CODEWITHSUNDEEP
rforeachdata.tableapply
MSM
MSM

Reputation: 85

Looping through each row and each column at minimum computational time

I have a dataset with 79003 rows and 97 columns. My dataset looks like as follows:

col1    col2          1  2         3          4         5        6        7        8         
str_11  str_44 0.2064191 0 0.6061358 0.92798677 2.7899374 1.098612 1.395511 0.000000 
str_11  str_22 0.9044563 0 1.7917595 0.00000000 1.1412787 1.504077 1.008228 0.000000 
str_11  str_18 0.8266786 0 0.5389965 0.81676114 0.2787134 0.000000 3.295837 0.000000 
str_11  str_1 0.8176492 0 5.0673306 4.45461768 0.8664189 6.549293 1.686399 2.079442

I am trying to iterate through each row and each column. I want to calculate the minimum and maximum value of column-wise and do the following calculation:

for (i in 1:nrow(log_trans2)){
    for (j in 3:ncol(log_trans2)){
        log_trans2[i, j] = log_trans2[i, ..j] -
          min(log_trans2[i, 3:ncol(log_trans2)]) / 
          (max(log_trans2[i, 3:ncol(log_trans2)]) - min(log_trans2[i, 3:ncol(log_trans2)]))
       }}

I added ..j after getting the error as

"Error in [.data.table(log_trans2, i, j) : j (the 2nd argument inside [...]) is a single symbol but column name 'j' is not found. Perhaps you intended DT[, ..j]. This difference to data.frame is deliberate and explained in FAQ 1.1

.

but it took more execution (like hours) . How do I reduce the timing with foreach or apply function?

The formula:

=(r-min(col))/(max(col)-min(col))

The expected outcome would be

    col1    col2    1   2   3   4   5   6   7   8
   Str_11   Str_44  0.029847796820572   0   0.080259104746805   0.11295123566895    0.405795371744574   0.138441206009843   0.167481921848205   0
   Str_11   Str_22  0.130782597207229   0   0.237248831160936   0   0.165998575836442   0.189535736027761   0.121002270272179   0
   Str_11   Str_18  0.119536094709514   0   0.071369116220582   0.099413248590107   0.040538762756246   0   0.39554907557078    0
   Str_11   Str_1   0.118230460268521   0   0.670970792433184   0.54220015449667    0.126020313332003   0.825306647460321   0.202392768285567   0.251126401405454

Upvotes: 0

Views: 443

Answers (2)

ThomasIsCoding
ThomasIsCoding

Reputation: 102261

Maybe you can use the code below with base R.

The idea is to first get the range of each columns, and then treat the columns as a matrix for manipulation. I believe it would be much faster than using for loops (benefits from matrix treatment),i.e.,

r <- apply(d,2,range)
df[-c(1:2)] <- t((t(df[-c(1:2)]) - r[1,])/as.vector(diff(r)))

such that

> df
     ol1   col2         1   2          3         4         5         6         7 8
1 str_11 str_44 0.0000000 NaN 0.01482649 0.2083202 1.0000000 0.1677451 0.1692960 0
2 str_11 str_22 1.0000000 NaN 0.27664986 0.0000000 0.3434840 0.2296549 0.0000000 0
3 str_11 str_18 0.8885766 NaN 0.00000000 0.1833516 0.0000000 0.0000000 1.0000000 0
4 str_11  str_1 0.8756412 NaN 1.00000000 1.0000000 0.2340315 1.0000000 0.2964541 1

DATA

> dput(df)
structure(list(ol1 = structure(c(1L, 1L, 1L, 1L), .Label = "str_11", class = "factor"), 
    col2 = structure(4:1, .Label = c("str_1", "str_18", "str_22", 
    "str_44"), class = "factor"), `1` = c(0.2064191, 0.9044563, 
    0.8266786, 0.8176492), `2` = c(0L, 0L, 0L, 0L), `3` = c(0.6061358, 
    1.7917595, 0.5389965, 5.0673306), `4` = c(0.92798677, 0, 
    0.81676114, 4.45461768), `5` = c(2.7899374, 1.1412787, 0.2787134, 
    0.8664189), `6` = c(1.098612, 1.504077, 0, 6.549293), `7` = c(1.395511, 
    1.008228, 3.295837, 1.686399), `8` = c(0, 0, 0, 2.079442)), class = "data.frame", row.names = c(NA, 
-4L))

Upvotes: 2

Ronak Shah
Ronak Shah

Reputation: 389135

Here is one way to do this avoiding loops : 

#Exclude columns which are not required for calculation
temp <- as.matrix(df[, -c(1:2)])
#Get column-wise minimum
min_vals <- matrixStats::colMins(temp)
#Get column-wise maximum
max_vals <- matrixStats::colMaxs(temp)
#Subtract minimum value of column from each element
s1 <- sweep(temp, 2, min_vals, `-`)
#Divide it by max - min
sweep(s1, 2, (max_vals - min_vals), `/`)

Upvotes: 2

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