I have a data like this
df <-structure(list(label = structure(c(5L, 6L, 7L, 8L, 3L, 1L, 2L,
9L, 10L, 4L), .Label = c(" holand", " holandindia", " Holandnorway",
" USAargentinabrazil", "Afghanestan ", "Afghanestankabol", "Afghanestankabolindia",
"indiaAfghanestan ", "USA", "USAargentina "), class = "factor"),
value = structure(c(5L, 4L, 1L, 9L, 7L, 10L, 6L, 3L, 2L,
8L), .Label = c("1941029507", "2367321518", "2849255881",
"2913128511", "2927576083", "4550996370", "457707181.9",
"637943892.6", "796495286.2", "89291651.19"), class = "factor")), .Names = c("label",
"value"), class = "data.frame", row.names = c(NA, -10L))
I want to get the largest name (in letter) and then see how many smaller and similar names are and assign them to a group
then go for another next large name and assign them to another group
until no group left
at first I calculate the length of each so I will have the length of them
library(dplyr)
dft <- data.frame(names=df$label,chr=apply(df,2,nchar)[,1])
colnames(dft)[1] <- "label"
df2 <- inner_join(df, dft)
Now I can simply find which string is the longest
df2[which.max(df2$chr),]
Now I should see which other strings have the letters similar to this long string . we have these possibilities
Afghanestankabolindia
it can be
A
Af
Afg
Afgh
Afgha
Afghan
Afghane
.
.
.
all possible combinations but the order of letter should be the same (from left to right) for example it should be Afghand cannot be fAhg
so we have only two other strings that are similar to this one
Afghanestan
Afghanestankabol
it is because they should be exactly similar and not even a letter different (more than the largest string) to be assigned to the same group
The desire output for this is as follows:
label value group
Afghanestan 2927576083 1
Afghanestankabol 2913128511 1
Afghanestankabolindia 1941029507 1
indiaAfghanestan 796495286.2 2
Holandnorway 457707181.9 3
holand 89291651.19 3
holandindia 4550996370 3
USA 2849255881 4
USAargentina 2367321518 4
USAargentinabrazil 637943892.6 4
why indiaAfghanestan is a seperate group? because it does not completely belong to another name (it has partially name from one or another). it should be part of a bigger name
I tried to use this one Find similar strings and reconcile them within one dataframe which did not help me at all
I found something else which maybe helps
require("Biostrings")
pairwiseAlignment(df2$label[3], df2$label[1], gapOpening=0, gapExtension=4,type="overlap")
but still I don't know how to assign them into one group
You could try
library(magrittr)
df$label %>%
tolower %>%
trimws %>%
stringdist::stringdistmatrix(method = "jw", p = 0.1) %>%
as.dist %>%
`attr<-`("Labels", df$label) %>%
hclust %T>%
plot %T>%
rect.hclust(h = 0.3) %>%
cutree(h = 0.3) %>%
print -> df$group
df
# label value group
# 1 Afghanestan 2927576083 1
# 2 Afghanestankabol 2913128511 1
# 3 Afghanestankabolindia 1941029507 1
# 4 indiaAfghanestan 796495286.2 2
# 5 Holandnorway 457707181.9 3
# 6 holand 89291651.19 3
# 7 holandindia 4550996370 3
# 8 USA 2849255881 4
# 9 USAargentina 2367321518 4
# 10 USAargentinabrazil 637943892.6 4
See ?stringdist::'stringdist-metrics' for an overview of the string dissimilarity measures offered by stringdist.
Related
I have a dataset that was recorded by observation(each observation has its own row of data). I am looking to combine/condense these rows by the plant they were found on - currently a character variable. All other columns are numerical vales.
EX:
This is the raw data
|Sci_Name|Honeybee_count|Other_bee_Obsevrved|Stem_count|
|---|---|---|---|
|Zizia aurea|1|5|10|
|Asclepias viridiflora|15|1|3|
|Viola unknown|0|0|4|
|Zizia aurea|0|2|6|
|Zizia aurea|3|6|3|
|Asclepias viridiflora|8|2|17|
and I want:
Sci_Name
Honeybee_count
Other_bee_Obsevrved
Stem_count
Zizia aurea
4
13
19
Asclepias viridiflora
23
3
20
Viola unknown
0
0
4
I am currently pulling this data from a CSV already in table form. I have been attempting to create a new table/data frame with one entry of each plant species, and blanks/0s for each other variable, which I can then use to c-binding the two together. This, however, has been clunky at best and I am having trouble figuring out how to have each row check itself. I am open to any approach, let me know what you think!
Thanks :D
We can use the formula method in aggregate from base R. On the rhs of the ~, specify the grouping variable and on the lhs, use . for denoting the rest of the variables. Specify the FUN as sum and it will do the column wise sum by group
aggregate(. ~ Sci_Name, df1, sum)
-output
Sci_Name Honeybee_count Other_bee_Obsevrved Stem_count
1 Asclepias viridiflora 23 3 20
2 Viola unknown 0 0 4
3 Zizia aurea 4 13 19
data
df1 <- structure(list(Sci_Name = c("Zizia aurea", "Asclepias viridiflora",
"Viola unknown", "Zizia aurea", "Zizia aurea", "Asclepias viridiflora"
), Honeybee_count = c(1L, 15L, 0L, 0L, 3L, 8L), Other_bee_Obsevrved = c(5L,
1L, 0L, 2L, 6L, 2L), Stem_count = c(10L, 3L, 4L, 6L, 3L, 17L)),
class = "data.frame", row.names = c(NA,
-6L))
I have the following dataframe with a character variable that represents the number of lanes on a highway, can I replace this vector with a similar vector that has numbers instead of letter?
df<- structure(list(Blocked.Lanes = c("|RS|RS|ML|", "|RS|", "|RS|ML|ML|ML|ML|",
"|RS|", "|RS|RE|", "|ML|ML|ML|", "|RS|ML|", "|RS|", "|ML|ML|ML|ML|ML|ML|",
"|RS|ML|ML|"), Event.Id = c(240314L, 240381L, 240396L, 240796L,
240948L, 241089L, 241190L, 241225L, 241226L, 241241L)), row.names = c(NA,
10L), class = "data.frame")
The output should be something like df2 below:
df2<- structure(list(Blocked.Lanes = c(3L, 1L, 5L, 1L, 2L, 3L, 2L,
1L, 6L, 3L), Event.Id = c(240314L, 240381L, 240396L, 240796L,
240948L, 241089L, 241190L, 241225L, 241226L, 241241L)), class = "data.frame", row.names = c(NA,
-10L))
One way would be to count number of "|" in each string. We subtract it with - 1 since there is an additional "|".
stringr::str_count(df$Blocked.Lanes, '\\|') - 1
#[1] 3 1 5 1 2 3 2 1 6 3
In base R :
lengths(gregexpr("\\|", df$Blocked.Lanes)) - 1
Another way would to be count exact words in the string.
stringr::str_count(df$Blocked.Lanes, '\\w+')
lengths(gregexpr("\\w+", df$Blocked.Lanes))
Similar to Ronak's solution you could also do:
stringr:str_count(df$Blocked.Lanes, "\\b[A-Z]{2}\\b")
if the lanes are always 2 letters long, or
stringr:str_count(df$Blocked.Lanes, "\\b[A-Z]+\\b")
if the lanes are always at least one letter long.
stringr:str_count(df$Blocked.Lanes, "(?<=\\|)[A-Z]+(?=\\|)")
also works.
Not as succinct as #Ronak Shah's, but another method in Base R.
String split on string literal "|" and then count elements:
df2 <- transform(df, Blocked.Lanes = lengths(Map(function(x) x[x != ""],
strsplit(df$Blocked.Lanes, "|", fixed = TRUE))))
Am at beginner stage of R programming, please help me in below issue.
I have different desc values assigned to the same sol attribute in different rows. I want to make all desc values of sol attribute in single row as mentioned below
My data is as follows:
sol desc
1 fry, toast
1 frt,grt,gty
1 ytr,uyt,ytr
6 hyt, ytr,oiu
4 hyg,hyu,loi
4 opu,yut,yut
I want the output as follows :
sol desc
1 fry,toast,frt,grt,gty,ytr,uyt,yir
6 hyt, ytr,oiu
4 hyg,hyu,loi,opu,yut,yut
Note: you can input any values in desc as per your convenience.
aggregate() is what you are looking for. Try this:
aggregate(desc ~ sol, data = df, paste, collapse = ",")
sol desc
1 1 fry, toast,frt,grt,gty,ytr,uyt,ytr
2 4 hyg,hyu,loi,opu,yut,yut
3 6 hyt, ytr,oiu
Data
df <- structure(list(sol = c(1L, 1L, 1L, 6L, 4L, 4L), desc = c("fry, toast",
"frt,grt,gty", "ytr,uyt,ytr", "hyt, ytr,oiu", "hyg,hyu,loi",
"opu,yut,yut")), .Names = c("sol", "desc"), class = "data.frame", row.names = c(NA,
-6L))
I imagine that there's some way to do this with sqldf, though I'm not familiar with the syntax of that package enough to get this to work. Here's the issue:
I have two data frames, each of which describe genomic regions and contain some other data. I have to combine the two if the region described in the one df falls within the region of the other df.
One df, g, looks like this (though my real data has other columns)
start_position end_position
1 22926178 22928035
2 22887317 22889471
3 22876403 22884442
4 22862447 22866319
5 22822490 22827551
And another, l, looks like this (this sample has a named column)
name start end
101 GRMZM2G001024 11149187 11511198
589 GRMZM2G575546 24382534 24860958
7859 GRMZM2G441511 22762447 23762447
658 AC184765.4_FG005 26282236 26682919
14 GRMZM2G396835 10009264 10402790
I need to merge the two dataframes if the values from the start_position OR end_position columns in g fall within the start-end range in l, returning only the columns in l that have a match. I've been trying to get findInterval() to do the job, but haven't been able to return a merged DF. Any ideas?
My data:
g <- structure(list(start_position = c(22926178L, 22887317L, 22876403L,
22862447L, 22822490L), end_position = c(22928035L, 22889471L,
22884442L, 22866319L, 22827551L)), .Names = c("start_position",
"end_position"), row.names = c(NA, 5L), class = "data.frame")
l <- structure(list(name = structure(c(2L, 12L, 9L, 1L, 8L), .Label = c("AC184765.4_FG005",
"GRMZM2G001024", "GRMZM2G058655", "GRMZM2G072028", "GRMZM2G157132",
"GRMZM2G160834", "GRMZM2G166507", "GRMZM2G396835", "GRMZM2G441511",
"GRMZM2G442645", "GRMZM2G572807", "GRMZM2G575546", "GRMZM2G702094"
), class = "factor"), start = c(11149187L, 24382534L, 22762447L,
26282236L, 10009264L), end = c(11511198L, 24860958L, 23762447L,
26682919L, 10402790L)), .Names = c("name", "start", "end"), row.names = c(101L,
589L, 7859L, 658L, 14L), class = "data.frame")
I have the following table ordered group by first, second and name.
myData <- structure(list(first = c(120L, 120L, 126L, 126L, 126L, 132L, 132L), second = c(1.33, 1.33, 0.36, 0.37, 0.34, 0.46, 0.53),
Name = structure(c(5L, 5L, 3L, 3L, 4L, 1L, 2L), .Label = c("Benzene",
"Ethene._trichloro-", "Heptene", "Methylamine", "Pentanone"
), class = "factor"), Area = c(699468L, 153744L, 32913L,
4948619L, 83528L, 536339L, 105598L), Sample = structure(c(3L,
2L, 3L, 3L, 3L, 1L, 1L), .Label = c("PO1:1", "PO2:1", "PO4:1"
), class = "factor")), .Names = c("first", "second", "Name",
"Area", "Sample"), class = "data.frame", row.names = c(NA, -7L))
Within each group I want to extract the area that correspond to the specific sample. Several groups don´t have areas from the samples, so if the sample is´nt detected it should return "NA".Ideally, the final output should be a column for each sample.
I have tried the ifelse function to create one column to each sample:
PO1<-ifelse(myData$Sample=="PO1:1",myData$Area, "NA")
However this doesn´t takes into account the group distribution. I want to do this, but within the group. Within each group (a group as equal value for first, second and Name columns) if sample=PO1:1, Area, else NA.
For the first group:
structure(list(first = c(120L, 120L), second = c(1.33, 1.33),
Name = structure(c(1L, 1L), .Label = "Pentanone", class = "factor"),
Area = c(699468L, 153744L), Sample = structure(c(2L, 1L), .Label = c("PO2:1",
"PO4:1"), class = "factor")), .Names = c("first", "second", "Name",
"Area", "Sample"), class = "data.frame", row.names = c(NA, -2L))
The output should be:
structure(list(PO1.1 = NA, PO2.1 = 153744L, PO3.1 = NA, PO4.1 = 699468L), .Names =c("PO1.1", "PO2.1", "PO3.1", "PO4.1"), class = "data.frame", row.names = c(NA, -1L))
Any suggestion?
As in the example in the quesiton, I am assuming Sample is a factor. If this is not the case, consider making it such.
First, lets clean up the column Sample to make it a legal name, or else it might cause errors
levels(myData$Sample) <- make.names(levels(myData$Sample))
## DEFINE THE CUTS##
# Adjust these as necessary
#--------------------------
max.second <- 3 # max & nin range of myData$second
min.second <- 0 #
sprd <- 0.15 # with spread for each group
#--------------------------
# we will cut the myData$second according to intervals, cut(myData$second, intervals)
intervals <- seq(min.second, max.second, sprd*2)
# Next, lets create a group column to split our data frame by
myData$group <- paste(myData$first, cut(myData$second, intervals), myData$Name, sep='-')
groups <- split(myData, myData$group)
samples <- levels(myData$Sample) ## I'm assuming not all samples are present in the example. Manually adjusting with: samples <- sort(c(samples, "PO3.1"))
# Apply over each group, then apply over each sample
myOutput <-
t(sapply(groups, function(g) {
#-------------------------------
# NOTE: If it's possible that within a group there is more than one Area per Sample, then we have to somehow allow for thi. Hence the "paste(...)"
res <- sapply(samples, function(s) paste0(g$Area[g$Sample==s], collapse=" - ")) # allowing for multiple values
unlist(ifelse(res=="", NA, res))
## If there is (or should be) only one Area per Sample, then remove the two lines aboce and uncomment the two below:
# res <- sapply(samples, function(s) g$Area[g$Sample==s]) # <~~ This line will work when only one value per sample
# unlist(ifelse(res==0, NA, res))
#-------------------------------
}))
# Cleanup names
rownames(myOutput) <- paste("Group", 1:nrow(myOutput), sep="-") ## or whichever proper group name
# remove dummy column
myData$group <- NULL
Results
myOutput
PO1.1 PO2.1 PO3.1 PO4.1
Group-1 NA "153744" NA "699468"
Group-2 NA NA NA "32913 - 4948619"
Group-3 NA NA NA "83528"
Group-4 "536339" NA NA NA
Group-5 "105598" NA NA NA
You cannot really expect R to intuit that there is a fourth factor level between PO2 and PO4 , now can you.
> reshape(inp, direction="wide", idvar=c('first','second','Name'), timevar="Sample")
first second Name Area.PO4:1 Area.PO2:1 Area.PO1:1
1 120 1.3 Pentanone 699468 153744 NA
3 126 0.4 Heptene 32913 NA NA
4 126 0.4 Heptene 4948619 NA NA
5 126 0.3 Methylamine 83528 NA NA
6 132 0.5 Benzene NA NA 536339
7 132 0.5 Ethene._trichloro- NA NA 105598