Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
menu search
person
Welcome To Ask or Share your Answers For Others

Categories

I already made a similar question but now I want just to restrict the new values of NA.

I have some data like this:

Date 1   Date 2    Date 3    Date 4    Date 5   Date 6
A  NA       0.1       0.2       NA        0.3    0.2
B  0.1      NA        NA        0.3       0.2    0.1
C  NA       NA        NA        NA        0.3    NA
D  0.1      0.2       0.3       NA        0.1    NA
E  NA       NA        0.1       0.2       0.1    0.3

I would like to change the NA values of my data based on the first date a value is registered. So for example for A, the first registration is Date 2. Then I want that before that registration the values of NA in A are 0, and after the first registration the values of NA become the mean of the nearest values (mean of date 3 and 5).

In case the last value is an NA, transform it into the last registered value (as in C and D). In the case of E all NA values will become 0.

Get something like this:

Date 1   Date 2    Date 3    Date 4    Date 5   Date 6 
A  0       0.1       0.2        0.25      0.3    0.2
B  0.1     0.2       0.2        0.3       0.2    0.1
C  0       0         0          0         0.3    0.3
D  0.1     0.2       0.3        0.2       0.1    0.1
E  0       0         0.1        0.2       0.1    0.3

Can you help me? I'm not sure how to do it in R.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
572 views
Welcome To Ask or Share your Answers For Others

1 Answer

Here is a way using na.approx from the zoo package and apply with MARGIN = 1 (so this is probably not very efficient but get's the job done).

library(zoo)
df1 <- as.data.frame(t(apply(dat, 1, na.approx, method = "constant", f = .5, na.rm = FALSE)))

This results in

df1
#   V1  V2  V3   V4  V5
#A  NA 0.1 0.2 0.25 0.3
#B 0.1 0.2 0.2 0.30 0.2
#C  NA  NA  NA   NA 0.3
#E  NA  NA 0.1 0.20 0.1

Replace NAs and rename columns.

df1[is.na(df1)] <- 0
names(df1) <- names(dat)
df1
#  Date_1 Date_2 Date_3 Date_4 Date_5
#A    0.0    0.1    0.2   0.25    0.3
#B    0.1    0.2    0.2   0.30    0.2
#C    0.0    0.0    0.0   0.00    0.3
#E    0.0    0.0    0.1   0.20    0.1

explanation

Given a vector

x <- c(0.1, NA, NA, 0.3, 0.2)
na.approx(x)

returns x with linear interpolated values

#[1] 0.1000000 0.1666667 0.2333333 0.3000000 0.2000000

But OP asked for constant values so we need the argument method = "constant" from the approx function.

na.approx(x, method = "constant") 
# [1] 0.1 0.1 0.1 0.3 0.2

But this is still not what OP asked for because it carries the last observation forward while you want the mean for the closest non-NA values. Therefore we need the argument f (also from approx)

na.approx(x, method = "constant", f = .5)
# [1] 0.1 0.2 0.2 0.3 0.2 # looks good

From ?approx

f : for method = "constant" a number between 0 and 1 inclusive, indicating a compromise between left- and right-continuous step functions. If y0 and y1 are the values to the left and right of the point then the value is y0 if f == 0, y1 if f == 1, and y0*(1-f)+y1*f for intermediate values. In this way the result is right-continuous for f == 0 and left-continuous for f == 1, even for non-finite y values.

Lastly, if we don't want to replace the NAs at the beginning and end of each row we need na.rm = FALSE.

From ?na.approx

na.rm : logical. If the result of the (spline) interpolation still results in NAs, should these be removed?

data

dat <- structure(list(Date_1 = c(NA, 0.1, NA, NA), Date_2 = c(0.1, NA, 
NA, NA), Date_3 = c(0.2, NA, NA, 0.1), Date_4 = c(NA, 0.3, NA, 
0.2), Date_5 = c(0.3, 0.2, 0.3, 0.1)), .Names = c("Date_1", "Date_2", 
"Date_3", "Date_4", "Date_5"), class = "data.frame", row.names = c("A", 
"B", "C", "E"))

EDIT

If there are NAs in the last column we can replace these with the last non-NAs before we apply na.approx as shown above.

dat$Date_6[is.na(dat$Date_6)] <- dat[cbind(1:nrow(dat),
                                           max.col(!is.na(dat), ties.method = "last"))][is.na(dat$Date_6)]

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
...