I'd like to get a function that sorts data frame in the way that equal numbers or 0 (/optionally NA) are in the same column.
It should look similar to this:
1 0 3 4 5
0 2 0 0 5
1 2 0 0 0
1 0 0 0 0
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Can you rollback to your previous update – akrun Aug 26 '19 at 01:04
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KonradX: thanks for deleting your other question, (I believe) that's the right thing to do. As suggested in its comments, though, it would be very useful to (1) un-accept @akrun's answer and explain why you are deselecting it; then (2) edit your question to provide sample unambiguous data, code attempted, and expected output. Asking questions in a way that SO deals with *well* is not always intuitive, and most definitely takes extra effort on your part. You will be rewarded by developing this skill, I'm confident. (https://stackoverflow.com/questions/5963269) – r2evans Aug 27 '19 at 21:10
2 Answers
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An option is to convert to data.frame and bind with map_df
library(purrr)
library(dplyr)
map_df(z, as.data.frame) %>%
mutate_all(replace_na, 0)
akrun
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Could you help me as well with repositioning this data frame from "triangular matrix" form to the shape I wanted? I've tried to do it by myself, but I failed :/ – Konrad X Aug 26 '19 at 00:31
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@KonradX The above gives a rectangular format as you intendeed. I guess the values you mentioned are the `rnorm` values, right – akrun Aug 26 '19 at 00:34
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@KonradX I am not fully getting the logic for expected output esp when you showed only 'x1', 'x2', 'x3', 'x4' – akrun Aug 26 '19 at 00:35
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Can you please edit your post and update as the format is not cleear in the comments – akrun Aug 26 '19 at 00:39
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I want to have in every column only 0 and the same number from vector "numbers". The attached image shows what I got, not what I wanted – Konrad X Aug 26 '19 at 00:44
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@KonradX I saw your image update. Isnt it the same format from my output – akrun Aug 26 '19 at 00:49
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Using just base R you could do this.
do.call(rbind, Map(function(z) {x <- t(apply(z, 1, `length<-`, l));x[is.na(x)] <- 0;x}, z))
# [,1] [,2] [,3] [,4]
# [1,] 1.3709584 0.0000000 0.0000000 0.0000000
# [2,] -0.5646982 0.0000000 0.0000000 0.0000000
# [3,] 0.3631284 0.0000000 0.0000000 0.0000000
# [4,] 0.6328626 0.0000000 0.0000000 0.0000000
# [5,] 1.3709584 -0.5646982 0.0000000 0.0000000
# [6,] 1.3709584 0.3631284 0.0000000 0.0000000
# [7,] 1.3709584 0.6328626 0.0000000 0.0000000
# [8,] -0.5646982 0.3631284 0.0000000 0.0000000
# [9,] -0.5646982 0.6328626 0.0000000 0.0000000
# [10,] 0.3631284 0.6328626 0.0000000 0.0000000
# [11,] 1.3709584 -0.5646982 0.3631284 0.0000000
# [12,] 1.3709584 -0.5646982 0.6328626 0.0000000
# [13,] 1.3709584 0.3631284 0.6328626 0.0000000
# [14,] -0.5646982 0.3631284 0.6328626 0.0000000
# [15,] 1.3709584 -0.5646982 0.3631284 0.6328626
Explanation: Essentially this is an rbind problem where the matrices in list z have different number of columns. Since the needed number of columns is known by l we can take each single row of a matrix as a vector with apply() and prolong it to l with length<-. Because this yields NAs we need to convert these to the desired zeroes. Map applies that to the whole list. Finally wrapping do.call(rbind..) around it binds the listed matrices into a single one.
Data
z <- list(structure(c(1.37095844714667, -0.564698171396089, 0.363128411337339,
0.63286260496104), .Dim = c(4L, 1L)), structure(c(1.37095844714667,
1.37095844714667, 1.37095844714667, -0.564698171396089, -0.564698171396089,
0.363128411337339, -0.564698171396089, 0.363128411337339, 0.63286260496104,
0.363128411337339, 0.63286260496104, 0.63286260496104), .Dim = c(6L,
2L)), structure(c(1.37095844714667, 1.37095844714667, 1.37095844714667,
-0.564698171396089, -0.564698171396089, -0.564698171396089, 0.363128411337339,
0.363128411337339, 0.363128411337339, 0.63286260496104, 0.63286260496104,
0.63286260496104), .Dim = 4:3), structure(c(1.37095844714667,
-0.564698171396089, 0.363128411337339, 0.63286260496104), .Dim = c(1L,
4L)))
l <- 4L
