I'm not very experienced R user, so seek advice how to optimize what I've build and in which direction to move on.
I have one reference data frame, it contains four columns with integer values and one ID.
df <- matrix(ncol=5,nrow = 10)
colnames(df) <- c("A","B","C","D","ID")
# df
for (i in 1:10){
df[i,1:4] <- sample(1:5,4, replace = TRUE)
}
df <- data.frame(df)
df$ID <- make.unique(rep(LETTERS,length.out=10),sep='')
df
A B C D ID
1 2 4 3 5 A
2 5 1 3 5 B
3 3 3 5 3 C
4 4 3 1 5 D
5 2 1 2 5 E
6 5 4 4 5 F
7 4 4 3 3 G
8 2 1 5 5 H
9 4 4 1 3 I
10 4 2 2 2 J
Second data frame has manual input, it's user input, I want to turn it into shiny app later on, that's why also I'm asking for optimization, because my code doesn't seem very neat to me.
df.man <- data.frame(matrix(ncol=5,nrow=1))
colnames(df.man) <- c("A","B","C","D","ID")
df.man$ID <- c("man")
df.man$A <- 4
df.man$B <- 4
df.man$C <- 3
df.man$D <- 4
df.man
A B C D ID
4 4 3 4 man
I want to filter rows from reference sequentially, following the rules:
If there is exact match in a whole row between reference table and manual than extract this(those) from reference and show me that row, if not then reduce number of matching columns from right to left until there is a match but not between less then two variables(columns A,B).
So with my limited knowledge I've wrote this:
# subtraction manual from reference
df <- df %>% dplyr::mutate(Adiff=A-df.man$A)%>%
dplyr::mutate(Bdiff=B-df.man$B)%>%
dplyr::mutate(Cdiff=C-df.man$C) %>%
dplyr::mutate(Ddiff=D-df.man$D)
# check manually how much in a row has zero difference and filter those
ifelse(nrow(df%>%filter(Adiff==0 & Bdiff==0 & Cdiff==0 & Ddiff==0)) != 0,
df0<-df%>%filter(Adiff==0 & Bdiff==0 & Cdiff==0 & Ddiff==0),
ifelse(nrow(df%>%filter(Adiff==0 & Bdiff==0 & Cdiff==0)) != 0,
df0<-df%>%filter(Adiff==0 & Bdiff==0 & Cdiff==0),
ifelse(nrow(df%>%filter(Adiff==0 & Bdiff==0)) != 0,
df0<-df%>%filter(Adiff==0 & Bdiff==0),
"less then two exact match")
))
tbl_df(df0[,1:5])
# A tibble: 1 x 5
A B C D ID
<int> <int> <int> <int> <chr>
1 4 4 3 3 G
It works and found ID G but looks ugly to me. So the first question is - What would be recommended way to improve this? Are there any functions, packages or smth I'm missing?
Second question - I want to complicate condition.
Imagine we have reference data set.
A B C D ID
2 4 3 5 A
5 1 3 5 B
3 3 5 3 C
4 3 1 5 D
2 1 2 5 E
5 4 4 5 F
4 4 3 3 G
2 1 5 5 H
4 4 1 3 I
4 2 2 2 J
Manual input is
A B C D ID
4 4 2 2 man
Filtering rules should be following:
If there is exact match in a whole row between reference table and manual than extract this(those) from reference and show me that row, if not then reduce number of matching columns from right to left until there is a match but not between less then two variables(columns A,B).
From those rows where I have only two variable matches filter those which has ± 1 difference in columns to the right. So I should have filtered case G and I from reference table from the example above.
keep going the way I did above, I would do the following:
ifelse(nrow(df0%>%filter(Cdiff %in% (-1:1) & Ddiff %in% (-1:1)))>0,
df01 <- df0%>%filter(Cdiff %in% (-1:1) & Ddiff %in% (-1:1)),
ifelse(nrow(df0%>%filter(Cdiff %in% (-1:1)))>0,
df01<- df0%>%filter(Cdiff %in% (-1:1)),
"NA"))
It will be about 11 columns at the end, but I assume it doesn't matter so much.
Keeping in mind this objective - how would you suggest to proceed?
Thanks!
This is a lot to sort through, but I have some ideas that might be helpful.
First, you could keep your df a matrix, and use row names for your letters. Something like:
set.seed(2)
df
A B C D
A 5 1 5 1
B 4 5 1 2
C 3 1 3 2
D 3 1 1 4
E 3 1 5 3
F 1 5 5 2
G 2 3 4 3
H 1 1 5 1
I 2 4 5 5
J 4 2 5 5
And for demonstration, you could use a vector for manual as this is input:
# Complete match example
vec.man <- c(3, 1, 5, 3)
To check for complete matches between the manual input and reference (all 4 columns), with all numbers, you can do:
df[apply(df, 1, function(x) all(x == vec.man)), ]
A B C D
3 1 5 3
If you don't have a complete match, would calculate differences between df and vec.man:
# Change example vec.man
vec.man <- c(3, 1, 5, 2)
df.diff <- sweep(df, 2, vec.man)
A B C D
A 2 0 0 -1
B 1 4 -4 0
C 0 0 -2 0
D 0 0 -4 2
E 0 0 0 1
F -2 4 0 0
G -1 2 -1 1
H -2 0 0 -1
I -1 3 0 3
J 1 1 0 3
The diffs that start with and continue with 0 will be your best matches (same as looking from right to left iteratively). Then, your best match is the column of the first non-zero element in each row:
df.best <- apply(df.diff, 1, function(x) which(x!=0)[1])
A B C D E F G H I J
1 1 3 3 4 1 1 1 1 1
You can see that the best match is E which was non-zero in the 4th column (last column did not match). You can extract rows that have 4 in df.best as your best matches:
df.match <- df[which(df.best == max(df.best, na.rm = T)), ]
A B C D
3 1 5 3
Finally, if you want all the rows with closest match +/- 1 if only 2 match, you could check for number of best matches (should be 3). Then, compare differences with vector c(0,0,1) which would imply 2 matches then 3rd column off by +/- 1:
# Example vec.man with only 2 matches
vec.man <- c(3, 1, 6, 9)
> df.match
A B C D
C 3 1 3 2
D 3 1 1 4
E 3 1 5 3
if (max(df.best, na.rm = T) == 3) {
vec.alt = c(0, 0, 1)
df[apply(df.diff[,1:3], 1, function(x) all(abs(x) == vec.alt)), ]
}
A B C D
3 1 5 3
This should be scalable for 11 columns and 4 matches.
To generalize for different numbers of columns, #IlyaT suggested:
n.cols <- max(df.best, na.rm=TRUE)
vec.alt <- c(rep(0, each=n.cols-1), 1)
Given the following df:
a=c('a','b','c')
b=c(1,2,5)
c=c(2,3,4)
d=c(2,1,6)
df=data.frame(a,b,c,d)
a b c d
1 a 1 2 2
2 b 2 3 1
3 c 5 4 6
I'd like to apply a function that normally takes a vector (and returns a vector) like cummax row by row to the columns in position b to d.
Then, I'd like to have the output back in the df, either as a vector in a new column of the df, or replacing the original data.
I'd like to avoid writing it as a for loop that would iterate every row, pull out the content of the cells into a vector, do its thing and put it back.
Is there a more efficient way? I've given the apply family functions a go, but I'm struggling to first get a good way to vectorise content of columns by row and get the right output.
the final output could look something like that (imagining I've applied a cummax() function).
a b c d
1 a 1 2 2
2 b 2 3 3
3 c 5 5 6
or
a b c d output
1 a 1 2 2 (1,2,2)
2 b 2 3 1 (2,3,3)
3 c 5 4 6 (5,5,6)
where output is a vector.
Seems this would just be a simple apply problem that you want to cbind to df:
> cbind(df, apply(df[ , 4:2] # work with columns in reverse order
, 1, # do it row-by-row
cummax) )
a b c d 1 2 3
d a 1 2 2 2 1 6
c b 2 3 1 2 3 6
b c 5 4 6 2 3 6
Ouch. Bitten by failing to notice that this would be returned in a column oriented matrix and need to transpose that result; Such a newbie mistake. But it does show the value of having a question with a reproducible dataset I suppose.
> cbind(df, t(apply(df[ , 4:2] , 1, cummax) ) )
a b c d d c b
1 a 1 2 2 2 2 2
2 b 2 3 1 1 3 3
3 c 5 4 6 6 6 6
To destructively assign the result to df you would just use:
df <- # .... that code.
This does the concatenation with commas (and as a result no longer needs to be transposed:
> cbind(df, output=apply(df[ , 4:2] , 1, function(x) paste( cummax(x), collapse=",") ) )
a b c d output
1 a 1 2 2 2,2,2
2 b 2 3 1 1,3,3
3 c 5 4 6 6,6,6
I have a data frame in R which is similar to the follows. Actually my real ’df’ dataframe is much bigger than this one here but I really do not want to confuse anybody so that is why I try to simplify things as much as possible.
So here’s the data frame.
id <-c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3)
a <-c(3,1,3,3,1,3,3,3,3,1,3,2,1,2,1,3,3,2,1,1,1,3,1,3,3,3,2,1,1,3)
b <-c(3,2,1,1,1,1,1,1,1,1,1,2,1,3,2,1,1,1,2,1,3,1,2,2,1,3,3,2,3,2)
c <-c(1,3,2,3,2,1,2,3,3,2,2,3,1,2,3,3,3,1,1,2,3,3,1,2,2,3,2,2,3,2)
d <-c(3,3,3,1,3,2,2,1,2,3,2,2,2,1,3,1,2,2,3,2,3,2,3,2,1,1,1,1,1,2)
e <-c(2,3,1,2,1,2,3,3,1,1,2,1,1,3,3,2,1,1,3,3,2,2,3,3,3,2,3,2,1,3)
df <-data.frame(id,a,b,c,d,e)
df
Basically what I would like to do is to get the occurrences of numbers for each column (a,b,c,d,e) and for each id group (1,2,3) (for this latter grouping see my column ’id’).
So, for column ’a’ and for id number ’1’ (for the latter see column ’id’) the code would be something like this:
as.numeric(table(df[1:10,2]))
##The results are:
[1] 3 7
Just to briefly explain my results: in column ’a’ (and regarding only those records which have number ’1’ in column ’id’) we can say that number '1' occured 3 times and number '3' occured 7 times.
Again, just to show you another example. For column ’a’ and for id number ’2’ (for the latter grouping see again column ’id’):
as.numeric(table(df[11:20,2]))
##After running the codes the results are:
[1] 4 3 3
Let me explain a little again: in column ’a’ and regarding only those observations which have number ’2’ in column ’id’) we can say that number '1' occured 4 times, number '2' occured 3 times and number '3' occured 3 times.
So this is what I would like to do. Calculating the occurrences of numbers for each custom-defined subsets (and then collecting these values into a data frame). I know it is not a difficult task but the PROBLEM is that I’m gonna have to change the input ’df’ dataframe on a regular basis and hence both the overall number of rows and columns might change over time…
What I have done so far is that I have separated the ’df’ dataframe by columns, like this:
for (z in (2:ncol(df))) assign(paste("df",z,sep="."),df[,z])
So df.2 will refer to df$a, df.3 will equal df$b, df.4 will equal df$c etc. But I’m really stuck now and I don’t know how to move forward…
Is there a proper, ”automatic” way to solve this problem?
How about -
> library(reshape)
> dftab <- table(melt(df,'id'))
> dftab
, , value = 1
variable
id a b c d e
1 3 8 2 2 4
2 4 6 3 2 4
3 4 2 1 5 1
, , value = 2
variable
id a b c d e
1 0 1 4 3 3
2 3 3 3 6 2
3 1 4 5 3 4
, , value = 3
variable
id a b c d e
1 7 1 4 5 3
2 3 1 4 2 4
3 5 4 4 2 5
So to get the number of '3's in column 'a' and group '1'
you could just do
> dftab[3,'a',1]
[1] 4
A combination of tapply and apply can create the data you want:
tapply(df$id,df$id,function(x) apply(df[id==x,-1],2,table))
However, when a grouping doesn't have all the elements in it, as in 1a, the result will be a list for that id group rather than a nice table (matrix).
$`1`
$`1`$a
1 3
3 7
$`1`$b
1 2 3
8 1 1
$`1`$c
1 2 3
2 4 4
$`1`$d
1 2 3
2 3 5
$`1`$e
1 2 3
4 3 3
$`2`
a b c d e
1 4 6 3 2 4
2 3 3 3 6 2
3 3 1 4 2 4
$`3`
a b c d e
1 4 2 1 5 1
2 1 4 5 3 4
3 5 4 4 2 5
I'm sure someone will have a more elegant solution than this, but you can cobble it together with a simple function and dlply from the plyr package.
ColTables <- function(df) {
counts <- list()
for(a in names(df)[names(df) != "id"]) {
counts[[a]] <- table(df[a])
}
return(counts)
}
results <- dlply(df, "id", ColTables)
This gets you back a list - the first "layer" of the list will be the id variable; the second the table results for each column for that id variable. For example:
> results[['2']]['a']
$a
1 2 3
4 3 3
For id variable = 2, column = a, per your above example.
A way to do it is using the aggregate function, but you have to add a column to your dataframe
> df$freq <- 0
> aggregate(freq~a+id,df,length)
a id freq
1 1 1 3
2 3 1 7
3 1 2 4
4 2 2 3
5 3 2 3
6 1 3 4
7 2 3 1
8 3 3 5
Of course you can write a function to do it, so it's easier to do it frequently, and you don't have to add a column to your actual data frame
> frequency <- function(df,groups) {
+ relevant <- df[,groups]
+ relevant$freq <- 0
+ aggregate(freq~.,relevant,length)
+ }
> frequency(df,c("b","id"))
b id freq
1 1 1 8
2 2 1 1
3 3 1 1
4 1 2 6
5 2 2 3
6 3 2 1
7 1 3 2
8 2 3 4
9 3 3 4
You didn't say how you'd like the data. The by function might give you the output you like.
by(df, df$id, function(x) lapply(x[,-1], table))