Convert a DataFrame into Adjacency/Weights Matrix in R
7
I have a DataFrame, df.
n is a column denoting the number of groups in the x column.x is a column containing the comma-separated groups.
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
> df
n x
2 a, b
3 a, c, d
2 c, d
2 d, b
I would like to convert this DataFrame into a weights matrix where the row and column names are the unique values of the groups in df$c, and the elements represent the number of times each of the groups appear together in df$c.
The output should look like this:
m <- matrix(c(0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0), nrow = 4, ncol = 4)
rownames(m) <- letters[1:4]; colnames(m) <- letters[1:4]
> m
a b c d
a 0 1 1 1
b 1 0 0 1
c 1 0 0 2
d 1 1 2 0
r matrix adjacency-matrix
asked 4 hours ago
Rich PaulooRich Pauloo
2,188930
add a comment |
7
I have a DataFrame, df.
n is a column denoting the number of groups in the x column.x is a column containing the comma-separated groups.
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
> df
n x
2 a, b
3 a, c, d
2 c, d
2 d, b
I would like to convert this DataFrame into a weights matrix where the row and column names are the unique values of the groups in df$c, and the elements represent the number of times each of the groups appear together in df$c.
The output should look like this:
m <- matrix(c(0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0), nrow = 4, ncol = 4)
rownames(m) <- letters[1:4]; colnames(m) <- letters[1:4]
> m
a b c d
a 0 1 1 1
b 1 0 0 1
c 1 0 0 2
d 1 1 2 0
r matrix adjacency-matrix
asked 4 hours ago
Rich PaulooRich Pauloo
2,188930
1
your question is unclear. I can't seecindf. it only hasnandx
– YOLO
4 hours ago
c is one of the x values. Its a frequency table of how often different letters appear in the same line in x
– RAB
4 hours ago
Do you mean df$x instead of df$c in the bolded part of the question?
– mikoontz
2 hours ago
add a comment |
7
7
7
I have a DataFrame, df.
n is a column denoting the number of groups in the x column.x is a column containing the comma-separated groups.
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
> df
n x
2 a, b
3 a, c, d
2 c, d
2 d, b
I would like to convert this DataFrame into a weights matrix where the row and column names are the unique values of the groups in df$c, and the elements represent the number of times each of the groups appear together in df$c.
The output should look like this:
m <- matrix(c(0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0), nrow = 4, ncol = 4)
rownames(m) <- letters[1:4]; colnames(m) <- letters[1:4]
> m
a b c d
a 0 1 1 1
b 1 0 0 1
c 1 0 0 2
d 1 1 2 0
r matrix adjacency-matrix
asked 4 hours ago
Rich PaulooRich Pauloo
2,188930
I have a DataFrame, df.
n is a column denoting the number of groups in the x column.x is a column containing the comma-separated groups.
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
> df
n x
2 a, b
3 a, c, d
2 c, d
2 d, b
I would like to convert this DataFrame into a weights matrix where the row and column names are the unique values of the groups in df$c, and the elements represent the number of times each of the groups appear together in df$c.
The output should look like this:
m <- matrix(c(0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0), nrow = 4, ncol = 4)
rownames(m) <- letters[1:4]; colnames(m) <- letters[1:4]
> m
a b c d
a 0 1 1 1
b 1 0 0 1
c 1 0 0 2
d 1 1 2 0
r matrix adjacency-matrix
r matrix adjacency-matrix
asked 4 hours ago
Rich PaulooRich Pauloo
2,188930
asked 4 hours ago
Rich PaulooRich Pauloo
2,188930
asked 4 hours ago
Rich PaulooRich Pauloo
2,188930
asked 4 hours ago
Rich PaulooRich Pauloo
2,188930
asked 4 hours ago
Rich PaulooRich Pauloo
2,188930
2,188930
1
your question is unclear. I can't seecindf. it only hasnandx
– YOLO
4 hours ago
c is one of the x values. Its a frequency table of how often different letters appear in the same line in x
– RAB
4 hours ago
Do you mean df$x instead of df$c in the bolded part of the question?
– mikoontz
2 hours ago
add a comment |
1
your question is unclear. I can't seecindf. it only hasnandx
– YOLO
4 hours ago
c is one of the x values. Its a frequency table of how often different letters appear in the same line in x
– RAB
4 hours ago
Do you mean df$x instead of df$c in the bolded part of the question?
– mikoontz
2 hours ago
1
1
your question is unclear. I can't see
c in df. it only has n and x– YOLO
4 hours ago
your question is unclear. I can't see
c in df. it only has n and x– YOLO
4 hours ago
c is one of the x values. Its a frequency table of how often different letters appear in the same line in x
– RAB
4 hours ago
c is one of the x values. Its a frequency table of how often different letters appear in the same line in x
– RAB
4 hours ago
Do you mean df$x instead of df$c in the bolded part of the question?
– mikoontz
2 hours ago
Do you mean df$x instead of df$c in the bolded part of the question?
– mikoontz
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
4
Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations.
library(tidyverse)
library(combinat)
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
df %>%
## Parse entries in x into distinct elements
mutate(split = map(x, str_split, pattern = ', '),
flat = flatten(split)) %>%
## Construct 2-element subsets of each set of elements
mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>%
unnest(combn) %>%
## Construct permutations of the 2-element subsets
mutate(perm = map(combn, permn)) %>%
unnest(perm) %>%
## Parse the permutations into row and column indices
mutate(row = map_chr(perm, 1),
col = map_chr(perm, 2)) %>%
count(row, col) %>%
## Long to wide representation
spread(key = col, value = nn, fill = 0) %>%
## Coerce to matrix
column_to_rownames(var = 'row') %>%
as.matrix()
answered 3 hours ago
Dan HicksDan Hicks
1876
add a comment |
2
Using Base R, you could do something like below
a = strsplit(as.character(df$x),', ')
b = unique(unlist(a))
d = unlist(sapply(a,combn,2,toString))
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
g = xtabs(V3~V1+V2,f)
g[lower.tri(g)] = t(g)[lower.tri(g)]
g
V2
V1 a b c d
a 0 1 1 1
b 1 0 0 0
c 1 0 0 2
d 1 0 2 0
answered 2 hours ago
OnyambuOnyambu
15.5k1520
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations.
library(tidyverse)
library(combinat)
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
df %>%
## Parse entries in x into distinct elements
mutate(split = map(x, str_split, pattern = ', '),
flat = flatten(split)) %>%
## Construct 2-element subsets of each set of elements
mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>%
unnest(combn) %>%
## Construct permutations of the 2-element subsets
mutate(perm = map(combn, permn)) %>%
unnest(perm) %>%
## Parse the permutations into row and column indices
mutate(row = map_chr(perm, 1),
col = map_chr(perm, 2)) %>%
count(row, col) %>%
## Long to wide representation
spread(key = col, value = nn, fill = 0) %>%
## Coerce to matrix
column_to_rownames(var = 'row') %>%
as.matrix()
answered 3 hours ago
Dan HicksDan Hicks
1876
add a comment |
4
Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations.
library(tidyverse)
library(combinat)
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
df %>%
## Parse entries in x into distinct elements
mutate(split = map(x, str_split, pattern = ', '),
flat = flatten(split)) %>%
## Construct 2-element subsets of each set of elements
mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>%
unnest(combn) %>%
## Construct permutations of the 2-element subsets
mutate(perm = map(combn, permn)) %>%
unnest(perm) %>%
## Parse the permutations into row and column indices
mutate(row = map_chr(perm, 1),
col = map_chr(perm, 2)) %>%
count(row, col) %>%
## Long to wide representation
spread(key = col, value = nn, fill = 0) %>%
## Coerce to matrix
column_to_rownames(var = 'row') %>%
as.matrix()
answered 3 hours ago
Dan HicksDan Hicks
1876
add a comment |
4
4
4
Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations.
library(tidyverse)
library(combinat)
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
df %>%
## Parse entries in x into distinct elements
mutate(split = map(x, str_split, pattern = ', '),
flat = flatten(split)) %>%
## Construct 2-element subsets of each set of elements
mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>%
unnest(combn) %>%
## Construct permutations of the 2-element subsets
mutate(perm = map(combn, permn)) %>%
unnest(perm) %>%
## Parse the permutations into row and column indices
mutate(row = map_chr(perm, 1),
col = map_chr(perm, 2)) %>%
count(row, col) %>%
## Long to wide representation
spread(key = col, value = nn, fill = 0) %>%
## Coerce to matrix
column_to_rownames(var = 'row') %>%
as.matrix()
answered 3 hours ago
Dan HicksDan Hicks
1876
Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations.
library(tidyverse)
library(combinat)
df <- data.frame(n = c(2, 3, 2, 2),
x = c("a, b", "a, c, d", "c, d", "d, b"))
df %>%
## Parse entries in x into distinct elements
mutate(split = map(x, str_split, pattern = ', '),
flat = flatten(split)) %>%
## Construct 2-element subsets of each set of elements
mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>%
unnest(combn) %>%
## Construct permutations of the 2-element subsets
mutate(perm = map(combn, permn)) %>%
unnest(perm) %>%
## Parse the permutations into row and column indices
mutate(row = map_chr(perm, 1),
col = map_chr(perm, 2)) %>%
count(row, col) %>%
## Long to wide representation
spread(key = col, value = nn, fill = 0) %>%
## Coerce to matrix
column_to_rownames(var = 'row') %>%
as.matrix()
answered 3 hours ago
Dan HicksDan Hicks
1876
answered 3 hours ago
Dan HicksDan Hicks
1876
answered 3 hours ago
Dan HicksDan Hicks
1876
answered 3 hours ago
Dan HicksDan Hicks
1876
1876
add a comment |
add a comment |
2
Using Base R, you could do something like below
a = strsplit(as.character(df$x),', ')
b = unique(unlist(a))
d = unlist(sapply(a,combn,2,toString))
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
g = xtabs(V3~V1+V2,f)
g[lower.tri(g)] = t(g)[lower.tri(g)]
g
V2
V1 a b c d
a 0 1 1 1
b 1 0 0 0
c 1 0 0 2
d 1 0 2 0
answered 2 hours ago
OnyambuOnyambu
15.5k1520
add a comment |
2
Using Base R, you could do something like below
a = strsplit(as.character(df$x),', ')
b = unique(unlist(a))
d = unlist(sapply(a,combn,2,toString))
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
g = xtabs(V3~V1+V2,f)
g[lower.tri(g)] = t(g)[lower.tri(g)]
g
V2
V1 a b c d
a 0 1 1 1
b 1 0 0 0
c 1 0 0 2
d 1 0 2 0
answered 2 hours ago
OnyambuOnyambu
15.5k1520
add a comment |
2
2
2
Using Base R, you could do something like below
a = strsplit(as.character(df$x),', ')
b = unique(unlist(a))
d = unlist(sapply(a,combn,2,toString))
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
g = xtabs(V3~V1+V2,f)
g[lower.tri(g)] = t(g)[lower.tri(g)]
g
V2
V1 a b c d
a 0 1 1 1
b 1 0 0 0
c 1 0 0 2
d 1 0 2 0
answered 2 hours ago
OnyambuOnyambu
15.5k1520
Using Base R, you could do something like below
a = strsplit(as.character(df$x),', ')
b = unique(unlist(a))
d = unlist(sapply(a,combn,2,toString))
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
g = xtabs(V3~V1+V2,f)
g[lower.tri(g)] = t(g)[lower.tri(g)]
g
V2
V1 a b c d
a 0 1 1 1
b 1 0 0 0
c 1 0 0 2
d 1 0 2 0
answered 2 hours ago
OnyambuOnyambu
15.5k1520
answered 2 hours ago
OnyambuOnyambu
15.5k1520
answered 2 hours ago
OnyambuOnyambu
15.5k1520
answered 2 hours ago
OnyambuOnyambu
15.5k1520
15.5k1520
add a comment |
add a comment |
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1
your question is unclear. I can't see
cindf. it only hasnandx– YOLO
4 hours ago
c is one of the x values. Its a frequency table of how often different letters appear in the same line in x
– RAB
4 hours ago
Do you mean df$x instead of df$c in the bolded part of the question?
– mikoontz
2 hours ago