R Equality While Ignoring Nas

R Equality while ignoring NAs

1 == NA returns a logical NA rather than TRUE or FALSE. If you want to call NA FALSE, you could add a second conditional:

set.seed(1)
x <- 1:10
x[4] <- NA
y <- sample(1:10, 10)

x <= y
# [1]  TRUE  TRUE  TRUE    NA FALSE  TRUE  TRUE FALSE  TRUE FALSE

x <= y & !is.na(x)
# [1]  TRUE  TRUE  TRUE FALSE FALSE  TRUE  TRUE FALSE  TRUE FALSE

You could also use a second processing step to convert all the NA values from your equality test to FALSE.

foo <- x <= y
foo[is.na(foo)] <- FALSE
foo
# [1]  TRUE  TRUE  TRUE FALSE FALSE  TRUE  TRUE FALSE  TRUE FALSE

Also, for what its worth, NA == NA returns NA as does NA != NA.

How to treat NAs like values when comparing elementwise in R

Taking advantage of the fact that:

T & NA = NA
but
F & NA = F

and

F | NA = NA
but
T | NA = T

The following solution works, with carefully placed brackets:

(a != b | (is.na(a) & !is.na(b)) | (is.na(b) & !is.na(a))) & !(is.na(a) & is.na(b))

You could define:

`%!=na%` <- function(e1, e2) (e1 != e2 | (is.na(e1) & !is.na(e2)) | (is.na(e2) & !is.na(e1))) & !(is.na(e1) & is.na(e2))

and then use:

a %!=na% b

How to properly ignore NA when comparing two dates?

I think you can remove is.na() in your ifelse() statement, which will keep the missing information

df %>% 
group_by(id) %>% 
mutate(Result = ifelse(Date1 < Date2, "Yes", "No")) 

r, does not equal, nas are not included

We can include another condition in the filter function which will keep the NA values:

df %>%
   filter(a != "B" | is.na(a))

#      a
# 1    A
# 2    C
# 3 <NA>
# 4    C
# 5    A
# 6 <NA>
# 7    A

From ?NA

Logical computations treat NA as a missing `TRUE/FALSE value...

There's more to the explanation, but you can consult the help file.

Comparing Column Values With NA

To simplify things, let's first redefine the data frame with stringsAsFactors=FALSE:

df <- read.table(header = TRUE, text = "A B
                     NA  TEST
                    TEST TEST
                    Abaxasdas Test", stringsAsFactors=FALSE)

You can compare the columns in a NA-safe way using identical:

mapply(identical, df$A, df$B)

To get the output with "YES" and "NO" instead of TRUE and FALSE:

ifelse(mapply(identical, df$A, df$B), "YES", "NO")

Output

> df$Output <- ifelse(mapply(identical, df$A, df$B), "YES", "NO")
> df
          A    B Output
1      <NA> TEST     NO
2      TEST TEST    YES
3 Abaxasdas Test     NO

An alternative

As joran suggested in a comment, replacing NA's with a value would make the comparison easier. If you don't want to change the values in the data frame (but maybe you should!), you could use a helper function like this:

rna <- function(x) replace(x, is.na(x), "")
ifelse(rna(df$A)==rna(df$B), "YES", "NO")

Selecting rows where consecutive values change while ignoring NAs

I would recommend to change NA with other conventional representation of missing values, such as -9999. After this you can use your method which(x[-1] != x[-length(x)]) + 1, or try rle function from base R.

# Sample data
x = c(1, 2, NA, 3, 3, NA, 3, NA, NA, 4, 4)

# Replace missing values with -9999
x[is.na(x)] <- -9999

# Calculate position of non-equal consecutive values
cumsum(rle(x)$length) + 1

# NOTE: you will need to remove last element of the output

R: Using rnorm() ignoring NAs

There are NA elements in the columns. An option is to convert the NA to 0 and then then use that in the calculation for the mean value as any arithmetic operation (+) with NA returns NA. Also, the na.rm = TRUE in OP's code is doing nothing as it is an argument for the function mean and not for the parameter mean in rnorm

tmp <- replace(airquality, is.na(airquality), 0)
rnorm(n = nrow(tmp), mean = (50 + tmp$Ozone*1.2 + 
     tmp$Solar.R*0.5 + tmp$Wind*3 + tmp$Temp*0.2), sd = 5)

-output

[1] 230.54274 175.63684 191.09352 272.75306 108.56648 147.00571 263.74854 176.54437 138.34712 193.08597  91.91993 244.38106 246.75853 254.24543
 [15] 151.91976 278.17564 291.79558 166.25150 291.45414 124.37602  91.79503 285.40843 107.71572 180.18917 133.19906 230.15677  85.62659 135.79484
 [29] 290.33483 328.01940 268.59444 233.47771 240.75695 231.80327 182.09191 204.27378 239.15409 197.93795 222.74681 338.83640 305.89129 226.42353
 [43] 221.24226 190.21195 269.85521 260.70751 223.62116 319.93789 135.87652 171.82266 174.49938 159.01576 102.26742 128.54807 211.20653 147.96291
 [57] 147.09487 118.30181 137.67034 128.21919 155.11839 380.05157 276.10527 247.92156 142.26157 249.46309 308.70804 299.31816 344.92304 341.46871
 [71] 276.41410 162.72329 255.91151 224.55619 253.61126 139.57793 284.49149 277.79043 313.16401 273.82248 284.77515 109.29104 219.85529 249.98464
 [85] 331.40262 334.25997 155.60046 209.86921 308.24376 283.52677 287.17854 289.04658 173.97213 135.25243 148.04647 180.20284 135.46867 159.87413
 [99] 354.59991 317.06474 324.99885 207.30217 176.03234 243.66570 262.85251 251.59681 127.67214 161.26074 174.00305 177.45175 247.33476 246.72993
[113] 262.35549 136.78726 226.31141 259.75610 397.83067 293.34740 156.55135 289.65580 322.11876 301.68719 280.98863 297.88105 276.24682 251.60182
[127] 290.87347 188.69579 200.89883 246.30607 242.00302 232.00063 250.95042 279.32326 268.79634 235.51977 130.42808 170.68919 258.11289 241.38551
[141] 122.56174 239.76333 206.85573 230.98431 120.80214 214.62313 126.46720 135.36000 213.49774 178.97910 216.34848 175.17351 231.16300

---

Another option is to multiply the columns separately with the weights and then get the sum with rowSums where we can use na.rm = TRUE

mn <- rowSums(transform(airquality, Ozone = Ozone * 1.2,
   Solar.R = Solar.R * 0.5, Wind = Wind * 0.3, 
 Temp = Temp * 0.2)[c("Ozone", "Solar.R", "Wind", "Temp")], na.rm = TRUE)
rnorm(n = nrow(airquality), mean = mn, sd = 5)

Combining columns, while ignoring duplicates and NAs

if 'df1' is the output, then we remove the 'NA' that follows a - with sub

df1 %>% 
    mutate(Var3 = sub("-NA", "", Var3))
# A tibble: 8 x 4
#     id  Var1  Var2        Var3
#  <chr> <chr> <chr>       <chr>
#1     A    A1    A1          A1
#2     B    F2    A2       A2-F2
#3     C  <NA>    A3          A3
#4     D A4-E9    A4       A4-E9
#5     E    E5    A5       A5-E5
#6     F  <NA>  <NA>          NA
#7     G B2-R4 A3-B2    A3-B2-R4
#8     H B3-B4 E1-G5 B3-B4-E1-G5

---

We can also do this slightly differently with tidyverse by gather into 'long' format, then split the 'value' column using separate_rows, grouped by 'id', summarise the 'Var3' column by pasteing the sorted unique elements of 'Var3' and left_join with the original dataset 'df'

library(tidyverse)
gather(df, key, value, -id) %>%
       separate_rows(value)  %>%
       group_by(id) %>% 
       summarise(Var3 = paste(sort(unique(value)), collapse='-')) %>% 
       mutate(Var3 = replace(Var3, Var3=='', NA)) %>% 
       left_join(df, .)
#   id  Var1  Var2        Var3
#1  A    A1    A1          A1
#2  B    F2    A2       A2-F2
#3  C  <NA>    A3          A3
#4  D A4-E9    A4       A4-E9
#5  E    E5    A5       A5-E5
#6  F  <NA>  <NA>        <NA>
#7  G B2-R4 A3-B2    A3-B2-R4
#8  H B3-B4 E1-G5 B3-B4-E1-G5

NOTE: The %>% makes even a simple code to appear in multiple lines, but if required, we can put all those statements in a single line and term as one-liner

---

Here is a one-liner

library(data.table)
setDT(df)[, Var3 := paste(sort(unique(unlist(strsplit(unlist(.SD),"-")))), collapse="-"), id]

Logically compare columns in a data frame that contain NA?

This seems to work. Please let me know what you think!

mapply(identical, df_example$metgoal.ref, df_example$metgoal.tri)

Then you can use it to filter your original data. I added ! because you're interested in rows where these fields are NOT identical.

In tidy it might look like this

filter(df_example, !mapply(identical, df_example$metgoal.ref, df_example$metgoal.tri))

or in base

df_example[!mapply(identical, df_example$metgoal.ref, df_example$metgoal.tri),]

---

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