Process one list each element according to corresponding position of other same length list
1
2
Entering edit mode
2.8 years ago
1106518271 ▴ 60

I have two lists, lista and listb,

lista contains three elements, each of their class() is frame:
frame1, frame2, frame3. Each frame nrows = 50, ncols = 100

listb contains three elements, each of their class() is numeric:
vector1, vector2, vector3. Each vector length is 50

I want to do
vector1 replace the 100th col (last column) of frame1,
vector2 replace the 100th col of frame2,
vector3 replace the 100th col of frame3.

How can I realize is not by for cycle use R? Many thanks! I search this question for several hours，but seems no similar can be reference.

yes,for R, it's

lista[[1]][, 100] <- listb[[1]]
lista[[2]][, 100] <- listb[[2]]
lista[[3]][, 100] <- listb[[3]]


here 1,2,3 is toy example, what I will process more than 200

R • 471 views
1
Entering edit mode

I don't fully understand what you want to do, something like this ?

lista[1][, 100] <- listb[1]
lista[2][, 100] <- listb[2]
lista[3][, 100] <- listb[3]

0
Entering edit mode

Can you explain how this is related to bioinformatics? If not it will be closed and it would be better if you try on StackOverflow

0
Entering edit mode

Thank you for reminding me. Actually, the last col is coverage value of each position on chromosome, but I should mapping this value to same scale so each sample can be compared when they are visualized. Now is let the transfered value to replace raw value.

5
Entering edit mode
2.8 years ago
zx8754 10k

We can use mapply, see this example:

# example data
set.seed(1);lista <- list(frame1 = data.frame(x = letters[1:2], y = runif(2)),
frame2 = data.frame(x = letters[3:4], y = runif(2)),
frame3 = data.frame(x = letters[5:6], y = runif(2)))

listb <- list(c(1, 11), c(2, 22), c(3, 33))

lista
# $frame1 # x y # 1 a 0.2655087 # 2 b 0.3721239 # #$frame2
# x         y
# 1 c 0.5728534
# 2 d 0.9082078
#
# $frame3 # x y # 1 e 0.2016819 # 2 f 0.8983897 listb # [[1]] # [1] 1 11 # # [[2]] # [1] 2 22 # # [[3]] # [1] 3 33 mapply(function(x, y){ x[ ncol(x) ] <- y # Assign y to the last column, in your case ncol(x) will return 100. x}, lista, listb, SIMPLIFY = FALSE) #$frame1
# x  y
# 1 a  1
# 2 b 11
#
# $frame2 # x y # 1 c 2 # 2 d 22 # #$frame3
# x  y
# 1 e  3
# 2 f 33


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