How to change the values of table with column name
1
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
asked Nov 14 '18 at 7:00
Nausif
13210
add a comment |
1
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
asked Nov 14 '18 at 7:00
Nausif
13210
add a comment |
1
1
1
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
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
r data.table
asked Nov 14 '18 at 7:00
Nausif
13210
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
13210
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
2
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 |
1
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])
})
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 |
1
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
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 useas.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
active
oldest
votes
active
oldest
votes
2
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 |
2
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 |
2
2
2
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
28k51640
add a comment |
add a comment |
1
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])
})
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 |
1
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])
})
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 |
1
1
1
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])
})
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])
})
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
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
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
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
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
Edited based on your example
– Andreas
Nov 14 '18 at 7:34
add a comment |
1
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
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 useas.integer? (edited)
– zx8754
Nov 14 '18 at 8:11
add a comment |
1
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
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 useas.integer? (edited)
– zx8754
Nov 14 '18 at 8:11
add a comment |
1
1
1
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
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
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
29.1k76398
I don't have string instead a data table to begin with
– Nausif
Nov 14 '18 at 7:22
@Nausif just useas.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 useas.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
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
@Nausif just use
as.integer ? (edited)– zx8754
Nov 14 '18 at 8:11
add a comment |
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