根据值标准对数据框或矩阵进行子集化

Question

Suppose I have a matrix or a data frame and I want only those values that are greater than 15 and no values between 85 and 90 both inclusive

a<-matrix(1:100,nrow = 10,  ncol = 10)
rownames(a) <- LETTERS[1:10]
colnames(a) <- LETTERS[1:10]
   A  B  C  D  E  F  G  H  I   J
A  1 11 21 31 41 51 61 71 81  91
B  2 12 22 32 42 52 62 72 82  92
C  3 13 23 33 43 53 63 73 83  93
D  4 14 24 34 44 54 64 74 84  94
E  5 15 25 35 45 55 65 75 85  95
F  6 16 26 36 46 56 66 76 86  96
G  7 17 27 37 47 57 67 77 87  97
H  8 18 28 38 48 58 68 78 88  98
I  9 19 29 39 49 59 69 79 89  99
J 10 20 30 40 50 60 70 80 90 100

Note: You can convert it into dataframe if you know this kind of operation is possible in dataframe

Now I want My result in such a format that only those values that are greater than 5 and less than 85 retain and all else got deleted and replaced with blank space. My desired out is like below

   A  B  C  D  E  F  G  H  I   J
A    11 21 31 41 51 61 71 81  91
B    12 22 32 42 52 62 72 82  92
C    13 23 33 43 53 63 73 83  93
D    14 24 34 44 54 64 74 84  94
E  5 15 25 35 45 55 65 75 85  95
F  6 16 26 36 46 56 66 76     96
G  7 17 27 37 47 57 67 77     97
H  8 18 28 38 48 58 68 78     98
I  9 19 29 39 49 59 69 79     99
J 10 20 30 40 50 60 70 80    100

Is there any kind of function in R which can take my condition and produce the desired result. I want to change code according to problem . I searched it over stack flow but didn't find something like this. I don't want to format based on rows or column. I tried a[a> 5 & a!=c(85:90)] but this give me values and looses the structure.

Answer 1

Assuming that the 'a' is matrix , we can assign the values of 'a' %in% 86:90 or | less than 5 ( a < 5 ) to NA. Here, I am not assigning it to '' as it will change the class from numeric to character . Also, assigning to NA would be useful for later processing.

a[a %in% 86:90 | a<5] <- NA

However, if we need it to be ''

a[a %in% 86:90 | a<5] <- ""

If we are using a data.frame

a1 <- as.data.frame(a)
a1[] <- lapply(a1, function(x) replace(x, x %in% 86:90| x <5, ""))
a1
#   A  B  C  D  E  F  G  H  I   J
#A    11 21 31 41 51 61 71 81  91
#B    12 22 32 42 52 62 72 82  92
#C    13 23 33 43 53 63 73 83  93
#D    14 24 34 44 54 64 74 84  94
#E  5 15 25 35 45 55 65 75 85  95
#F  6 16 26 36 46 56 66 76     96
#G  7 17 27 37 47 57 67 77     97
#H  8 18 28 38 48 58 68 78     98
#I  9 19 29 39 49 59 69 79     99
#J 10 20 30 40 50 60 70 80    100

NOTE: This changes the class of each column to character

---

In the OP's code, a!=c(85:90) will not work as intended as the 85:90 will recycle to the length of the 'a' and the comparison will be between the corresponding values in the recycled value and 'a'. Instead, we need to use %in% for a vector with length > 1.