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I'm having trouble figuring out how to do the opposite of the answer to this question (and in R not python).
Count the amount of times value A occurs with value B
Basically I have a dataframe with a lot of combinations of pairs of columns like so:
df <- data.frame(id1 = c("1","1","1","1","2","2","2","3","3","4","4"),
id2 = c("2","2","3","4","1","3","4","1","4","2","1"))
I want to count, how often all the values in column A occur in the whole dataframe without the values from column B. So the results for this small example would be the output of:
df_result <- data.frame(id1 = c("1","1","1","2","2","2","3","3","4","4"),
id2 = c("2","3","4","1","3","4","1","4","2","1"),
count = c("4","5","5","3","5","4","2","3","3","3"))
The important criteria for this, is that the final results dataframe is collapsed by the pairs (so in my example rows 1 and 2 are duplicates, and they are collapsed and summed by the total frequency 1 is observed without 2). For tallying the count of occurances, it's important that both columns are examined. I.e. order of columns doesn't matter for calculating the frequency - if column A has 1 and B has 2, this counts the same as if column A has 2 and B has 1.
I can do this very slowly by filtering for each pair, but it's not really feasible for my real data where I have many many different pairs.
Any guidance is greatly appreciated.
First paste the two id columns together to id12 for later matching. Then use sapply to go through all rows to see the records where id1 appears in id12 but id2 doesn't. sum that value and only output the distinct records. Finally, remove the id12 column.
library(dplyr)
df %>% mutate(id12 = paste0(id1, id2),
count = sapply(1:nrow(.),
function(x)
sum(grepl(id1[x], id12) & !grepl(id2[x], id12)))) %>%
distinct() %>%
select(-id12)
Or in base R completely:
id12 <- paste0(df$id1, df$id2)
df$count <- sapply(1:nrow(df), function(x) sum(grepl(df$id1[x], id12) & !grepl(df$id2[x], id12)))
df <- df[!duplicated(df),]
Output
id1 id2 count
1 1 2 4
2 1 3 5
3 1 4 5
4 2 1 3
5 2 3 5
6 2 4 4
7 3 1 2
8 3 4 3
9 4 2 3
10 4 1 3
A full tidyverse version:
library(tidyverse)
df %>%
mutate(id = paste(id1, id2),
count = map(cur_group_rows(), ~ sum(str_detect(id, id1[.x]) & str_detect(id, id2[.x], negate = T))))
A more efficient approach would be to work on a tabulation format:
tab = crossprod(table(rep(seq_len(nrow(df)), ncol(df)), c(df$id1, df$id2)))
#tab
#
# 1 2 3 4
# 1 7 3 2 2
# 2 3 6 1 2
# 3 2 1 4 1
# 4 2 2 1 5
So, now, we have the times each value appears with another (irrespectively of their order in the two columns). Here on, we need a way to subset the above table by each pair and subtract the value of their cooccurence from the value of each id's total appearance.
Make a grid of all combinations:
gr = expand.grid(id1 = colnames(tab), id2 = rownames(tab), stringsAsFactors = FALSE)
Create 2-column matrices to subset the table:
id1.ij = cbind(match(gr$id1, colnames(tab)),
match(gr$id1, rownames(tab)))
id2.ij = cbind(match(gr$id1, colnames(tab)),
match(gr$id2, rownames(tab)))
Subtract the respective values:
cbind(gr, count = tab[id1.ij] - tab[id2.ij])
# id1 id2 count
#1 1 1 0
#2 2 1 3
#3 3 1 2
#4 4 1 3
#5 1 2 4
#6 2 2 0
#7 3 2 3
#8 4 2 3
#9 1 3 5
#10 2 3 5
#11 3 3 0
#12 4 3 4
#13 1 4 5
#14 2 4 4
#15 3 4 3
#16 4 4 0
Of course, if we do not need the full grid of values, we can set:
gr = unique(df)
which results in:
# id1 id2 count
#1 1 2 4
#3 1 3 5
#4 1 4 5
#5 2 1 3
#6 2 3 5
#7 2 4 4
#8 3 1 2
#9 3 4 3
#10 4 2 3
#11 4 1 3
There is a injury score called ISS score
I have a table of injury data in rows according to pt ID.
I would like to obtain the top three values for the 6 injury columns.
Column values range from 0-5.
pt_id head face abdo pelvis Extremity External
1 4 0 0 1 0 3
2 3 3 5 0 3 2
3 0 0 2 1 1 1
4 2 0 0 0 0 1
5 5 0 0 2 0 1
My output for the above example would be
pt-id n1 n2 n3
1 4 3 1
2 5 3 3
3 2 1 1
4 2 1 0
5 5 2 1
values can be in a list or in new columns as calculating the score is simple from that point on.
I had thought that I would be able to create a list for the 6 injury columns and then apply a sort to each list taking the top three values. My code for that was:
ais$ais_list <- setNames(split(ais[,2:7], seq(nrow(ais))), rownames(ais))
But I struggled to apply the sort to the lists within the data frame as unfortunately some of the data in my data set includes NA values
We could use apply row-wise and sort the dataframe and take only first three values in each row.
cbind(df[1], t(apply(df[-1], 1, sort, decreasing = TRUE)[1:3, ]))
# pt_id 1 2 3
#1 1 4 3 1
#2 2 5 3 3
#3 3 2 1 1
#4 4 2 1 0
#5 5 5 2 1
As some values may contain NA it is better we apply sort using anonymous function and then take take top 3 values using head.
cbind(df[1], t(apply(df[-1], 1, function(x) head(sort(x, decreasing = TRUE), 3))))
A tidyverse option is to first gather the data, arrange it in descending order and for every row select only first three values. We then replace the injury column with the column names which we want and finally spread the data back to wide format.
library(tidyverse)
df %>%
gather(injury, value, -pt_id) %>%
arrange(desc(value)) %>%
group_by(pt_id) %>%
slice(1:3) %>%
mutate(injury = 1:3) %>%
spread(injury, value)
# pt_id `1` `2` `3`
# <int> <int> <int> <int>
#1 1 4 3 1
#2 2 5 3 3
#3 3 2 1 1
#4 4 2 1 0
#5 5 5 2 1
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$ID,])
ID x y
2 2 a 5
3 3 a 2
4 4 b 3
1 1 a 0
5 5 b 0
I have 2 dataframes where im trying to compare the value in one with another
If the value matches in both table 1 and 2, then a third value from table 2 is inserted into Table one.
Example Table My DF
words number
1 it 1
2 was 2
3 the 3
4 LTD QTY 4
5 end 5
6 of 6
7 winter 7
Table x.sub
lev_dist Var1 Var2
31 1 LTD QTY LTD QTY
What i want to say is, if Var1 in x.sub is equal to words in MyDF then insert x.sub.lev_dist in a third column next to the word in mydf
My attempt is below but keeps producing 3 in the results instead of the lev_value
mydf$lev_dist <- ifelse(test = (mydf$words == x.sub$Var1),x.sub$Var1,0)
Results:
words number lev_dist
1 it 1 0
2 was 2 0
3 the 3 0
4 LTD QTY 4 3
5 end 5 0
6 of 6 0
7 winter 7 0
Can anyone help
The x.sub$Var1 is a factor column. So, when we do the ifelse, we get the numeric levels of the factor. Replace x.sub$Var1 with as.character(x.sub$Var1) in the ifelse
mydf$lev_dist <- ifelse(mydf$words == as.character(x.sub$Var1)),
x.sub$lev_dist,0)
This could have avoided if the columns were of character class. Using stringsAsFactors=FALSE in the read.csv/read.table or data.frame would ensure that all the character columns are of character class.
You can also use merge:
x.sub = setNames(x.sub,c('lev_dist','words','Var2'))
df_ = merge(df, x.sub[,1:2], by='words', all=T)
df_[is.na(df_)]=0
# >df_
# words number lev_dist
#1 end 5 0
#2 it 1 0
#3 LTD QTY 4 1
#4 of 6 0
#5 the 3 0
#6 was 2 0
#7 winter 7 0
Let's say I have a data frame with 10 numeric variables V1-V10 (columns) and multiple rows (cases).
What I would like R to do is: For each case, give me the number of occurrences of a certain value in a set of variables.
For example the number of occurrences of the numeric value 99 in that single row for V2, V3, V6, which obviously has a minimum of 0 (none of the three have the value 99) and a maximum of 3 (all of the three have the value 99).
I am really looking for an equivalent to the SPSS function COUNT: "COUNT creates a numeric variable that, for each case, counts the occurrences of the same value (or list of values) across a list of variables."
I thought about table() and library plyr's count(), but I cannot really figure it out. Vectorized computation preferred. Thanks a lot!
If you need to count any particular word/letter in the row.
#Let df be a data frame with four variables (V1-V4)
df <- data.frame(V1=c(1,1,2,1,L),V2=c(1,L,2,2,L),
V3=c(1,2,2,1,L), V4=c(L, L, 1,2, L))
For counting number of L in each row just use
#This is how to compute a new variable counting occurences of "L" in V1-V4.
df$count.L <- apply(df, 1, function(x) length(which(x=="L")))
The result will appear like this
> df
V1 V2 V3 V4 count.L
1 1 1 1 L 1
2 1 L 2 L 2
3 2 2 2 1 0
4 1 2 1 2 0
I think that there ought to be a simpler way to do this, but the best way that I can think of to get a table of counts is to loop (implicitly using sapply) over the unique values in the dataframe.
#Some example data
df <- data.frame(a=c(1,1,2,2,3,9),b=c(1,2,3,2,3,1))
df
# a b
#1 1 1
#2 1 2
#3 2 3
#4 2 2
#5 3 3
#6 9 1
levels=unique(do.call(c,df)) #all unique values in df
out <- sapply(levels,function(x)rowSums(df==x)) #count occurrences of x in each row
colnames(out) <- levels
out
# 1 2 3 9
#[1,] 2 0 0 0
#[2,] 1 1 0 0
#[3,] 0 1 1 0
#[4,] 0 2 0 0
#[5,] 0 0 2 0
#[6,] 1 0 0 1
Try
apply(df,MARGIN=1,table)
Where df is your data.frame. This will return a list of the same length of the amount of rows in your data.frame. Each item of the list corresponds to a row of the data.frame (in the same order), and it is a table where the content is the number of occurrences and the names are the corresponding values.
For instance:
df=data.frame(V1=c(10,20,10,20),V2=c(20,30,20,30),V3=c(20,10,20,10))
#create a data.frame containing some data
df #show the data.frame
V1 V2 V3
1 10 20 20
2 20 30 10
3 10 20 20
4 20 30 10
apply(df,MARGIN=1,table) #apply the function table on each row (MARGIN=1)
[[1]]
10 20
1 2
[[2]]
10 20 30
1 1 1
[[3]]
10 20
1 2
[[4]]
10 20 30
1 1 1
#desired result
Here is another straightforward solution that comes closest to what the COUNT command in SPSS does — creating a new variable that, for each case (i.e., row) counts the occurrences of a given value or list of values across a list of variables.
#Let df be a data frame with four variables (V1-V4)
df <- data.frame(V1=c(1,1,2,1,NA),V2=c(1,NA,2,2,NA),
V3=c(1,2,2,1,NA), V4=c(NA, NA, 1,2, NA))
#This is how to compute a new variable counting occurences of value "1" in V1-V4.
df$count.1 <- apply(df, 1, function(x) length(which(x==1)))
The updated data frame contains the new variable count.1 exactly as the SPSS COUNT command would do.
> df
V1 V2 V3 V4 count.1
1 1 1 1 NA 3
2 1 NA 2 NA 1
3 2 2 2 1 1
4 1 2 1 2 2
5 NA NA NA NA 0
You can do the same to count how many time the value "2" occurs per row in V1-V4. Note that you need to select the columns (variables) in df to which the function is applied.
df$count.2 <- apply(df[1:4], 1, function(x) length(which(x==2)))
You can also apply a similar logic to count the number of missing values in V1-V4.
df$count.na <- apply(df[1:4], 1, function(x) sum(is.na(x)))
The final result should be exactly what you wanted:
> df
V1 V2 V3 V4 count.1 count.2 count.na
1 1 1 1 NA 3 0 1
2 1 NA 2 NA 1 1 2
3 2 2 2 1 1 3 0
4 1 2 1 2 2 2 0
5 NA NA NA NA 0 0 4
This solution can easily be generalized to a range of values.
Suppose we want to count how many times a value of 1 or 2 occurs in V1-V4 per row:
df$count.1or2 <- apply(df[1:4], 1, function(x) sum(x %in% c(1,2)))
A solution with functions from the dplyr package would be the following:
Using the example data set from LechAttacks answer:
df <- data.frame(V1=c(1,1,2,1,NA),V2=c(1,NA,2,2,NA),
V3=c(1,2,2,1,NA), V4=c(NA, NA, 1,2, NA))
Count the appearances of "1" and "2" each and both combined:
df %>%
rowwise() %>%
mutate(count_1 = sum(c_across(V1:V4) == 1, na.rm = TRUE),
count_2 = sum(c_across(V1:V4) == 2, na.rm = TRUE),
count_12 = sum(c_across(V1:V4) %in% 1:2, na.rm = TRUE)) %>%
ungroup()
which gives the table:
V1 V2 V3 V4 count_1 count_2 count_12
1 1 1 1 NA 3 0 3
2 1 NA 2 NA 1 1 2
3 2 2 2 1 1 3 4
4 1 2 1 2 2 2 4
5 NA NA NA NA 0 0 0
In my effort to find something similar to Count from SPSS in R is as follows:
`df <- data.frame(a=c(1,1,NA,2,3,9),b=c(1,2,3,2,NA,1))` #Dummy data with NAs
`df %>%
dplyr::mutate(count = rowSums( #this allows calculate sum across rows
dplyr::select(., #Slicing on .
dplyr::one_of( #within select use one_of by clarifying which columns your want
c('a','b'))), na.rm = T)) #once the columns are specified, that's all you need, na.rm is cherry on top
That's how the output looks like
>df
a b count
1 1 1 2
2 1 2 3
3 NA 3 3
4 2 2 4
5 3 NA 3
6 9 1 10
Hope it helps :-)