Get Row and Column Indices of Matches Using 'Which()'

Get row and column indices of matches using `which()`

For your first question you need to also pass arr.ind= TRUE to which:

> which(m == 1, arr.ind = TRUE)
     row col
[1,]   3   1
[2,]   2   2
[3,]   1   3
[4,]   4   3

Get indices of matches with a column in a second data.table

Using .EACHI and adding the resulting list column by reference.

dt2[ , res := dt1[ , i := .I][.SD, on = .(firstName), .(.(i)), by = .EACHI]$V1]
#    lid firstName res
# 1:   1     Maria  NA
# 2:   2       Jim 1,4
# 3:   3      Jack  NA
# 4:   4      Anne 3,5

Get column index from data frame that matches numeric vector?

Here's a base R approach, which compares every column in dat with testVec to see if they are identical. Use which to output the column index if they're identical.

which(sapply(1:ncol(dat), function(x) identical(dat[,x], testVec)))
[1] 3

UPDATE
@nicola has provided a better syntax to my original code (you can see it in the comment under this answer):

which(sapply(dat, identical, y = testVec))
z 
3 

get top 3 values per row and keep the column index

You can use order to get the indices, and head(x, 3) to get the top 3 values.

head(apply(m, 1, function(x) order(x, decreasing = TRUE)), 3)
     M001_0.6 M002_0.6 M004_0.6 M012_0.6 M013_0.6
[1,]        1        5        5        2        3
[2,]        5        2        3        3        4
[3,]        2        4        2        1        2

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Then you could get something like:

list(index = t(head(apply(m, 1, function(x) order(x, decreasing = TRUE)), 3)),
     values = t(head(apply(m, 1,function(x) sort(x, decreasing = TRUE)), 3)))

$index
         [,1] [,2] [,3]
M001_0.6    1    5    2
M002_0.6    5    2    4
M004_0.6    5    3    2
M012_0.6    2    3    1
M013_0.6    3    4    2

$values
                 [,1]         [,2]         [,3]
M001_0.6 0.0016572027 0.0015840909 0.0015597202
M002_0.6 0.0005361538 0.0004630419 0.0004630419
M004_0.6 0.0017303146 0.0014134965 0.0012916433
M012_0.6 0.0107961884 0.0107230765 0.0105768528
M013_0.6 0.0018277971 0.0017546853 0.0015840909

Subset of a data frame according to the combination of rows and column indices

You could use diag:

diag(as.matrix(df[r.idx, c.idx]))
#[1] 21 35 21 32 26

Pandas: Comparing each row's value with index and replacing adjacent column's value

You need to use indexing lookup, for this you first need to ensure that the names in A match the column names (0 -> 'ans_0'):

idx, cols = pd.factorize('ans_'+df['A'].astype(str))

import numpy as np
df['B'] = (df.reindex(cols, axis=1).to_numpy()
             [np.arange(len(df)), idx]
           )

output:

                     A   B  ans_0  ans_3  ans_4
timestamp                                      
2022-05-09 09:28:00  0  20     20    200    100
2022-05-09 09:28:01  3  80     10     80     50
2022-05-09 09:28:02  4  10     30     60     10

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