如何使用基于现有数据框的值创建新数据框，并使用R中的数字向量创建范围。

Question

I have a 96 x 48 dataframe df. The first column is an identifying field (char), columns 2 - 48 are numeric values. I also have two numeric vectors with 96 elements each, consisting of upper and lower bounds that correspond to each row.

I would like to create a new dataframe with an identical column 1, but for columns 2-48 I would like to see if the value is between the values in the two vectors for each row. Then I'd like to have 1 in the new data frame if it is, 0 if it is not (a boolean of sorts).

example:

df:

1  2  3  4 .. 48
    a  7  11 15   58
    b  6  9  13   46
    c  8  14 20   73

vectors:

upper: 24, 35, 22, 63
    lower: 10, 11, 12, 11

return:

1  2  3  4 .. 48  
    a  0  1  1    0   (between upper[1] and lower[1])
    b  0  0  1    0   (between upper[2] and lower[2])
    c  0  1  1    0   ...

I'd like to do this without a loop since I'm pretty sure there's a way to do this, but I can't seem to find it.

Answer 1

One method using dplyr:

# Data
df <- data.frame(id=letters[1:3], col2=c(7,6,8), col3=c(11,9,14), col4=c(15,13,20), col48=c(58,46,73))

# chain of operations
library(dplyr)
df %>%
  mutate(upper = c(24, 35, 22), lower = c(10, 11, 12)) %>%
  mutate_at(paste0("col", c(2:4, 48)), funs(.>=lower & .<=upper)) %>%
  mutate_at(paste0("col", c(2:4, 48)), as.integer) %>%
  select(-lower, -upper)

Output:

  col1 col2 col3 col4 col48
1    a    0    1    1     0
2    b    0    0    1     0
3    c    0    1    1     0

Answer 2

since you said that the other variables are numeric, then we can do:

ifelse(t(upper.bounds-t(df[-1])>0&lower.bounds-t(df[-1])<0),1,0)
     c2 c3 c4 c48
[1,]  0  0  1   0
[2,]  0  0  1   0
[3,]  0  1  1   0

There is no need of lapply or forloop where the data:

df=read.table(text=" c1  c2  c3  c4 c48
    a  7  11 15   58
            b  6  9  13   46
            c  8  14 20   73 
            ",h=T)

Answer 3

You can avoid an explicit for loop by using an implicit loop via lappy that loops over all columns. I think that loop is not critical from a performance point-of-view if you loop over the columns but only if you loop over the rows (since R stores the elements of a column as vector in continuous memory locations so that the performance is optimal but the elements of each row are spreaded over the memory locations which causes a performance penalty to loop over rows 1 by 1):

df <- data.frame(c1 = c(7, 6, 8), c2 = c(11, 9, 14), c3 = c(15, 13, 20), c48 = c(58, 46, 73))
df

lower.bounds <- c(10, 11, 12) # , 11)
upper.bounds <- c(24, 35, 22) # , 63)

res <- lapply(df, function(col) {ifelse(col >= lower.bounds & col <= upper.bounds, 1, 0)})
as.data.frame(res)
# c1 c2 c3 c48
# 1  0  1  1   0
# 2  0  0  1   0
# 3  0  1  1   0

Answer 4

Another option is to just use apply over columns. I think it is pretty simple and clean.

df <- data.frame(V2=c(7,6,8), V3=c(11,9,14), V4=c(15,13,20), V48=c(58,46,73))

upper <- c(24, 35, 22)
lower <- c(10, 11, 12)

data.frame(apply(df,2,function(x)((upper>=x)*(x>=lower))))
  V2 V3 V4 V48
  1  0  1  1   0
  2  0  0  1   0
  3  0  1  1   0

---

EDIT: After MKR comment, I became curious and had to test performance. If there is any suggestion on how to measure it in a different way, please comment.

df <- data.frame(V2=c(7,6,8), V3=c(11,9,14), V4=c(15,13,20), V48=c(58,46,73))

upper <- c(24, 35, 22)
lower <- c(10, 11, 12)

 start.time <- Sys.time()
 data.frame(apply(df,2,function(x)((upper>=x)*(x>=lower))))
  #V2 V3 V4 V48
  #1  0  1  1   0
  #2  0  0  1   0
  #3  0  1  1   0
 Sys.time()-start.time
  #Time difference of 0.0146079 secs

 start.time <- Sys.time()
 data.frame(apply(df,2,function(x)(as.numeric((upper>=x)&(x>=lower)))))
  #V2 V3 V4 V48
  #1  0  1  1   0
  #2  0  0  1   0
  #3  0  1  1   0
 Sys.time()-start.time
  #Time difference of 0.0124476 secs

 start.time <- Sys.time()
 data.frame(ifelse(upper > df[] & lower < df[], 1, 0))
  #V2 V3 V4 V48
  #1  0  1  1   0
  #2  0  0  1   0
  #3  0  1  1   0
 Sys.time()-start.time
  #Time difference of 0.008914948 secs

Answer 5

Another possible simpler solution could be:

    df <- data.frame(c1 = c(7, 6, 8), 
                     c2 = c(11, 9, 14), 
                     c3 = c(15, 13, 20), 
                     c48 = c(58, 46, 73))

    lower.bounds <- c(10, 11, 12)
    upper.bounds <- c(24, 35, 22)

    ifelse(upper.bounds > df[] & lower.bounds < df[], 1, 0)
  # Result:
  #       c1 c2 c3 c48
  #  [1,]  0  1  1   0
  #  [2,]  0  0  1   0
  #  [3,]  0  1  1   0

OR

    as.data.frame(ifelse(upper.bounds > df[] & lower.bounds < df[], 1, 0))
  # Result:
  # 
  #    c1 c2 c3 c48
  #  1  0  1  1   0
  #  2  0  0  1   0
  #  3  0  1  1   0