user3354212 - 1 year ago 83

R Question

I have a dataframe:

`df = read.table(text="group X1 X2 X3 X4 X5 X6 X7`

P1 H H H H H H H

P1 C D C D B C C

P1 D C B A C D H

P1 D C B A C D D

P1 C D C D B C D

P2 C D B D C D C

P2 H H H H H H H

P2 D C C A B C D

P3 C D C D B C C

P3 H H H H H H H

P3 C D C D B C C

P3 D C B A C D D", header=T, stringsAsFactors=F)

I have another dataframe:

`df2 = read.table(text="Group col R S`

P1 'X2 X4 X7' 'C A D' 'D D C'

P2 'X2 X3 X4 X6' 'C C A C' 'D B D D'

P3 'X3 X5 X6 X7' 'B C D D' 'C B C C'", header=T, stringsAsFactors=F)

I would like to add a column named "assign" to hold the assignment which is based on df2. For example, if df$group=="P1", then only concatenate columns in df shown in df2$col "P1" row, if all columns have the same letter "H", then assign "H" to the "assign" column; if match the string in df2$R column, assign "R"; if match the string in df2$S column, assign "S"; if not match any three cases as mentioned, then assign "U".

I have tested my script in the group "P1", but I don't know how to return the assigned value to the df and go through the loop. Appreciate any helps.

I expect the result as:

`df = read.table(text="group 1 2 3 4 5 6 7 assign`

P1 H H H H H H H H

P1 C D C D B C C S

P1 D C B A C D D U

P1 D C B A C D D R

P1 C D C D B C D U

P2 C D B D C D C S

P2 H H H H H H H H

P2 D C C A B C D R

P3 C D C D B C C S

P3 H H H H H H H H

P3 C D C D B C C S

P3 D C B A C D D R

", header=T, stringsAsFactors=F)

Answer Source

You can use `data.table`

and solve your problem in three steps:

merge the

`data.table`

sthis is the key step, build a pattern to match later, the cool thing is that we can use a flexible number of

`.SDcols`

across the`by`

groups in the`data.table`

build the

`assign`

variable

Here is the code:

```
# data
require(data.table)
dt = data.table(df)
dt2 = data.table(df2)
# add col_int, a list(!) of col indices, to dt2 for each Group
dt3 = dt2[, list(col_name = strsplit(col, ' ')[[1]]), by = Group]
dt3 = dt3[, col_idx := match(col_name, names(dt))]
dt3 = dt3[, list(col_idx = list(col_idx)), by = Group]
dt2 = merge(dt2, dt3, by = 'Group')
# solution
dt = merge(x = dt,
y = dt2,
by = 'Group')
idx_matching_table = names(dt)
# a: using strings
dt[,
j = pattern := {
.SD[, do.call('paste', c(.SD)), .SDcols = strsplit(col, ' ')[[1]]]
},
by = list(Group, col)]
# b: using indices
dt[,
j = pattern_2 := {
# .SD has less cols (compared to dt), therefore find out what the integer index of col_idx in .SD is:
col_idx_sd = match(idx_matching_table[col_idx[[1]]], names(.SD))
.SD[, do.call('paste', c(.SD)), .SDcols = col_idx_sd]
},
by = list(Group, col)]
dt[, identical(pattern, pattern_2)] # TRUE
dt[, assign := 'U']
dt[pattern %like% '[H ]+H', assign := 'H']
dt[pattern == R, assign := 'R']
dt[pattern == S, assign := 'S']
```

**EDIT** I replaced `apply(.SD, 1, paste, collapse = ' ')`

with `do.call('paste', c(.SD))`

to avoid coercion to `matrix`

.