I got a df such as
structure(list(id = c(15305, 15305, 15305, 6224, 6224), transfer = c(0,
1, 0, 1, 0), hosp = c(2182, 2452, 2846, 1474, 1476), out = c(2183,
NA, 2857, NA, 1486), Insti = c(NA, NA, NA, NA, NA)), class = "data.frame", row.names = c(NA,
-5L))
And I want to insert NA in the leading "hosp" column if the lagging "out" and lagging "Insti" columns are NA AND the "transfer" column == 1
I want the df to look like this
structure(list(id2 = c(15305, 15305, 15305, 6224, 6224), transfer2 = c(0,
1, 0, 1, 0), hosp2 = c(2182, 2452, NA, 1474, NA), out2 = c(2183,
NA, 2857, NA, 1486), Insti2 = c(NA, NA, NA, NA, NA)), class = "data.frame", row.names = c(NA,
-5L))
You can use the following solution:
library(dplyr)
df %>%
mutate(hosp = case_when(
is.na(lag(out)) & is.na(lag(Insti)) & lag(transfer) == 1 ~ NA_real_,
TRUE ~ hosp
))
id transfer hosp out Insti
1 15305 0 2182 2183 NA
2 15305 1 2452 NA NA
3 15305 0 NA 2857 NA
4 6224 1 1474 NA NA
5 6224 0 NA 1486 NA
To get the "lag" you may remove last value and add NA as first value. Here a base R solution using ifelse.
transform(df,
hosp=ifelse(is.na(c(NA, out[-nrow(df)])) & is.na(c(NA, Insti[-nrow(df)])) &
c(NA, Insti[-nrow(df)]) == 1, NA, hosp))
# id transfer hosp out Insti
# 1 15305 0 NA 2183 NA
# 2 15305 1 2452 NA NA
# 3 15305 0 NA 2857 NA
# 4 6224 1 1474 NA NA
# 5 6224 0 NA 1486 NA
I have two incomplete dataframes (df_a, df_b): Columns are missing or NA values. "by" is the merge index and df_a has "priority" over df_b.
df_a = structure(list(Datum = structure(c(1635163200, 1635166800, 1635170400, 1635174000), class = c("POSIXct", "POSIXt")), Vorhersage = c(10.297922, 10.155121, 10.044135, 9.699513), Export = c("10.912", "10.47", NA, NA), color = c("rgb(0,128,0)", "rgb(0,128,0)", NA, NA), Status = c("ok", "ok", NA, NA), Plausibilität = c("4", "4", NA, NA), min = c(7.93000000000001, 9.4, 8.7, 8.3), max = c(12.31715325, 12.42822725, 12.51326325, 12.28620625)), row.names = c(NA, -4L), class = "data.frame")
df_b = structure(list(Datum = structure(c(1632510000, 1632513600, 1632517200, 1632520800), class = c("POSIXct", "POSIXt")), Vorhersage = c(14.821988, 14.832919, 14.706179, 14.573527), Referenz = c(16.6, 16.2, 15.9, 16), DWD_Name = c("Elpersbüttel", "Elpersbüttel", "Elpersbüttel", "Elpersbüttel"), Export = c(17.198, 16.713, 16.378, 16.358), color = c("rgb(0,128,0)", "rgb(0,128,0)", "rgb(0,128,0)", "rgb(0,128,0)"), Status = c("ok", "ok", "ok", "ok"), Plausibilität = c(4, 4, 4, 4), min = c(13.05, 12.808, 11.631891, 12.312), max = c(17, 17, 16.9, 16.7)), row.names = c(NA, -4L), class = "data.frame")
desired output is:
Datum Vorhersage Export color Status Plausibilität min max Referenz
1 2021-10-25 14:00:00 10.3 10.912 rgb(0,128,0) ok 4 7.9 12 NA
2 2021-10-25 15:00:00 10.2 10.47 rgb(0,128,0) ok 4 9.4 12 NA
3 2021-10-25 16:00:00 10.0 <NA> <NA> <NA> <NA> 8.7 13 NA
4 2021-10-25 17:00:00 9.7 <NA> <NA> <NA> <NA> 8.3 12 NA
5 2021-09-24 21:00:00 14.8 17.198 rgb(0,128,0) ok 4 13.1 17 17
6 2021-09-24 22:00:00 14.8 16.713 rgb(0,128,0) ok 4 12.8 17 16
7 2021-09-24 23:00:00 14.7 16.378 rgb(0,128,0) ok 4 11.6 17 16
8 2021-09-25 00:00:00 14.6 16.358 rgb(0,128,0) ok 4 12.3 17 16
DWD_Name
1 <NA>
2 <NA>
3 <NA>
4 <NA>
5 Elpersbüttel
6 Elpersbüttel
7 Elpersbüttel
8 Elpersbüttel
# for rebuild:
structure(list(Datum = structure(c(1635163200, 1635166800, 1635170400,
1635174000, 1632510000, 1632513600, 1632517200, 1632520800), class = c("POSIXct",
"POSIXt")), Vorhersage = c(10.297922, 10.155121, 10.044135, 9.699513,
14.821988, 14.832919, 14.706179, 14.573527), Export = c("10.912",
"10.47", NA, NA, "17.198", "16.713", "16.378", "16.358"), color = c("rgb(0,128,0)",
"rgb(0,128,0)", NA, NA, "rgb(0,128,0)", "rgb(0,128,0)", "rgb(0,128,0)",
"rgb(0,128,0)"), Status = c("ok", "ok", NA, NA, "ok", "ok", "ok",
"ok"), Plausibilität = c("4", "4", NA, NA, "4", "4", "4", "4"
), min = c(7.93000000000001, 9.4, 8.7, 8.3, 13.05, 12.808, 11.631891,
12.312), max = c(12.31715325, 12.42822725, 12.51326325, 12.28620625,
17, 17, 16.9, 16.7), Referenz = c(NA, NA, NA, NA, 16.6, 16.2,
15.9, 16), DWD_Name = c(NA, NA, NA, NA, "Elpersbüttel", "Elpersbüttel",
"Elpersbüttel", "Elpersbüttel")), row.names = c(NA, -8L), class = "data.frame")
Thanks to the help of #r2evans I tried the following:
by = "Datum"
library(data.table)
colnms <- setdiff(intersect(names(df_a), names(df_b)), by)
setDT(df_a)
setDT(df_b)
merge(df_a, df_b, by = by, all = TRUE
)[, (colnms) := lapply(colnms, function(nm) fcoalesce(.SD[[paste0(nm, ".x")]], .SD[[paste0(nm, ".y")]]))
][, c(outer(colnms, c(".x", ".y"), paste0)) := NULL ][]
but I get the following error:
Error in fcoalesce(.SD[[paste0(nm, ".x")]], .SD[[paste0(nm, ".y")]]) :
Item 2 is type double but the first item is type character. Please coerce
Most of the other answers are good, but many either over-complicate the result (in my opinion) or they perform a left or right join, not the full join as expected in the OP.
Here's a quick solution that uses dynamic column names.
library(data.table)
colnms <- setdiff(intersect(names(df_a), names(df_b)), "by")
colnms
# [1] "a"
setDT(df_a)
setDT(df_b)
merge(df_a, df_b, by = "by", all = TRUE
)[, (colnms) := lapply(colnms, function(nm) fcoalesce(.SD[[paste0(nm, ".x")]], .SD[[paste0(nm, ".y")]]))
][, c(outer(colnms, c(".x", ".y"), paste0)) := NULL ][]
# by b c a
# <num> <num> <num> <num>
# 1: 1 1 NA 1
# 2: 2 NA 2 2
# 3: 3 3 3 3
# 4: 4 NA 4 4
Notes:
the normal data.table::[ merge is a left-join only, so we need to use data.table::merge in order to be able to get a full-join with all=TRUE;
because it's using merge, the repeated columns get the .x and .y suffixes, something we can easily capitalize on;
the canonical and most-performant way when using (colnms) := ... is to also include .SDcols=colnms, but that won't work as well here since we need the suffixed columns, not the colnms columns themselves; this is a slight performance penalty but certainly not an anti-pattern (I believe) given what we need to do; and since we could have more than one duplicate column, we have to be careful to do it with each pair at a time, not all of them at once;
the last [-block (using outer) is for removing the duplicate columns; without it, the output would have column names c("by", "a.x", "b", "a.y", "c", "a"). It uses outer because that's a straight-forward way to get 1-or-more colnms and combine .x and .y to each of them; it then uses data.table's := NULL shortcut for removing one-or-more columns.
This isn't the most elegant, but you can make a function that applies your rule to coalesce the values if they occur in both data frames.
# find the unique column names (not called "by")
cols <- union(names(df_a),names(df_b))
cols <- cols[!(cols == "by")]
# merge the data sets
df_merge <- merge(df_a, df_b, by = "by", all = TRUE)
# function to check for the base column names that now have a '.x' and
# a '.y' version. for the columns, fill in the NAs from '.x' with the
# value from '.y'
col_val <- function(col_base, df) {
x <- names(df)
if (all(paste0(col_base, c(".x", ".y")) %in% x)) {
na.x <- is.na(df[[paste0(col_base, ".x")]])
df[[paste0(col_base, ".x")]][na.x] <- df[[paste0(col_base, ".y")]][na.x]
df[[paste0(col_base, ".x")]]
} else {
df[[col_base]]
}
}
# apply this function to every column
cbind(df_merge["by"], sapply(cols, col_val, df = df_merge))
This will give the following result.
by a b c
1 1 1 1 NA
2 2 2 NA 2
3 3 3 3 3
4 4 4 NA 4
I know you specified base, by the natural_join() function is worth mentioning.
library(rqdatatable)
natural_join(df_a, df_b, by = "by", jointype = "FULL")
This gives exactly what you want.
by a b c
1 1 1 1 NA
2 2 2 NA 2
3 3 3 3 3
4 4 4 NA 4
Not the answer with R base. But one possible solution with the package data.table
library(data.table)
setDT(df_a)
setDT(df_b)
df_a <- rbind(df_a, list(4, NA, NA))
df_b <- rbind(list(1, NA, NA), df_b)
df_a[df_b, `:=` (a = fifelse(is.na(a), i.a, a), c = c), on = .(by)][]
#> by a b c
#> 1: 1 1 1 NA
#> 2: 2 2 NA 2
#> 3: 3 3 3 3
#> 4: 4 4 NA 4
Edit with the help of #r2evans, A much more elegant and efficient solution:
df_a[df_b, `:=` (a = fcoalesce(a, i.a), c = c), on = .(by)][]
#> by a b c
#> 1: 1 1 1 NA
#> 2: 2 2 NA 2
#> 3: 3 3 3 3
#> 4: 4 4 NA 4
Created on 2021-10-19 by the reprex package (v2.0.1)
here a dynamic solution.. not bad, but maybe someone knows how to speed it up.
get_complete_df<-function(df_a,df_b, by = "by"){
df_a = unique(df_a)
df_b = unique(df_b)
nam_a = names(df_a)[!(names(df_a) == by)]
nam_b = names(df_b)[!(names(df_b) == by)]
nums_a = unlist(lapply(df_a, is.numeric))
nums_b = unlist(lapply(df_b, is.numeric))
nums = unique(names(df_a)[nums_a],names(df_b)[nums_b])
## try to supplement NAs
x = df_b[[by]][df_b[[by]] %in% df_a[[by]]]
y = nam_b[nam_b %in% nam_a]
vna = is.na(df_a[df_a[,1] %in% x,y])
df_a[df_a[,1] %in% x ,y][vna] = df_b[df_b[,1] %in% x,y][vna]
## get complete df
all_names = c(nam_a,nam_b )
all_names = c(by, unique(all_names))
all_by = na.omit(unique(c(df_a[[by]],df_b[[by]]) ))
## build
df_o = as.data.frame(matrix(nrow = length(all_by),ncol = length(all_names)))
names(df_o) = all_names
df_o[[by]] = all_by
## fill in content
df_o[df_o[,1] %in% df_b[,1],names(df_b)] = df_b
df_o[df_o[,1] %in% df_a[,1],names(df_a)] = df_a ## df_a has priority!
# fix numeric:
# why did some(!) num fields changed to chr ?
df_o[,nums] = as.data.frame(apply(df_o[,nums], 2, as.numeric))
df_o
}
I have a data frame with up to 5 measurements (x) and their corresponding time:
df = structure(list(x1 = c(92.9595722286402, 54.2085219673818,
46.3227062573019,
NA, 65.1501442134141, 49.736451235317), time1 = c(43.2715277777778,
336.625, 483.975694444444, NA, 988.10625, 510.072916666667),
x2 = c(82.8368681534474, 53.7981639701784, 12.9993531230419,
NA, 64.5678816290574, 55.331442940348), time2 = c(47.8166666666667,
732, 506.747222222222, NA, 1455.25486111111, 958.976388888889
), x3 = c(83.5433119686794, 65.723072881366, 19.0147593408309,
NA, 65.1989838202356, 36.7000828457705), time3 = c(86.5888888888889,
1069.02083333333, 510.275, NA, 1644.21527777778, 1154.95694444444
), x4 = c(NA, 66.008102917677, 40.6243513885846, NA, 62.1694420909955,
29.0078249523063), time4 = c(NA, 1379.22986111111, 520.726388888889,
NA, 2057.20833333333, 1179.86805555556), x5 = c(NA, 61.0047472617535,
45.324715258421, NA, 59.862110645527, 45.883161439362), time5 = c(NA,
1825.33055555556, 523.163888888889, NA, 3352.26944444444,
1364.99513888889)), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -6L))
"NA" means that the person (row) didn't have a measurement.
I would like to calculate the difference between the last existing measurement and the first one.
So for the first one it would be x3 minus x1 (6.4), for the second it would be -6.8 and so on.
I tried something like this, which didnt work:
df$diff = apply(df %>% select(., contains("x")), 1, function(x) head(x,
na.rm = T) - tail(x, na.rm=T))
Any suggestions? Also, is apply/rowwise the most efficient way, or is there a vectorized function to do that?
A vectorized way would be using max.col where we get "first" and "last" non-NA value using ties.method parameter
#Get column number of first and last col
first_col <- max.col(!is.na(df[x_cols]), ties.method = "first")
last_col <- max.col(!is.na(df[x_cols]), ties.method = "last")
#subset the dataframe to include only `"x"` cols
new_df <- as.data.frame(df[grep("^x", names(df))])
#Subtract last non-NA value with the first one
df$new_calc <- new_df[cbind(1:nrow(df), last_col)] -
new_df[cbind(1:nrow(df), first_col)]
Using apply you could do
x_cols <- grep("^x", names(df))
df$new_calc <- apply(df[x_cols], 1, function(x) {
new_x <- x[!is.na(x)]
if (length(new_x) > 0)
new_x[length(new_x)] - new_x[1L]
else NA
})
We can use tidyverse methods on the tbl_df. Create a row names column (rownames_to_column), gather the 'x' columns to 'long' format while removing the NA elements (na.rm = TRUE), grouped by row name, get the difference of first and last 'val'ues and bind the extracted column with the original dataset 'df'
library(tidyverse)
rownames_to_column(df, 'rn') %>%
select(rn, starts_with('x')) %>%
gather(key, val, -rn, na.rm = TRUE) %>%
group_by(rn) %>%
summarise(Diff = diff(c(first(val), last(val)))) %>%
mutate(rn = as.numeric(rn)) %>%
complete(rn = min(rn):max(rn)) %>%
pull(Diff) %>%
bind_cols(df, new_col = .)
# A tibble: 6 x 11
# x1 time1 x2 time2 x3 time3 x4 time4 x5 time5 new_col
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 93.0 43.3 82.8 47.8 83.5 86.6 NA NA NA NA -9.42
#2 54.2 337. 53.8 732 65.7 1069. 66.0 1379. 61.0 1825. 6.80
#3 46.3 484. 13.0 507. 19.0 510. 40.6 521. 45.3 523. -0.998
#4 NA NA NA NA NA NA NA NA NA NA NA
#5 65.2 988. 64.6 1455. 65.2 1644. 62.2 2057. 59.9 3352. -5.29
#6 49.7 510. 55.3 959. 36.7 1155. 29.0 1180. 45.9 1365. -3.85
I am new to R, and could not find specific help for my question on this site.
I have (among others) ten character variables in my dataframe $grant_database, country_1 through country_10. Each contains either a country code, for example E20, F27 or G10, or an NA. Each case is a grant to a project. The ten country variables specify which country/countries a grant is benefitting. In my dataframe, most, but not all cases will have at least one country code, first marked in country_1, many will have one for country_2 as well, and some even for country_3 to _10. All empty fields are marked with an NA.
id country_1 country_2 country_3 country_4 country_5 country_6 ...new_binaryvar
1 F20 NA NA NA NA NA 0
2 E12 E17 E52 NA NA NA 0
3 O62 O33 NA NA NA NA 0
4 E21 E20 NA NA NA NA 1
5 NA NA NA NA NA NA 0
...
I wish to create a new factor flagging grants which benefit a defined subset of countries. This binary "dummy" variable should give the value "1" to each case that in at least one of the ten country variables corresponds with a list of country codes. It should give "0" to each case/grant that does not have a corresponding country code in any of its ten country variables. Let this subset of country codes to be flagged be: E20, F27 and G10 (in reality, there are about 40 to be flagged, from 150+).
Would you help me out by suggesting a way to program this? Thank you very much for your help!
Assuming that you wanted to check whether a subset of "countrycodes" are there in each of the "country" variables with the condition that if atleast one of the "countrycode" is present in a particular row, that row will get "1", or else "0". The idea is to create a vector (v1) of "countrycodes" that needs to be checked. Convert the dataset (df) to matrix after removing the "id" column (as.matrix(df[,-1])) and then create a logical vector by comparing with "v1" (%in%). The vector can be changed back to "matrix" by assigning the dimensions (dim<-) to dimension of df[,-1] ie (c(5,7)). Do the rowSums, double negate (!!), finally add 0 to get the binary dummy variable.
v1 <- c('E20', 'F27', 'G10')
(!!rowSums(`dim<-`(as.matrix(df[,-1]) %in% v1, c(5,7))))+0
#[1] 0 0 0 1 0
newdata
df <- structure(list(id = 1:5, country_1 = c("F20", "E12", "O62", "E21",
NA), country_2 = c(NA, "E17", "O33", "E20", NA), country_3 = c(NA,
"E52", NA, NA, NA), country_4 = c(NA, NA, NA, NA, NA), country_5 = c(NA,
NA, NA, NA, NA), country_6 = c(NA, NA, NA, NA, NA), country_7 = c(NA,
NA, NA, NA, NA)), .Names = c("id", "country_1", "country_2",
"country_3", "country_4", "country_5", "country_6", "country_7"
), class = "data.frame", row.names = c(NA, -5L))