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

132110

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

132110

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

132110

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

132110

edited Nov 14 '18 at 7:20

asked Nov 14 '18 at 7:00

Nausif

132110

edited Nov 14 '18 at 7:20

asked Nov 14 '18 at 7:00

Nausif

132110

asked Nov 14 '18 at 7:00

Nausif

132110

asked Nov 14 '18 at 7:00

Nausif

132110

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

30.5k51640

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

2,00331220

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

30.1k763100

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

30.5k51640

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

30.5k51640

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

30.5k51640

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

30.5k51640

answered Nov 14 '18 at 7:22

Sotos

30.5k51640

answered Nov 14 '18 at 7:22

Sotos

30.5k51640

answered Nov 14 '18 at 7:22

Sotos

30.5k51640

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

2,00331220

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

2,00331220

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

2,00331220

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

2,00331220

edited Nov 14 '18 at 7:33

answered Nov 14 '18 at 7:06

Andreas

2,00331220

answered Nov 14 '18 at 7:06

Andreas

2,00331220

answered Nov 14 '18 at 7:06

Andreas

2,00331220

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

30.1k763100

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

30.1k763100

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

30.1k763100

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

30.1k763100

edited Nov 14 '18 at 8:11

answered Nov 14 '18 at 7:11

zx8754

30.1k763100

answered Nov 14 '18 at 7:11

zx8754

30.1k763100

answered Nov 14 '18 at 7:11

zx8754

30.1k763100

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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