Merge multiple data frames with partially matching rows - r

I have data frames with lists of elements such as NAMES. There are different names in dataframes, but most of them match together. I'd like to combine all of them in one list in which I'd see whether some names are missing from any of df.
DATA sample for df1:
X x
1 1 rh_Structure/Focus_S
2 2 rh_Structure/Focus_C
3 3 lh_Structure/Focus_S
4 4 lh_Structure/Focus_C
5 5 RH_Type-Function-S
6 6 RH_REFERENT-S
and for df2
X x
1 1 rh_Structure/Focus_S
2 2 rh_Structure/Focus_C
3 3 lh_Structure/Focus_S
4 4 lh_Structure/Focus_C
5 5 UCZESTNIK
6 6 COACH
and expected result would be:
NAME. df1 df2
1 COACH NA 6
2 lh_Structure/Focus_C 4 4
3 lh_Structure/Focus_S 3 3
4 RH_REFERENT-S 6 NA
5 rh_Structure/Focus_C 2 2
6 rh_Structure/Focus_S 1 1
7 RH_Type-Function-S 5 NA
8 UCZESTNIK NA 5
I can do that with merge.data.frame(df1,df2,by = "x", all=T),
but the I can't do it with more df with similar structure. Any help would be appreciated.

It might be easier to work with this in a long form. Just rbind all the datasets below one another with a flag for which dataset they came from. Then it's relatively straightforward to get a tabulation of all the missing values (and as an added bonus, you can see if you have any duplicates in any of the source datasets):
dfs <- c("df1","df2")
dfall <- do.call(rbind, Map(cbind, mget(dfs), src=dfs))
table(dfall$x, dfall$src)
# df1 df2
# COACH 0 1
# lh_Structure/Focus_C 1 1
# lh_Structure/Focus_S 1 1
# RH_REFERENT-S 1 0
# rh_Structure/Focus_C 1 1
# rh_Structure/Focus_S 1 1
# RH_Type-Function-S 1 0
# UCZESTNIK 0 1

Related

R - How to create multiple datasets based on levels of factor in multiple columns?

I'm kinda new to R and still looking for ways to make my code more elegant. I want to create multiple datasets in a more efficient way, each based on a particular value over different columns.
This is my dataset:
df<-data.frame(A=c(1,2,2,3,4,5,1,1,2,3),
B=c(4,4,2,3,4,2,1,5,2,2),
C=c(3,3,3,3,4,2,5,1,2,3),
D=c(1,2,5,5,5,4,5,5,2,3),
E=c(1,4,2,3,4,2,5,1,2,3),
dummy1=c("yes","yes","no","no","no","no","yes","no","yes","yes"),
dummy2=c("high","low","low","low","high","high","high","low","low","high"))
And I need each column to be a factor:
df[colnames(df)] <- lapply(df[colnames(df)], factor)
Now, what I want to obtain is one dataframe called "Likert_rank_yes" that contains all the observations that in the column "dummy1" have "yes", one dataframe called "Likert_rank_no" that contains all the observations that in the column "dummy1" have "no", one dataframe called "Likert_rank_high" that contains all the observations that in the column "dummy2" have "high" and so on for all my other dummies.
I want to loop or streamline the process in some way, so that there are few commands to run to get all the datasets I need.
The first two dataframes should look something like this:
Dataframe called "Likert_rank_yes" that contains all the observations that in the column "dummy1" have "yes"
Dataframe called "Likert_rank_no" that contains all the observations that in the column "dummy1" have "no"
I have to do this with several dummies with multiple levels and would like to automate/loop the process or make it more efficient, so that I don't have to subset and rename every dataframe for each dummy level. Ideally I would also need to drop the last column in each df created (the one containing the dummy considered).
I tried splitting like below but it seems it is not possible using multiple values, I just get 4 dfs (yes AND high observations, yes AND low obs, no AND high obs etc.) like so:
Splitting with a list of columns doesn't work
list_df <- split(df[c(1:5)], list(df$dummy1,df$dummy2), sep=".")
Can you help? Thanks in advance!
You need two lapplys:
vals <- colnames(df)[1:5]
dummies <- colnames(df)[-(1:5)]
step1 <- lapply(dummies, function(x) df[, c(vals, x)])
step2 <- lapply(step1, function(x) split(x, x[, 6]))
names(step2) <- dummies
step2
# $dummy1
# $dummy1$no
# A B C D E dummy1
# 3 2 2 3 5 2 no
# 4 3 3 3 5 3 no
# 5 4 4 4 5 4 no
# 6 5 2 2 4 2 no
# 8 1 5 1 5 1 no
#
# $dummy1$yes
# A B C D E dummy1
# 1 1 4 3 1 1 yes
# 2 2 4 3 2 4 yes
# 7 1 1 5 5 5 yes
# 9 2 2 2 2 2 yes
# 10 3 2 3 3 3 yes
#
#
# $dummy2
# $dummy2$high
# A B C D E dummy2
# 1 1 4 3 1 1 high
# 5 4 4 4 5 4 high
# 6 5 2 2 4 2 high
# 7 1 1 5 5 5 high
# 10 3 2 3 3 3 high
#
# $dummy2$low
# A B C D E dummy2
# 2 2 4 3 2 4 low
# 3 2 2 3 5 2 low
# 4 3 3 3 5 3 low
# 8 1 5 1 5 1 low
# 9 2 2 2 2 2 low
For the first data set ("dummy1" and "no") use step2$dummy1$no or step2[[1]][[1]] or step2[["dummy1"]][["no"]].
For programming purposes it is usually better to keep the list intact since it makes it simple to write code that processes all of the data frames in the list without having to specify them individually.
You are very close:
tbls <- unlist(step2, recursive=FALSE)
list2env(tbls, envir=.GlobalEnv)
ls()
# [1] "df" "dummies" "dummy1.no" "dummy1.yes" "dummy2.high" "dummy2.low" "step1" "step2" "tbls" "vals"
This will create the same set of tables.

Add rows to dataframe from another dataframe, based on a vector

I'd like to add rows to a dataframe based on a vector within the dataframe. Here are the dataframes (df2 is the one I'd like to add rows to; df1 is the one I'd like to take the rows from):
ID=c(1:5)
x=c(rep("a",3),rep("b",2))
y=c(rep(0,5))
df1=data.frame(ID,x,y)
df2=df1[2:4,1:2]
df2$y=c(5,2,3)
df1
ID x y
1 1 a 0
2 2 a 0
3 3 a 0
4 4 b 0
5 5 b 0
df2
ID x y
2 2 a 5
3 3 a 2
4 4 b 3
I'd like to add to df2 any rows that aren't in df1, based on the ID vector. so my output dataframe would look like this:
ID x y
1 b 0
5 b 0
2 a 5
3 a 2
4 b 3
Can anyone see a way of doing this neatly, please? I need to do it for a lot of dataframes, all with different numbers of rows. I've tried using merge or rbind but I haven't been able to work out how to do it based on the vector.
Thank you!
A solution with dplyr:
bind_rows(df2,anti_join(df1,df2,by="ID"))
# ID x y
#1 2 a 5
#2 3 a 2
#3 4 b 3
#4 1 a 0
#5 5 b 0
You can do the following:
missingIDs <- which(!df1$ID %in% df2$ID) #check which df1 ID's are not in df2, see function is.element()
df.toadd <- df1[missingIDs,] #define the data frame to add to df2
result <- rbind(df.toadd, df2) #use rbind to add it
result
ID x y
1 1 a 0
5 5 b 0
2 2 a 5
3 3 a 2
4 4 b 3
What about this one-liner?
rbind(df2, df1[!df1$ID %in% df2$ID,])
ID x y
2 2 a 5
3 3 a 2
4 4 b 3
1 1 a 0
5 5 b 0

Order/Sort/Rank a table

I have a table like this
table(mtcars$gear, mtcars$cyl)
I want to rank the rows by the ones with more observations in the 4 cylinder. E.g.
4 6 8
4 8 4 0
5 2 1 2
3 1 2 12
I have been playing with order/sort/rank without much success. How could I order tables output?
We can convert table to data.frame and then order by the column.
sort_col <- "4"
tab <- as.data.frame.matrix(table(mtcars$gear, mtcars$cyl))
tab[order(-tab[sort_col]), ]
# OR tab[order(tab[sort_col], decreasing = TRUE), ]
# 4 6 8
#4 8 4 0
#5 2 1 2
#3 1 2 12
If we don't want to convert it into data frame and want to maintain the table structure we can do
tab <- table(mtcars$gear, mtcars$cyl)
tab[order(-tab[,dimnames(tab)[[2]] == sort_col]),]
# 4 6 8
# 4 8 4 0
# 5 2 1 2
# 3 1 2 12
Could try this. Use sort for the relevant column, specifying decreasing=TRUE; take the names of the sorted rows and subset using those.
table(mtcars$gear, mtcars$cyl)[names(sort(table(mtcars$gear, mtcars$cyl)[,1], dec=T)), ]
4 6 8
4 8 4 0
5 2 1 2
3 1 2 12
In the same scope as Milan, but using the order() function, instead of looking for names() in a sort()-ed list.
The [,1] is to look at the first column when ordering.
table(mtcars$gear, mtcars$cyl)[order(table(mtcars$gear, mtcars$cyl)[,1], decreasing=T),]

R two table merge

I have two data.frame df1 and df2 .
df1=data.frame(id=c(1,2,2),var1=c(3,5,5),var3=c(2,3,4))
df2=data.frame(id=c(1,1,2,2),var1=c('NONE','NONE','NONE','NONE'),var3=c(2,4,6,5))
now I want to merge to one data.frame. First, I should change the df2$var1. re encoding the df2$var1 with df1$var1 when df2$id match with df1$id. For example, df1$id=1 df1$var1=3 then df2$id=1 and df2$var1=3, so the result should like this:
df1=data.frame(id=c(1,2,2),var1=c(3,5,5),var3=c(2,3,4)).
df2=data.frame(id=c(1,1,2,2),var1=c(3,3,5,5),var3=c(2,4,6,5))
secondly, I want to merge two data.frame and delete the same one.the result should like this:
df=data.frame(id=c(1,1,2,2,2,2),var1=c(3,3,5,5,5,5),var2=c(2,4,3,4,6,5))
Sorry, it's my first to use stackoverflow. And most importantly,English isn't my native language.
library(dplyr)
union_all(df1, df2) %>%
distinct() %>%
arrange(id, var1)
id var1 var3
1 1 3 2
2 1 3 4
3 2 5 3
4 2 5 4
5 2 5 6
6 2 5 5
First,I use dplyr::union,then I found that the order is disrupted.
So,finally I use union_all, then rank it
I think this is what you want.
library(sqldf)
sqldf("select b.id, a.var1, b.var3 from df1 a left join df2 b on a.id = b.id")
id var1 var3
1 1 3 2
2 1 3 4
3 2 5 5
4 2 5 6
5 2 5 5
6 2 5 6
This is the same as the example you gave of your desired result, except for the 3rd column of the 3rd and 4th row. I believe that is due to a typo in your example, however if I am mistaken about this please let me know (and just explain why those values would be different and I will update my answer accordingly).
By the way, there are multiple ways to do this, but I find this one to be quick and easy.
with merge:
df2$var1 <- df1[df2$id,'var1'];
df2
id var1 var3
1 1 3 2
2 1 3 4
3 2 5 6
4 2 5 5
df <- merge(df1, df2, by='id')[-2:-3]
df
id var1.y var3.y
1 1 3 2
2 1 3 4
3 2 5 6
4 2 5 5
5 2 5 6
6 2 5 5

Select max or equal value from several columns in a data frame

I'm trying to select the column with the highest value for each row in a data.frame. So for instance, the data is set up as such.
> df <- data.frame(one = c(0:6), two = c(6:0))
> df
one two
1 0 6
2 1 5
3 2 4
4 3 3
5 4 2
6 5 1
7 6 0
Then I'd like to set another column based on those rows. The data frame would look like this.
> df
one two rank
1 0 6 2
2 1 5 2
3 2 4 2
4 3 3 3
5 4 2 1
6 5 1 1
7 6 0 1
I imagine there is some sort of way that I can use plyr or sapply here but it's eluding me at the moment.
There might be a more efficient solution, but
ranks <- apply(df, 1, which.max)
ranks[which(df[, 1] == df[, 2])] <- 3
edit: properly spaced!

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