handling matrix in R -

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I have such data 

  SB1 SB2 SB 3 db 1 db2 db 3 dB 4 dB 5 fb fb 2 fb 3 fb 4 sb 1 ... -45 32 -4 10 -13-9 12 -45 -4 -43-2 -2 sb 2 45 ... 30 70 55 60 16 20 22 48 0 8 SB 3 -32 -30 ... -26 - 40 -10 -14 31 -10 -82-97 -4 dB 1 4 -706 ... -5 2 7 32 -4 0 -45 0 DB 2 -10-55 40 5 ... 2 3 26 -43 0 - 43 0 DB3 13 -60 10 -2 -2 ... -3 29 20 -4 -6 0 DB 49-61 14-7 - 3 3 ... 20 19 -30 -8 0 DB 5 -12 -20 -31 -32 -26 -29 -20 ... 0 -20 -23 -20 FB 1 45 -22 10 4 43-20-19 0 ... -20 -20 -20FB 2 4-48 82 0 0 4 30 20 20 ... 0 FB 3 43 0 97 45 43 6 8 23 20 0 ... 0 FB 4 2-84 020 20 0 0 ...

What do I need The negative number is removed from the matrix and can be done with this command: 

  Apply (difference, 2, function (x) {ifelse (x <0,0 , X)})

However, this code removes both negative "N ..." which I want for further calculation. How can I remove negative but still preserve "..."? I have tried this code 

  applied (inter base -2.2, function (x) {ifelse (x> 0% x% in% "...", x, 0 }})

but it does not work and R gave me this warning 

  error in applying (inter 1, 2, function (x Please give a positive length

Please give me some advice and if you can, please tell me which best option to hand over data in the matrix form

Dhan Thanks.

Some things:

  1. Is this Class data. Frame is a purpose? Because then each column will be only one atomic type, which means that each element of your matrix is ​​of type character due to "...".

  2. If you really want to math, why do dots? You can not use them in a mathematical function in a meaningful way. Your function Not only works because the ">" operator compares the value of the alphabet if the letter Remove because their alphabet value is highe r in minus R.

  3. In the end, the argument with the applied () function MARGIN = 2 uses the full column as the input for the applied column, while every one what you want to do There are so many R functions vectors to apply a function for the element, but the work you want should be written with this in mind.

If you really want dots to stay the way, then you have to do a check for the first. One way to do this is: 

  checkform & lt; - Function (x) {if (x == "...") Returns ("...") and {x & lt; - as.integer (x) x & lt; - ifelse (x> 0,0, x) returns (x)} data.frame (lapply (x = inter, fUN = function (x) {sapply (x = x, explanation: checkerfun) First element compare to dots If the element is something then it adds it to the integer and applies your replacement. Checkerphone is a function that expects the same element as input, not a vector such that it is of a type In the form of input in the methodology *. In the previous row, I have a spell in the nest Lapli occurs in: The difference runs the work on each difference, which is a list of atomic vectors. Therefore, the labor symbol of lapply () requires another sapply () function, on each element of each given vector The checker runs.

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