

})

edited Nov 14 '18 at 7:33

answered Nov 14 '18 at 7:06

Andreas

1,7861718

I have updated the example, now how will I do it?
– Nausif
Nov 14 '18 at 7:08

Can you also post the code for data initialization?
– Andreas
Nov 14 '18 at 7:10

I have updated the code initialization
– Nausif
Nov 14 '18 at 7:21

Edited based on your example
– Andreas
Nov 14 '18 at 7:34

add a comment |

Using lapply, but there must be a better "data.table-way":

library(data.table)



x <- fread("8456 8720 5780

1    2    3

3    3    2

2    1    1", header = TRUE)





as.data.table(lapply(x, function(i) as.integer(colnames(x)[ i ])))

#    8456 8720 5780

# 1: 8456 8720 5780

# 2: 5780 5780 8720

# 3: 8720 8456 8456

edited Nov 14 '18 at 8:11

answered Nov 14 '18 at 7:11

zx8754

29.1k76398

I don't have string instead a data table to begin with
– Nausif
Nov 14 '18 at 7:22

@Nausif just use as.integer ? (edited)
– zx8754
Nov 14 '18 at 8:11

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Here is another idea,

m1 <- matrix(names(x)[unlist(x)], ncol = ncol(x))



#tidy up

setNames(data.frame(m1), names(x))

#  8456 8720 5780

#1 8456 8720 5780

#2 5780 5780 8720

#3 8720 8456 8456

answered Nov 14 '18 at 7:22

Sotos

28k51640

add a comment |

Here is another idea,

m1 <- matrix(names(x)[unlist(x)], ncol = ncol(x))



#tidy up

setNames(data.frame(m1), names(x))

#  8456 8720 5780

#1 8456 8720 5780

#2 5780 5780 8720

#3 8720 8456 8456

answered Nov 14 '18 at 7:22

Sotos

28k51640

add a comment |

Here is another idea,

m1 <- matrix(names(x)[unlist(x)], ncol = ncol(x))



#tidy up

setNames(data.frame(m1), names(x))

#  8456 8720 5780

#1 8456 8720 5780

#2 5780 5780 8720

#3 8720 8456 8456

answered Nov 14 '18 at 7:22

Sotos

28k51640

Here is another idea,

m1 <- matrix(names(x)[unlist(x)], ncol = ncol(x))



#tidy up

setNames(data.frame(m1), names(x))

#  8456 8720 5780

#1 8456 8720 5780

#2 5780 5780 8720

#3 8720 8456 8456

answered Nov 14 '18 at 7:22

Sotos

28k51640

answered Nov 14 '18 at 7:22

Sotos

28k51640

answered Nov 14 '18 at 7:22

Sotos

28k51640

answered Nov 14 '18 at 7:22

Sotos

28k51640

add a comment |

You can use lapply to the DataFrame df, so that based on the value, it will retrive the column name in that index

df <-  lapply(df, function(x) {

    return(colnames(df)[x])

})

edited Nov 14 '18 at 7:33

answered Nov 14 '18 at 7:06

Andreas

1,7861718

I have updated the example, now how will I do it?
– Nausif
Nov 14 '18 at 7:08

Can you also post the code for data initialization?
– Andreas
Nov 14 '18 at 7:10

I have updated the code initialization
– Nausif
Nov 14 '18 at 7:21

Edited based on your example
– Andreas
Nov 14 '18 at 7:34

add a comment |

You can use lapply to the DataFrame df, so that based on the value, it will retrive the column name in that index

df <-  lapply(df, function(x) {

    return(colnames(df)[x])

})

edited Nov 14 '18 at 7:33

answered Nov 14 '18 at 7:06

Andreas

1,7861718

I have updated the example, now how will I do it?
– Nausif
Nov 14 '18 at 7:08

Can you also post the code for data initialization?
– Andreas
Nov 14 '18 at 7:10

I have updated the code initialization
– Nausif
Nov 14 '18 at 7:21

Edited based on your example
– Andreas
Nov 14 '18 at 7:34

add a comment |

You can use lapply to the DataFrame df, so that based on the value, it will retrive the column name in that index

df <-  lapply(df, function(x) {

    return(colnames(df)[x])

})

edited Nov 14 '18 at 7:33

answered Nov 14 '18 at 7:06

Andreas

1,7861718

You can use lapply to the DataFrame df, so that based on the value, it will retrive the column name in that index

df <-  lapply(df, function(x) {

    return(colnames(df)[x])

})

edited Nov 14 '18 at 7:33

answered Nov 14 '18 at 7:06

Andreas

1,7861718

edited Nov 14 '18 at 7:33

answered Nov 14 '18 at 7:06

Andreas

1,7861718

answered Nov 14 '18 at 7:06

Andreas

1,7861718

answered Nov 14 '18 at 7:06

Andreas

1,7861718

I have updated the example, now how will I do it?
– Nausif
Nov 14 '18 at 7:08

Can you also post the code for data initialization?
– Andreas
Nov 14 '18 at 7:10

I have updated the code initialization
– Nausif
Nov 14 '18 at 7:21

Edited based on your example
– Andreas
Nov 14 '18 at 7:34

add a comment |

I have updated the example, now how will I do it?
– Nausif
Nov 14 '18 at 7:08

Can you also post the code for data initialization?
– Andreas
Nov 14 '18 at 7:10

I have updated the code initialization
– Nausif
Nov 14 '18 at 7:21

Edited based on your example
– Andreas
Nov 14 '18 at 7:34

I have updated the example, now how will I do it?
– Nausif
Nov 14 '18 at 7:08

Can you also post the code for data initialization?
– Andreas
Nov 14 '18 at 7:10

I have updated the code initialization
– Nausif
Nov 14 '18 at 7:21

Edited based on your example
– Andreas
Nov 14 '18 at 7:34

add a comment |

Using lapply, but there must be a better "data.table-way":

library(data.table)



x <- fread("8456 8720 5780

1    2    3

3    3    2

2    1    1", header = TRUE)





as.data.table(lapply(x, function(i) as.integer(colnames(x)[ i ])))

#    8456 8720 5780

# 1: 8456 8720 5780

# 2: 5780 5780 8720

# 3: 8720 8456 8456

edited Nov 14 '18 at 8:11

answered Nov 14 '18 at 7:11

zx8754

29.1k76398

I don't have string instead a data table to begin with
– Nausif
Nov 14 '18 at 7:22

@Nausif just use as.integer ? (edited)
– zx8754
Nov 14 '18 at 8:11

add a comment |

Using lapply, but there must be a better "data.table-way":

library(data.table)



x <- fread("8456 8720 5780

1    2    3

3    3    2

2    1    1", header = TRUE)





as.data.table(lapply(x, function(i) as.integer(colnames(x)[ i ])))

#    8456 8720 5780

# 1: 8456 8720 5780

# 2: 5780 5780 8720

# 3: 8720 8456 8456

edited Nov 14 '18 at 8:11

answered Nov 14 '18 at 7:11

zx8754

29.1k76398

I don't have string instead a data table to begin with
– Nausif
Nov 14 '18 at 7:22

@Nausif just use as.integer ? (edited)
– zx8754
Nov 14 '18 at 8:11

add a comment |

Using lapply, but there must be a better "data.table-way":

library(data.table)



x <- fread("8456 8720 5780

1    2    3

3    3    2

2    1    1", header = TRUE)





as.data.table(lapply(x, function(i) as.integer(colnames(x)[ i ])))

#    8456 8720 5780

# 1: 8456 8720 5780

# 2: 5780 5780 8720

# 3: 8720 8456 8456

edited Nov 14 '18 at 8:11

answered Nov 14 '18 at 7:11

zx8754

29.1k76398

Using lapply, but there must be a better "data.table-way":

library(data.table)



x <- fread("8456 8720 5780

1    2    3

3    3    2

2    1    1", header = TRUE)





as.data.table(lapply(x, function(i) as.integer(colnames(x)[ i ])))

#    8456 8720 5780

# 1: 8456 8720 5780

# 2: 5780 5780 8720

# 3: 8720 8456 8456

edited Nov 14 '18 at 8:11

answered Nov 14 '18 at 7:11

zx8754

29.1k76398

edited Nov 14 '18 at 8:11

answered Nov 14 '18 at 7:11

zx8754

29.1k76398

answered Nov 14 '18 at 7:11

zx8754

29.1k76398

answered Nov 14 '18 at 7:11

zx8754

29.1k76398

I don't have string instead a data table to begin with
– Nausif
Nov 14 '18 at 7:22

@Nausif just use as.integer ? (edited)
– zx8754
Nov 14 '18 at 8:11

add a comment |

I don't have string instead a data table to begin with
– Nausif
Nov 14 '18 at 7:22

@Nausif just use as.integer ? (edited)
– zx8754
Nov 14 '18 at 8:11

I don't have string instead a data table to begin with
– Nausif
Nov 14 '18 at 7:22

@Nausif just use as.integer ? (edited)
– zx8754
Nov 14 '18 at 8:11

add a comment |

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