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I want to calculate the shortest path between two proteins using two dataframes. For example, I want to calculate the shortest path of first from the first list and the first from the seconds list, the first from the first list and the second from the second list, etc.
structure(list(LAS1L = c("FKBP4", "RBM6", "UPF1", "SLC25A5",
"DHX33", "ELAC2", "CCDC124", "RPS20", "CSDE1", "AKAP8L", "UTP18",
"PTBP1", "DCN", "MATR3", "SAMD4A", "AQR", "STRAP", "SEC63", "BCLAF1",
"TFB1M", "GRN", "ZCCHC8", "NSUN2", "SKIV2L2", "STAU2", "CTNNA1",
"YTHDC2", "POLR2B", "TPR", "MAP4", "NOP16", "FAM120A", "R3HDM1",
"PTCD2", "RRP12", "MRTO4", "THRAP3", "NOP58", "USP36", "MLL3",
"PUM2", "MRPL43", "ZFR", "RC3H2", "ZC3H11A", "PARP12", "ALDH18A1",
"CSDA", "CCAR1")), class = "data.frame", row.names = c(NA, -49L
))
structure(list(GNL3L = c("FMR1", "FRAXA", "UBA1", "CSTF2", "MECP2",
"PHF6", "RBM10", "GSPT2", "SLC25A5", "EIF1AX", "NKRF", "RPS4X",
"RBMX2", "HTATSF1", "LAS1L", "MBNL3", "HUWE1", "RPL10", "RPL15",
"RBMX", "NONO", "RPGR", "UPF3B", "RBM3", "HNRNPH2", "UTP14A",
"DKC1", "MEX3C", "DDX3X", "FLNA", "FAM120C")), class = "data.frame", row.names = c(NA,
-31L))
So far, I just come out with this.
sp<-shortest_path[protein1[,1],protein2[,1]]
dput for shortest_path:
structure(c(0, 4, 6, 4, 4, 4, 4, 3, 3, 3, 5, 3, 5, 3, 3, 3, 4,
3, 3, 3, 4, 0, 5, 4, 4, 4, 4, 3, 3, 3, 5, 3, 5, 3, 3, 3, 4, 3,
3, 3, 6, 5, 0, 6, 4, 6, 5, 5, 5, 5, 7, 5, 6, 5, 5, 3, 6, 5, 5,
5, 4, 4, 6, 0, 3, 3, 3, 3, 3, 3, 4, 3, 5, 3, 3, 3, 4, 3, 3, 3,
4, 4, 4, 3, 0, 4, 3, 3, 3, 3, 4, 3, 5, 3, 3, 3, 4, 3, 3, 3, 4,
4, 6, 3, 4, 0, 3, 3, 3, 3, 5, 3, 5, 3, 3, 3, 4, 3, 3, 3, 4, 4,
5, 3, 3, 3, 0, 3, 3, 2, 3, 3, 5, 3, 3, 3, 4, 3, 3, 3, 3, 3, 5,
3, 3, 3, 3, 0, 2, 2, 4, 2, 4, 2, 2, 2, 3, 2, 2, 2, 3, 3, 5, 3,
3, 3, 3, 2, 0, 2, 4, 2, 4, 2, 2, 2, 3, 2, 2, 2, 3, 3, 5, 3, 3,
3, 2, 2, 2, 0, 2, 2, 4, 2, 2, 2, 3, 2, 2, 2, 5, 5, 7, 4, 4, 5,
3, 4, 4, 2, 0, 4, 6, 4, 4, 4, 5, 4, 4, 4, 3, 3, 5, 3, 3, 3, 3,
2, 2, 2, 4, 0, 4, 2, 2, 2, 3, 2, 2, 2, 5, 5, 6, 5, 5, 5, 5, 4,
4, 4, 6, 4, 0, 4, 4, 4, 5, 4, 4, 4, 3, 3, 5, 3, 3, 3, 3, 2, 2,
2, 4, 2, 4, 0, 2, 2, 3, 2, 2, 2, 3, 3, 5, 3, 3, 3, 3, 2, 2, 2,
4, 2, 4, 2, 0, 2, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 4,
2, 4, 2, 2, 0, 3, 2, 2, 2, 4, 4, 6, 4, 4, 4, 4, 3, 3, 3, 5, 3,
5, 3, 3, 3, 0, 3, 3, 3, 3, 3, 5, 3, 3, 3, 3, 2, 2, 2, 4, 2, 4,
2, 2, 2, 3, 0, 1, 2, 3, 3, 5, 3, 3, 3, 3, 2, 2, 2, 4, 2, 4, 2,
2, 2, 3, 1, 0, 2, 3, 3, 5, 3, 3, 3, 3, 2, 2, 2, 4, 2, 4, 2, 2,
2, 3, 2, 2, 0), .Dim = c(20L, 20L), .Dimnames = list(c("1810055G02Rik",
"2810046L04Rik", "4922501C03Rik", "4930572J05Rik", "9830001H06Rik",
"A1CF", "A2M", "AAGAB", "AATF", "ABCA1", "ABCA13", "ABCA2", "ABCA4",
"ABCB1", "ABCB7", "ABCC2", "ABCC8", "ABCD1", "ABCD3", "ABCD4"
), c("1810055G02Rik", "2810046L04Rik", "4922501C03Rik", "4930572J05Rik",
"9830001H06Rik", "A1CF", "A2M", "AAGAB", "AATF", "ABCA1", "ABCA13",
"ABCA2", "ABCA4", "ABCB1", "ABCB7", "ABCC2", "ABCC8", "ABCD1",
"ABCD3", "ABCD4")))
Thanks in advance!
Maybe you can try the code below
outer(protein1$LAS1L, protein2$GNL3L, FUN = function(x, y) shortest_path[x, y])
I'm currently trying to develop a similar result as this link. I have a significant number of columns and several different labels for the x-axis.
col1 <- c(2, 4, 1, 2, 5, 1, 2, 0, 1, 4, 4, 3, 5, 2, 4, 3, 3, 6, 5, 3, 6, 4, 3, 4, 4, 3, 4,
2, 4, 3, 3, 5, 3, 5, 5, 0, 0, 3, 3, 6, 5, 4, 4, 1, 3, 3, 2, 0, 5, 3, 6, 6, 2, 3,
3, 1, 5, 3, 4, 6)
col2 <- c(2, 4, 4, 0, 4, 4, 4, 4, 1, 4, 4, 3, 5, 0, 4, 5, 3, 6, 5, 3, 6, 4, 4, 2, 4, 4, 4,
1, 1, 2, 2, 3, 3, 5, 0, 3, 4, 2, 4, 5, 5, 4, 4, 2, 3, 5, 2, 6, 5, 2, 4, 6, 3, 3,
3, 1, 4, 3, 5, 4)
col3 <- c(2, 5, 4, 1, 4, 2, 3, 0, 1, 3, 4, 2, 5, 1, 4, 3, 4, 6, 3, 4, 6, 4, 1, 3, 5, 4, 3,
2, 1, 3, 2, 2, 2, 4, 0, 1, 4, 4, 3, 5, 3, 2, 5, 2, 3, 3, 4, 2, 4, 2, 4, 5, 1, 3,
3, 3, 4, 3, 5, 4)
col4 <- c(2, 5, 2, 1, 4, 1, 3, 4, 1, 3, 5, 2, 4, 3, 5, 3, 4, 6, 3, 4, 6, 4, 3, 2, 5, 5, 4,
2, 3, 2, 2, 3, 3, 4, 0, 1, 4, 3, 3, 5, 4, 4, 4, 3, 3, 5, 4, 3, 5, 3, 6, 6, 4, 2,
3, 3, 4, 4, 4, 6)
data2 <- data.frame(col1,col2,col3,col4)
data2[,1:4] <- lapply(data2[,1:4], as.factor)
colnames(data2)<- c("A","B","C", "D")
> x.axis.list
[[1]]
expression(beta[paste(1, ",", 1L)])
[[2]]
expression(beta[paste(1, ",", 2L)])
[[3]]
expression(beta[paste(1, ",", 3L)])
[[4]]
expression(beta[paste(1, ",", 4L)])
myplots <- vector('list', ncol(data2))
for (i in seq_along(data2)) {
message(i)
myplots[[i]] <- local({
i <- i
p1 <- ggplot(data2, aes(x = data2[[i]])) +
geom_histogram(fill = "lightgreen") +
xlab(x.axis.list[[i]])
print(p1)
})
}
In the past, I've been able to do something similar to this where I can just put x.axis.list[[i]] in my loop and change the symbols. However, I continue to get the term expression on the axis. So the symbol for Beta is correct as well as the subscript but the word "expression" remains. I'm not sure exactly what I'm doing wrong, for a moment, I was able to produce a plot without "expression" but it has since stayed in the ggplot.
I want to be able to produce this plot, or one with the title on the y-axis without the word "expression".
My image currently looks . I'm not worried about this example data and the result of the plot, I'm wondering how to get rid of "expression" so only the math symbol shows.
Thanks in advance.
You can do:
for (i in seq_along(data2)) {
df <- data2[i]
names(df)[1] <- "x"
myplots[[i]] <- local({
p1 <- ggplot(df, aes(x = x)) +
geom_bar(fill = "lightgreen", stat = "count") +
xlab(x.axis.list[[i]])
})
}
And we can show all the plots together:
library(patchwork)
(myplots[[1]] + myplots[[2]]) / (myplots[[3]] + myplots[[4]])
Note I created the expression list like this:
x.axis.list <- lapply(1:4, function(i){
parse(text = paste0("beta[paste(1, \",\", ", i, ")]"))
})
I want to produce a 3D scatterplot and add a surface fitted with a linear regression, using plotly. My data:
structure(list(political_trust = c(1, 6, 7, 5, 0, 2, 1, 3, 5,
0, 2, 5, 5, 6, 6, 3, 3, 2, 5, 8, 3, 7, 3, 4, 5, 4, 5, 0, 0, 4,
6, 1, 0, 4, 0, 5, 5, 6, 7, 3, 5, 4, 5, 2, 4, 4, 7, 6, 7, 5, 4,
6, 7, 5, 7, 3, 3, 3, 2, 5, 2, 7, 3, 2, 7, 2, 3, 0, 7, 5, 7, 3,
0, 7, 2, 6, 3, 8, 7, 2, 2, 5, 0, 1, 6, 3, 6, 5, 1, 3, 4, 4, 5,
3, 3, 0, 2, 4, 9, 6, 3, 3, 2, 3, 4, 5, 8, 0, 4, 1, 5, 0, 4, 0,
5, 6, 3, 2, 7, 5, 4, 3, 8, 3, 4, 0, 3, 6, 7, 7, 2, 3, 5, 5, 5,
0, 3, 2, 1, 7, 5, 0, 4, 0, 2, 7, 3, 0, 8, 3, 2, 4, 5, 5, 3, 2,
3, 8, 6, 5, 6, 7, 0, NA, 7, 7, 2, 0, 3, 4, 7, 2, 1, 2, 0, 0,
4, 3, 3, 6, 6, 1, 4, 0, 4, 0, 0, 7, 6, 4, 4, 6, 5, 4, 3, 3, 0,
NA, 2, 5), political_interest = c(2, 0, 3, 3, 2, 1, 2, 2, 2,
2, 2, 2, 3, 3, 3, 3, 2, 2, 3, 2, 1, 2, 2, 2, 2, 0, 2, 1, 3, 1,
1, 1, 1, 1, 2, 3, 2, 2, 2, 1, 3, 3, 2, 3, 2, 1, 3, 2, 0, 3, 1,
1, 2, 1, 2, 2, 1, 3, 3, 2, 3, 2, 3, 2, 2, 1, 2, 0, 3, 1, 2, 2,
1, 3, 2, 2, 1, 2, 2, 0, 3, 2, 2, 1, 2, 1, 1, 3, 1, 1, 3, 2, 0,
2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 0, 1, 1, 2, 2, 2, 2,
2, 0, 0, 2, 3, 2, 2, 2, 3, 3, 0, 3, 3, 1, 2, 1, 1, 1, 2, 3, 2,
2, 2, 0, 2, 2, 2, 1, 2, 3, 3, 1, 2, 0, 1, 1, 0, 2, 2, 1, 2, 2,
2, 2, 3, 2, 1, 2, 2, 0, 0, 3, 2, 2, 2, 1, 2, 3, 0, 1, 2, 3, 2,
2, 2, 1, 3, 1, 1, 2, 2, 3, 3, 1, 2, 2, 2, 2, 2, 1, 0, 1, 1, 0,
3, 3), education_level = c(0, 2, 1, 5, 5, 0, 4, 4, 0, 0, 3, 2,
3, 4, 0, 4, 4, 4, 4, 3, 0, NA, 4, 0, 4, 3, 4, 1, 5, 2, NA, 0,
0, 4, 3, 3, 5, 3, 4, 0, 4, 4, 0, 4, 5, 4, 2, 2, 0, 5, 3, 0, 4,
1, 5, 4, 0, 4, 4, 5, 5, 4, 4, 4, 5, 2, 3, 2, 4, 0, 4, 0, 5, 4,
4, 4, 4, 4, 4, 2, 4, 5, 3, 4, 3, 0, 4, 4, 4, 3, 4, 4, 0, 3, 4,
2, 3, 3, 0, 4, 4, 4, 5, 4, 0, 4, 4, 4, 0, 3, 1, 4, NA, 4, 0,
1, 2, 4, 0, 2, 1, 4, 4, 4, 3, NA, 5, 2, 1, 0, 0, 4, 3, 3, 4,
3, 0, 3, NA, 4, 0, 0, 4, 5, 4, 5, 2, 2, 0, 3, 4, 3, 1, 3, 2,
3, 5, 0, 4, 5, 0, 5, 2, 0, 3, NA, NA, 2, 4, 3, 4, 3, 2, 2, 4,
4, 3, 0, 4, 0, 4, 4, 3, 0, 4, 4, 3, 5, 0, 3, 0, 4, 3, 0, 3, 3,
3, 4, 5, 1)), row.names = c(NA, -200L), class = "data.frame")
I start by defining a list of relevant variables - this is not necessary but basically a consequence of using the code in a Shiny up:
input <- list()
input$x <- "education_level"
input$y <- "political_trust"
input$z <- "political_interest"
Next, creating the surface data:
# Regressing "political_interest" on "education_level" and "political_trust":
lm <- lm(as.formula(paste0(input$z, " ~ ", input$x, " + ", input$y)), data)
# Defining range of values that outcome will be predicted for
axis_x <- seq(min(data[, input$x], na.rm = T),
max(data[, input$x], na.rm = T), by = 0.2)
axis_y <- seq(min(data[, input$y], na.rm = T),
max(data[, input$y], na.rm = T), by = 0.2)
# Predicting outcome, and getting data into surface format
lm_surface <- expand.grid(x = axis_x, y = axis_y, KEEP.OUT.ATTRS = F)
colnames(lm_surface) <- c(input$x, input$y)
lm_surface <- acast(lm_surface, as.formula(paste0(input$x, " ~ ", input$y)),
value.var = input$z)
Last, plotting this with plotly:
data %>%
filter(!is.na(get(input$z))) %>%
filter(!is.na(get(input$x))) %>%
filter(!is.na(get(input$y))) %>%
plot_ly(., x = ~jitter(get(input$x), factor = 2.5),
y = ~jitter(get(input$y), factor = 2.5),
z = ~jitter(get(input$z), factor = 2.5),
type = "scatter3d", mode = "markers",
marker = list(size = 2, color = "#cccccc")) %>%
add_surface(., z = lm_surface,
x = axis_x,
y = axis_y,
type = "surface")
This gives me the following. As you can see, the surface does not cover the full range of the y-dimension. Note also that the surface plotted is "quadratic" - i.e. same length in x and y - although it should have non-quadratic dimensions.
I can bring plotly to draw larger surface area, e.g. by changing the range of values like below, but it always stays quadratic.
axis_x <- seq(0, 10, by = 0.2)
axis_y <- seq(0, 10, by = 0.2)
Ok, question solved.
It's important which dimension of the surface matrix (lm_surface) is which. Swapping x and y when applying acast fixes the issue:
lm_surface <- acast(lm_surface, as.formula(paste0(input$y, " ~ ", input$x)),
value.var = input$z)
I have the following three dimensional array:
dput(a)
structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 2, 1, 1, 1, 2, 2,
2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 6, 2, 7, 6, 2, 7, 6, 2, 7, 4, 2, 4, 4, 2, 6, 4, 2, 4, 6, 2,
7, 4, 2, 6, 4, 2, 6, 4, 2, 6, 4, 2, 4, 4, 2, 6, 4, 2, 4, 4, 2,
6, 4, 2, 6, 4, 2, 6, 6, 2, 7, 4, 2, 6, 4, 2, 6, 4, 2, 4, 2, 3,
1, 2, 3, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 3, 7, 2, 3,
7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3,
7, 2, 3, 7, 1, 2, 5, 2, 3, 7, 1, 2, 4, 2, 3, 7, 2, 3, 7, 2, 3,
7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 3, 7, 2, 6, 3, 2, 6, 3, 2, 6,
3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6,
3, 1, 1, 1, 2, 6, 3, 1, 5, 5, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6,
3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 13,
2, 3, 13, 2, 3, 5, 2, 3, 5, 2, 15, 17, 2, 15, 17, 2, 15, 17,
2, 3, 5, 2, 15, 17, 2, 3, 13, 2, 15, 17, 2, 15, 17, 2, 3, 13,
2, 3, 5, 2, 15, 17, 2, 15, 17, 2, 3, 5, 2), .Dim = c(3L, 20L,
6L), .Dimnames = list(c("cl.tmp", "cl.tmp", "cl.tmp"), NULL,
NULL))
The dimension of this array (a) is 3x20x6 (after edits).
I wanted to count the proportion of times that a[,i,] matches a[,j,] element-by-element in the matrix. Basically, I wanted to get mean(a[,i,] == a[,j,]) for all i, j, and I would like to do this fast but in R.
It occurred to me that the outer function might be a possibility but I am not sure how to specify the function. Any suggestions, or any other alternative ways?
The output would be a 20x20 symmetric matrix of nonnegative elements with 1 on the diagonals.
The solution given below works (thanks!) but I have one further question (sorry).
I would like to display the coordinates above in a heatmap. I try the following:
n<-dim(a)[2]
xx <- matrix(apply(a[,rep(1:n,n),]==a[,rep(1:n,each=n),],2,sum),nrow=n)/prod(dim(a)[-2])
image(1:20, 1:20, xx, xlab = "", ylab = "")
This gives me the following heatmap.
However, I would like to display (reorder the coordinate) such that I get all the coordinates that have high-values amongst each other together. However, I would not like to bias the results by deciding on the number of groups myself. I tried
hc <- hclust(as.dist(1-xx), method = "single")
but I can not decide how to cut the resulting tree to decide on bunching the coordinates together. Any suggestions? Bascically, in the figure, I would like the coordinate pairs in the top left (and bottom right off-diagonal blocks) to be as low-valued (in this case as red) as possible.
Looking around on SO, I found that there exists a function heatmap which might do this,
heatmap(xx,Colv=T,Rowv=T, scale='none',symm = T)
and I get the following:
which is all right, but I can not figure out how to get rid of the dendrograms on the sides or the axes labels. It does work if I extract out and do the following:
yy <- heatmap(xx,Colv=T,Rowv=T, scale='none',symm = T,keep.dendro=F)
image(1:20, 1:20, xx[yy$rowInd,yy$colInd], xlab = "", ylab = "")
so I guess that is what I will stick with. Here is the result:
Try this:
n<-dim(a)[2]
matrix(apply(a[,rep(1:n,n),]==a[,rep(1:n,each=n),],2,sum),nrow=n)/prod(dim(a)[-2])
It has to be stressed that the memory usage of this method goes with n^2 so you might have trouble to use it with larger arrays.
I have a 5 x 5 scatterplot matrix that I created using ggplot. I made histograms for X and Y axis, but I needed an additional histogram for the diagonals of the matrix as well.
Edited for data
data <- structure(c(5, 5, 5, 3, 4, 4, 2, 4, 4, 4, 5, 4, 5, 4, 5, 1, 4,
3, 5, 4, 5, 2, 3, 3, 3, 4, 2, 5, 2, 4, 3, 3, 3, 3, 5, 4, 3, 4,
4, 4, 3, 3, 5, 3, 1, 3, 4, 5, 5, 3, 2, 4, 5, 4, 4, 5, 3, 5, 1,
3, 4, 5, 3, 2, 4, 3, 4, 1, 4, 3, 5, 2, 3, 3, 4, 5, 5, 5, 4, 3,
1, 1, 4, 2, 5, 4, 4, 1, 5, 3, 4, 2, 4, 3, 4, 4, 5, 4, 5, 1, 4,
5, 5, 5, 3, 4, 4, 2, 4, 4, 4, 5, 4, 5, 4, 5, 1, 4, 3, 5, 4, 5,
2, 3, 3, 3, 4, 2, 5, 2, 4, 3, 3, 3, 3, 5, 4, 3, 4, 4, 4, 3, 3,
5, 3, 1, 3, 4, 5, 5, 3, 2, 4, 5, 4, 4, 5, 3, 5, 1, 3, 3, 5, 2,
1, 1, 4, 5, 4, 5, 1, 1, 5, 4, 5, 3, 1, 3, 5, 5, 5, 5, 2, 1, 1,
1, 2, 3, 5, 1, 2, 5, 3, 5, 4, 5, 2, 2, 5, 2, 3, 5), .Dim = c(101L,
2L))
Here is the code
library(ggplot2)
library(gridExtra)
data <- as.data.frame(data)
x <- data$V2
y <- data$V1
xhist <- qplot(x, geom="histogram", binwidth = 0.5)
yhist <- qplot(y, geom="histogram", binwidth = 0.5) + coord_flip()
none <- ggplot()+geom_point(aes(1,1), colour="white") +
theme(axis.ticks=element_blank(), panel.background=element_blank(),
axis.text.x=element_blank(), axis.text.y=element_blank(),
axis.title.x=element_blank(), axis.title.y=element_blank())
g1 <- ggplot(data, aes(x,y)) +
geom_point(size = 1, position = position_jitter(w=0.3, h=0.3))
grid.arrange(yhist, g1, none, xhist, ncol=2, nrow=2, widths=c(1, 4), heights=c(4,1))
Is there a way to directly plot z-axis histogram from this data alone? What I want is to remove the panel of 'none', and instead place a histogram for data points across the diagonal.