How to subset rows with only odd numbers in 1 column in R [closed]
The question is quite badly formulated but if I understand it correctly it would mean doing something like that: col=3 #col is the # of the column where the numbers should be odd dat[dat[,col]%%2==1,] #dat is the data frame or matrix containing your data Or in the example you give...
Subset a dataframe based on time variable
You can change the 'Time' column to 'POSIXct' class and then subset datasubcolrow$Time <- as.POSIXct(datasubcolrow$Time, format='%d/%m/%Y %H:%M:%OS') subset(datasubcolrow, Time < as.POSIXct('15/05/2015 13:30:15.417', format='%d/%m/%Y %H:%M:%OS')) data datasubcolrow <- structure(list(Time = c("15/05/2015 13:30:07.291", "15/05/2015 13:30:08.307", "15/05/2015 13:30:09.323", "15/05/2015 13:30:10.338", "15/05/2015 13:30:11.354", "15/05/2015 13:30:12.370", "15/05/2015 13:30:13.386", "15/05/2015 13:30:14.402", "15/05/2015 13:30:15.417",...
subset data from matching package
If you just want to keep, say, the first row of treat for each value of ID, then you can use slice: DATA_clean <- treat %>% group_by(ID) %>% slice(1) Your original code didn't work because n() returns to the total number of rows for each value of ID. If every...
Unexpected behavior in subsetting aggregate function in R
r,logic,aggregate,subset,subsetting
The problem is that == is alternating between the values of "honda" and "harley" and comparing with the value in the relevant position of your "manufacturer" variable. On the other hand, %in% (as suggested by MrFlick) and | are checking across the entire "manufacturer" variable before deciding which values to...
R keep rows where all values are < a threshold
This should work: dfOutput <- dfInput[apply(dfInput[,3:19]>0.00000001 & dfInput[,3:19]<0.300, 1, all, na.rm=TRUE), ] And now for a reproducible example, I'll explain what is going on: # data df <- data.frame(x = c(1:3, NA, 3:1), y=c(NA, NA, NA, 3, 3, 2, 3)) # this returns a matrix! df[, 1:2] > 2 #...
How to subset data with several conditions in R?
You could try dat <- subset(wg, Year > 2007 & Year < 2010 & hour == 23 & mint %in% c(30, 32, 39, 40, 41, 49, 31)) ...
efficient way to remove duplicates from list of custom objects in python
python,list,filtering,subset,subsetting
You need realize methods hash and eq on object class A: def __init__(self, a): self.attr1 = a def __hash__(self): return hash(self.attr1) def __eq__(self, other): return self.attr1 == other.attr1 def __repr__(self): return str(self.attr1) Example: l = [A(5), A(4), A(4)] print list(set(l)) print list(set(l))[0].__class__ # ==> __main__.A. It's a object of class...
Subset rows that contain some sequence of characters in one variable
Try df[grep('1{3,}', df$history),] ...
How to subset two different periods in a text file?
#one way is to use `filter` from `dplyr` package (and assuming Date is already in Date format) library(dplyr) wg %>% filter(year %in% c(2008,2009) & months(Date) %in% c("July","August","September") #If you want to stick to subset, replace second & with |: subset(wg, date >= "2008-07-01" & date <= "2008-09-30" | date >=...
Calculating the slope of each row in a large data set using R
r,data.frame,subset,linear-regression
First, you might want to write a function that can calculate the slope for three consecutive values, like this: slope <- function(x){ if(all(is.na(x))) # if x is all missing, then lm will throw an error that we want to avoid return(NA) else return(coef(lm(I(1:3)~x))[2]) } Then you can use the apply()...
How to subset a data.frame?
You may try with data.table. Here, we convert the 'data.frame' to 'data.table' (setDT(a)), grouped by 'var1', we get a logical index for 'var2' elements that are greater than or equal to corresponding 'var2' elements for which 'var3' is TRUE and subset the dataset .SD. library(data.table) setDT(a)[,.SD[var2 >= var2[var3]], var1] #...
R: Data frame operations: filtering common rows and removing rows of several data frames
this should work: library(dplyr) inner_join(dfA, dfB) %>% anti_join(dfC) which gives: Efficiency Value 1 8 7 2 2 4 ...
Convert a character variable into a logical expression in order to use it later inside the subset argument of the subset() function
Try this : subset(raw_data,eval(parse(text=keep_rows))) Test : keep_rows <- "Blok>1" raw_data<- data.frame(Blok=c(1,2,3,0)) subset(raw_data,eval(parse(text=keep_rows))) Blok 2 2 3 3 ...
Subset of a data set
one way: iris$Sepal.Width[ iris$Species %in% "virginica"] Probably best to google subsetting though as this is all easily available in tutorials everywhere, and this will have been asked elsewhere on SO....
Subsetting a data frame based on key spanning several columns in another (summary) data frame
If we already created 3 datasets and want to subset the first "a" based on the elements of "c/c1", one option is anti_join from dplyr library(dplyr) anti_join(a, c1, by=c('A', 'B', 'C')) Update Or we could use a base R option with interaction to paste the columns of interest together in...
Filtering a Dataset by another Dataset in R
Try this WantedData=Data2[Data2$ccession_number %in% SubsetData1$accession_number, ] ...
Subset data with a weighting factor
You'll want to use something like the following: new_data <- Data[sample(nrow(Data), N, prob = (1 - Data$Prob), replace = F),] ...
Reactive subset in ddply for rmarkdown shiny
First, you are trying to modify a reactive object outside the reactive expression. I would suggest to define column names inside the expression. Second, I don't think that modifying bc()$Yield is an authorized operation. So I would try do generate Yield also inside a reactive expression. Below is an edited...
R - How to change values in one Matrix based on elements in another Matrix
r,if-statement,matrix,subset,covariance
I think this does what you're asking, if I'm interpreting the question correctly. I've given you a couple solutions, pick your poison. The first relies on a nested for loop which could be slow and further optimized if you knew for sure your matrix was symmetric. m <- read.table(header=T, stringsAsFactors=F,...
CPLEX/OPL model - constraints with subset index
constraints,scheduling,subset,cplex,opl
It is not clear what your problem is, but I am guessing your problem is to do with modelling things like products(j) in constraint 2. Try using sets for these - so create an array of sets of products in each product family. There are examples of this in the...
R returns the same number of rows after subsetting but clearly deletes rows
There is no problem. Look at nrow(Sin). You should see that is has fewer rows after subsetting. The first column in the output is the "row name". It is not a cumulative index that tells you how many rows there are. Row names are preserved after subsetting (ie they will...
R: Avoid repeating lines of code using R subsets in scripts
With your example, it is now clearer what you want and I give it a second try. I use dataGL_all as defined in your question and the define stations <- rep(c("FLI","FBE"),each=2) directions <- rep(c("in","out"),times=length(stations)/2) You could also extract the stations and directions from your data frame. Using your example, the...
R to Python subsetting via vector
Based on your stated use of pandas colsToUse = ['col1', 'col2', 'col3'] rowsToUse = np.random.choice(range(len(df1)), 500) df2 = df1.ix[:, colsToUse] df3 = df1.ix[rowsToUse, :] There are also some other DataFrame helper functions for indexing: df1.loc, df1.iloc, and df1.xs. It's also helpful to look at the guide NumPy for MATLAB Users...
R: Handling subsets using dynlm
If you do the subsetting yourself via data = zooX[...,], then dynlm() doesn't see the full sample and hence has to lose two observations. If you supply the full data = zooX and then set end = 14 and start = 15 respectively, then dynlm() can first put together the...
how find all groups of subsets of set A? Set partitions in Python
python,algorithm,python-3.x,set,subset
Solution: def partitions(A): if not A: yield [] else: a, *R = A for partition in partitions(R): yield partition + [[a]] for i, subset in enumerate(partition): yield partition[:i] + [subset + [a]] + partition[i+1:] Explanation: The empty set only has the empty partition. For a non-empty set, take out one...
subsetting 1-column matrix deletes rownames [duplicate]
you can read about argument drop in the help page: ?'[' M[which(rownames(M) != "A"), ,drop=FALSE] ...
Data cleaning using subset with 2 conditions on same variable
just a little googling would have solved your problem, for example read this about logical operators, like this? ITEproduction_2014.2015<-subset(ITEproduction_2014.2015,Date.Difference>3 & Date.Difference<40) ...
MySQL: Query a subset of rows
You could try something like this: select user_id from user_assets where asset_id = 1 or asset_id = 2 ... group by user_id having count(distinct asset_id) = (number of assets you are looking for) demo here showing your required output. the distinct isn't necessary if (user_id, asset_id) is a unique key...
R How to mutate a subset of rows
Using data.table, we'd do: setDT(data)[colA == "ABC", ColB := "XXXX"] and the values are modified in-place, unlike if-else, which'd copy the entire column to replace just those rows where the condition satisfies. We call this sub-assign by reference. You can read more about it in the new HTML vignettes....
Slicing rows of pandas dataframe between
You can do this using a variant of the compare-cumsum-groupby pattern. Starting from >>> df["markers"].isin(["x","y"]) 0 False 1 False 2 True 3 False 4 False 5 False 6 True 7 False 8 False 9 True Name: markers, dtype: bool We can shift and take the cumulative sum to get: >>>...
Mongodb query documents where nested array is equal or subset of fixed array
c#,arrays,mongodb,subset,mongodb-csharp
Use mongo Set Operator using $setIsSubset in aggregation you will get your result, check following query : db.collectionName.aggregate({ "$project": { "Name": 1, "Tags": 1, "match": { "$setIsSubset": ["$Tags", ["A", "B", "C"]] //check Tags is subset of given array in your case array is ["A","B","C"] } } }, { "$match": {...
Include column after grouping using datatable
You can use := to create a new column ninefive[, .(zgrp=.N), by= .(cgrp, zip)][, V1:=100*(zgrp/sum(zgrp)), by=zip][, zgrp:=NULL] # cgrp zip V1 #1: 3 12007 19.35484 #2: 4 12007 48.38710 #3: 1 12007 32.25806 #4: 1 12008 57.89474 #5: 4 12008 31.57895 #6: 3 12008 10.52632 Or as @Frank commented, you...
Keep the second occurrence in a column in R
r,conditional,subset,find-occurrences
Here's another possible data.table solution library(data.table) setDT(df1)[, list(Value = c("uncensored", "censored"), Time = c(Time[match("uncensored", Value)], Time[(.N - match("uncensored", rev(Value))) + 2L])), by = ID] # ID Value Time # 1: 1 uncensored 3 # 2: 1 censored 5 # 3: 2 uncensored 2 # 4: 2 censored 5 Or similarly,...
Need to count the number of times a threshold value is met (or exceeded) per year (using R)
Adapting my comment into an answer, taking into account the presented data and OP's comments. Note, code is not checked as dput of data was not obtained. library("dplyr") data_summarised <- data %>% mutate(date = as.Date(paste(YR, MO, DA, sep = "-"))) %>% # concatenate YR MO DA into an ISO date,...
My recursion of getting subset doesn't print out the right answer
algorithm,recursion,combinations,subset
I think this can help you: void subset(vector<int> &input, vector<int> output, int current) { static int n=0; if (current == input.size()) { cout<<n++<<":\t {"; for(int i=0;i<output.size();i++) cout<<output[i]<<", "; cout<<"\b\b}\n"; } else { subset(input, output, current+1); //exclude current'th item output.push_back(input[current]); subset(input, output, current+1); //include current'th item } } and first time...
R - how to replace values in a subset of a vector
r,vector,replace,subset,readline
You can build the new set of file lines as follows: new.r.lines <- c(r.lines[1:27],r.lines[28:1027][grepl(var,r.lines[28:1027])],r.lines[1028:length(r.lines)]); This combines lines 1:27 with the subset of the following 28:1027 lines that match your search pattern, then further combines with lines 1028 to the end of the file. Thus, you can pass that to writeLines()...
R - Subsetting to group of matrices based on a condition
Use split and cumsum: ccc <- data.frame(ccc) split(ccc[ccc$aaa==1,], cumsum(ccc$aaa!=1)[ccc$aaa==1]) #$`0` # aaa bbb #1 1 4 #2 1 4 #3 1 4 # #$`2` # aaa bbb #6 1 3 # #$`3` # aaa bbb #8 1 3 #9 1 2 # #$`4` # aaa bbb #11 1 2 #12...
Subset specific rows with vector of identifiers - warning messages
Your commands depends on equality and this requires an atomic data, but instead you gave a vector. Instead of this, you can use operators as @davidArenburg has mentioned. You can do this by first forming a list of T/F and then retrieve the list according to corresponding value of the...
subset my df provided that each ID has >10 obs a month
We could create a column 'MonthYr' from the 'date' column after converting it to 'Date' class. Get the number of observations ('n') per group ('permno', 'MonthYr') and use that to remove the IDs ('permno') that have at least one 'n' less than 10. library(dplyr) res <- df1 %>% mutate(MonthYr=format(as.Date(date, format='%m/%d/%Y'),...
How to subset a longtable in R and create boxplots with ggplot
As you'd like to produce boxplots for each group and year in the same graph, I think your dataset is ready for that and you can do the following: p <- ggplot(tmp.data, aes(factor(year), fill=group, value)) p + geom_boxplot() ...
Sum of all subparts of an array of integers
arrays,algorithm,sum,big-o,subset
You can easily solve this problem with a simple recursive. def F(arr): if len(arr) == 1: return (arr[0], 1) else: r = F(arr[:-1]) return (11 * r[0] + (r[1] + 1) * arr[-1], 2 * r[1] + 1) So, how does it work? It is simple. Let say we want...
R, create a new sorted dataframe with use of dplyr?
library(ggplot2) diamonds %>% group_by(clarity) %>% summarise(mean_price = mean(price) , min_price =min(price) ,max_price = max(price) , median_price = median(as.numeric(price)), count = n()) %>% arrange(clarity) for arranging in descending order use arrange(desc(clarity)) instead of arrange(clarity)...
R how to retrieve data from data.frame with multiple conditions
You mentioned data.table, so here's two possible approaches for both requests library(data.table) For 1. setDT(df)[, .SD[all(V2 != "X")], by = V1] # V1 V2 V3 # 1: HIJ P 40 # 2: HIJ Y 41 For 2. df[, .SD[.N == 1L], by = V1] # V1 V2 V3 # 1:...
Creating data subset with a vector - why does data have to be sorted?
I think I spot a couple problems. In grep you don't want to set value to be TRUE. Setting value to be true returns the matched word instead of the index of the row. Also you are missing a comma (hence the undefinied columns error). Try This: LakeK_all[grep("^S", LakeK_all$Lake), ]...
How to filter dataframe with multiple conditions?
you can try this with(df, df[ (x==1 & y>15) | (x==2 & y>5), ]) x y 1 1 30 4 2 10 5 2 18 or with dplyr library(dplyr) filter(df, (x==1 & y>15) | (x==2 & y>5)) ...
Subqueries: What am I doing fundamentally wrong?
Perhaps my question was not formulated correctly, but this post had the solutions I was essentially looking for: http://stackoverflow.com/a/123481/2966951 http://stackoverflow.com/a/121435/2966951 Filtering out the most recent row was my problem. I was surprised that selecting from a subquery with a max value could yield anything other than that value....
Outputting various subsets from one data frame based on dates
I might do as following. Only end dates seem to be necessary as start dates are just 1 year before. Loop is made using lapply() which iterates over all end dates. Subsetting is done mainly with difftime() by filtering any non-zero time difference between the two dates. set.seed(24) df1 <-...
subset data.table keeping only elements greater than certain value applied to all columns
If you melt your data.table to long format, this is easy: library(reshape2) news1 <- melt(news, id.vars = "ID") news2 <- news1[abs(value) > 0.01,] # ID variable value #1: 8 diff.jan 0.101 #2: 202 diff.apr 10.000 #3: 203 diff.apr 11.000 #4: 50 diff.aug 0.221 dcast.data.table(news2, ID ~ variable) # ID diff.jan...
How to create a new variable with values from different variables if another variable equals a set value in R?
r,conditional,condition,subset
I was told the problem with my code is that I needed to put indexing on either side. Without the indexing on the right side, it does not know which row to apply the value from. So the correct code in this case would be: df$new[df$date=='a' & !is.na(df$date)] <- df$va[df$date=='a'...
Make two random teams form dataset and get overall team score in R
You could try sampling the indices of players to construct the first team instead of sampling the names. idx1 <- sample(1:nrow(players), 5) You can actually use these indices to grab all the information about each team: team1 <- players[idx1,] team2 <- players[-idx1,] The score for each team can be computed...
Extract series of observations from dataframe for complete sets of data
r,loops,data.frame,pattern-matching,subset
Your question boils down to searching for sequences of "ABC" within the sequences of the IDs: (matches <- gregexpr("ABC", paste(dat$ID, collapse=""))[[1]]) # [1] 8 # ... This indicates that the only match begins at row 8. You now know that the information for Sensor1 are at rows numbered matches, the...
removing and aggregating duplicates
You could do: library(dplyr) df %>% # create an hypothetical "customer.name" column mutate(customer.name = sample(LETTERS[1:10], size = n(), replace = TRUE)) %>% # group data by "Parcel.." group_by(Parcel..) %>% # apply sum() to the selected columns mutate_each(funs(sum(.)), one_of("X.11", "X.13", "X.15", "num_units")) %>% # likewise for mean() mutate_each(funs(mean(.)), one_of("Acres", "Ttl_sq_ft", "Mtr.Size"))...
How to get subset of two arrays into a different array using javascript?
I hope that you want; (renewed :) function subset() { var arr1 = [1, 9, 3, 5, 4, 8, 2, 6, 3, 4] var arr2 = [5, 2, 4, 8, 2, 6, 4] var arr3 = []; var minSize=2; // minimum 2 element must in intersection var findedMax = "";...
(R) [] / subset() returns an empty data frame
If there are lagging/leading spaces, this could occur. Remove those and it should work. library(stringr) data[,5] <- str_trim(data[,5]) Or data[,5] <- gsub('^\\s+|\\s+$', '', data[,5]) data[data[,5]=='Y',] Another option without removing the spaces would be grep data[grep('\\bY\\b', data[,5]),] ...
Is one serie of constraints a subset of the other?
The comment by @false is correct, as far as I can see: given two sets S and S': if S is a subset of S', then the intersection of the complement of S' and S should be the empty set (there are no elements outside of S' that are elements...
Combining the terms of two MuMIn subset of models to create new subset in R
It isn't a nicest solution, but does what you want. library(MuMIn) options(na.action = na.fail) fm1 <- lm(y ~ X1 + X2, Cement) m1 <- dredge(fm1) ms1 <- subset(m1, delta < 32) fm2 <- lm(y ~ X3 + X4, Cement) m2 <- dredge(fm2) ms2 <- subset(m2, delta < 20) a1 <-...
Filter rows in R using subset
Use a regular expression. For example: myd <- subset(df, grepl("-01$", Date_ID)) or myd <- df[grep("-01$", df$Date_ID),] ...
Subsetting rows by passing an argument to a function
The problem is that you pass the condition as a string and not as a real condition, so R can't evaluate it when you want it to. if you still want to pass it as string you need to parse and eval it in the right place for example: cond...
How to subset by distinct rows in a data frame or matrix?
r,matrix,filter,data.frame,subset
Try this: (I suspect will be faster than any apply approach) mat[ rowSums(mat == mat[,1])!=ncol(mat) , ] # ---with your object--- [,1] [,2] [,3] [1,] 1 2 3 [2,] 1 3 2 ...
R: Subsetting and plotting a SpatialPoints object
Subsetting can be done by using []. See the SpatialPolygons-class help (?'SpatialPolygons-class'): Methods [...]: [ : select subset of (sets of) polygons; NAs are not permitted in the row index" So using your data: library(sp) Sr1 = Polygon(cbind(c(2,4,4,1,2),c(2,3,5,4,2))) Sr2 = Polygon(cbind(c(5,4,2,5),c(2,3,2,2))) Sr3 = Polygon(cbind(c(4,4,5,10,4),c(5,3,2,5,5))) Sr4 = Polygon(cbind(c(5,6,6,5,5),c(4,4,3,3,4)), hole = TRUE)...
R use string to refer to column
Short answer is: do not use subset but something like employ.data[employ.data[salary_string]>23000,] ...
Filtering a list of integer in range, to exclude the subsets in python
You can solve this by sorting first: import operator ranges=[[0,1,2,3,4], [1,2], [0,1], [2,3,4], [3,4,5], [3,4,5,6], [4,5], [6,7], [5,6]] sorted_ranges = sorted(ranges,key=operator.itemgetter(-1),reverse=True) sorted_ranges = sorted(sorted_ranges,key=operator.itemgetter(0)) filtered = [] i,j = 0,0 while i < len(sorted_ranges): filtered.append(sorted_ranges[i]) j = i+1 while j < len(sorted_ranges) and sorted_ranges[i][-1] >= sorted_ranges[j][-1]: print "Remove " ,...
Dynamic Programing approach for a subset sum
java,algorithm,recursion,dynamic-programming,subset
Here's the super naive solution that simply generates a power set on your input array and then iterates over each set to see if the sum satisfies the given total. I hacked it together with code already available on StackOverflow. O(2n) in time and space. Gross. You can use the...
Subsetting by summing number of values in clustered data in R [duplicate]
Using lapply() student.count = 2 # depends on your choice out = do.call(rbind, lapply(split(df, f = df$Schools), function(x){ x$no.of.students = length(x$Students); x = subset(x, no.of.students > student.count) })) #> out # Schools Students no.of.students #SchA.1 SchA st1 5 #SchA.2 SchA st2 5 #SchA.3 SchA st3 5 #SchA.4 SchA st4 5...
Filtering rows by a criteria defined by variables
It sounds like you're looking for get: x[get(cname) > cutoff,] # val1 val2 # 1: 2 5 # 2: 3 4 # 3: 4 2 # 4: 5 4 # 5: 3 5 ...
Getting specific rows according to a subset in R
You can try subset(df, V1 %in% l) # V1 V2 #1 a54 hi #3 sdx637 hi intersect can be used to get the common elements intersect(df$V1, l) #[1] "a54" "sdx637" but this will not give a logical index to subset the data, df[intersect(df$V1, l),] # V1 V2 #NA <NA> <NA>...
How to create a new variable with values from different variables if another variable equals a set value in R?
You want the ifelse function, which is a vectorized conditional: > x <- c(1, 1, 0, 0, 1) > y <- c(1, 2, 3, 4, 5) > z <- c(6, 7, 8, 9, 10) > ifelse(x == 1, y, z) [1] 1 2 8 9 5 You will have to...
How do I query a mongo document containing subset of nested array
Use mongo aggregation like following : First use $unwind this will unwind stuff and then use $match to find elements greater than 4. After that $group data based on things.name and add required fields in $project. The query will be as following: db.collection.aggregate([ { $unwind: "$things" }, { $unwind: "$things.stuff"...
subseting data that starts with a letter
This will return only those rows beginning with a capital "S" using the substr()-ing function: dat[ substr( dat$City, 1 ,1) == "S" , ] Could also have used: dat[ grepl("^S", dat$City) , ] The second option is a very simple regular expression. Look at ?regex and ?grep....
Loop through various data subsets in lm() in R
I'm not sure it makes sense to copy your covariates into a new list like that. Here's a way to loop over columns and to dynimcally build formulas dat <- data.frame(y = rnorm(10), x1 = rnorm(10), x2 = rnorm(10), x3 = rnorm(10) ) dat1 <- dat[-9,] #x.list not used fit...
Subsetting / unmerging data frames in R based on a variable
One solution in base R: #using as.character since one$x and two$x are factors in this case > two[ as.character(one$x) != as.character(two$x), ] x y z 11 k 11 -0.6680130 12 l 12 -1.0501888 13 m 13 -1.0987269 14 n 14 1.0045557 15 o 15 -0.6002310 16 p 16 1.3162201 17...
Loop within dataframe subset
data = sub240 is an assignment statement. You can assign things on their own line or in function definitions and calls, but you can only provide logical statements in a while loop definition. If you want logical equality, you need ==. But unless data changes in the loop AND you...
Creating a subset in R using a double loop [closed]
r,loops,double,conditional,subset
Or simply with base R aggregate(Julian_Day ~., df, min) # Year Id Julian_Day # 1 1901 1 40 # 2 1968 1 200 # 3 1901 5 56 Or library(dplyr) df %>% group_by(Id, Year) %>% summarise(Julian_Day = min(Julian_Day)) # Source: local data frame [3 x 3] # Groups: Id #...
Calculating Function from Variable 1 after Eliminating Level(s) from Variable 2 Using R
You were close, try this mean(iris$Sepal.Length[which(iris$Species != 'setosa')]) or mean(iris$Sepal.Length[iris$Species != 'setosa']) or mean(iris[iris$Species!= "setosa", "Sepal.Length"]) ...
Extract grouped Subset with condition
With the sample data dd<-read.table(text="Group Count Value 1 1 1000 1 10 2000 2 6 1000 2 7 2000", header=T) you can do this with base R subset(dd, Count>.25*ave(Count, Group, FUN=sum)) or the dplyr library library(dplyr) dd %>% group_by(Group) %>% filter(Count > .25 * sum(Count)) perhaps you'll find one more...
How to select columns by values in a row in R
This will give you the desired output df. givenStr <- "that" row <- df[df$strs==givenStr,] df[,c(1,1+which(row[,-1]==1))] ...
Arguments for Subset within a function in R colon v. greater or equal to
You can use %in% instead of == subset(data, x %in% 1:3) In general, if we are comparing two vectors of unequal sizes, %in% would be used. There are cases where we can take advantage of the recycling (it can fail too) if the length of one of the vector is...
Maintain data frame rows after subet
I think this is what you want. I've done it using dplyr's group_by and summarize here. For each Batch/ID it calculates the number of observations, the number of observations where measurement is between 6 and 7 and the ratio of those two. library(dplyr) # example data set set.seed(10) Measurement <-...
Return duplicates in a list based on 2 criteria
We can use Reduce with intersect in base R lapply(my.list, function(x) x[with(x, Letters %in% Reduce(intersect, split(Letters, Numbers))),]) Or using dplyr library(dplyr) lapply(my.list, function(x) x %>% group_by(Letters) %>% filter(n_distinct(Numbers)==2)) Instead of having a list, it can be changed to a single dataset with an additional grouping column and then do the...
Multiple filter using grep and subset in R
First, if you want a dataframe, you should use data.frame, not c: df <- data.frame(id, problem, solution1, solution2) Then you can subset like this for instance (no need to use subset per se) df2 <- df[!(grepl("a", df$problem) & (grepl("eat", df$solution1) | grepl("eat", solution2))),] # id problem solution1 solution2 # 2...
split dataframe in groups before each non-NA
I think library("plyr") df <- mutate(df,ID=cumsum(!is.na(df$Height))) dfsum <- ddply(df,.(ID),summarise, stems=length(ID), avg_diameter = sqrt(sum((Diameter)^2))) head(dfsum) ## ID stems avg_diameter ## 1 1 1 7.480282 ## 2 2 1 4.774648 should work ... ? To "order[] the rows of each subset acc. to desc(Diameter)", ddply(df,.(ID), arrange,desc(Diameter)) ...
Subset based on a criteria
Creating the Test/Sample dataset data test; infile datalines dlm=','; input Over : 8. Ball : 8. Bowling : $15. Runs_scored : 8. Count : 8. ; datalines; 39,1,Ali,1,1 39,2,Ali,1,2 39,3,Ali,2,3 39,4,Ali,1,4 39,5,Ali,1,5 39,6,Ali,1,6 36,1,Anderson,1,1 36,2,Anderson,1,2 36,3,Anderson,1,3 36,4,Anderson,0,4 36,6,Anderson,0,6 ; run; Selecting the distinct overs(as I understand Cricket, each over would...
Selecting rows by offsetting
Here's another idea using dplyr: library(dplyr) my_df %>% filter(lead(let == "b", 5) | lag(let == "b", 5)) Or as per @akrun suggestion using the devel version of data.table: setDT(my_df)[shift(let == "b", 5) | shift(let == "b", type = "lead", 5)] Which gives: # num let #1 0.36723709 a #2 0.24743170...
Subset data frames inside of a list based on column classes
Another option is to use Filter within lapply lapply(list12, Filter, f = is.numeric) # [[1]] # x1 z1 # 1 1 0 # 2 2 1 # 3 3 0 # 4 4 1 # # [[2]] # x2 y2 # 1 0 0 # 2 1 1 # 3...
Subsetting Column from Matrix in R without specific column names specified
You can try a[,setdiff(colnames(a), S)] Or a[,!colnames(a) %in% S] ...
Subset duplicates based on two columns [duplicate]
You can try df1[duplicated(df1)|duplicated(df1, fromLast=TRUE),] # A B #2 1A 2 #3 1A 2 #5 2 4 #6 2 4 #7 3A 0 #8 3A 0 #9 4A 1 #10 4A 1 data df1 <- structure(list(A = c("1", "1A", "1A", "2", "2", "2", "3A", "3A", "4A", "4A", "5"), B =...
How to subtract a complete character vector with repeated characters from the other vector in R
You can use the function pmatch: x[-pmatch(y,x)] #[1] "A" "C" "A" "B" "D" Edit If your data can be strings of more than 1 character, here is an option to get what you want: xNew <- unlist(sapply(x[!duplicated(x)], function(item, tab1, tab2) { rep(item, tab1[item] - ifelse(item %in% names(tab2), tab2[item], 0)) },...
how to subset in r for this particular condition?
You can use package dplyr: library(dplyr) intersect(df1,df2) # a b #1 1 m #2 3 f Edit for the new data.frames with c column: you can use function semi_join (also from dplyr): semi_join(df1,df2,by=c("a","b")) # a b c #1 1 m df1 #2 3 f df1 Other option, in base R:...
combination of pair subsets from a list in lisp
Read up on mapcar et al: (defparameter a (list 1 2 3 4)) (mapcon (lambda (tail) (mapcar (lambda (x) (cons (car tail) x)) (cdr tail))) a) ==> ((1 . 2) (1 . 3) (1 . 4) (2 . 3) (2 . 4) (3 . 4)) ...
Subset of a table that contains at least one element of another table
Try foverlaps from data.table library(data.table) setkey(setDT(df1), Chromosome, start, end) setkey(setDT(df2), Chromosome, start, end) setnames(unique(foverlaps(df1, df2, nomatch=0)[, c(1,4:5), with=FALSE]), names(df1))[] # Chromosome start end #1: 1 1 450 #2: 2 3500 3585 #3: 2 7850 10000 Or as @Arun commented, we can use which=TRUE (to extract the indices) and subset 'df1'...
Why subset cut decimal part?
This is not a subsetting issue, it's a formatting/presentation issue. You're in the first circle of Burns's R Inferno ("[i]f you are using R and you think you’re in hell, this is a map for you"): another aspect of virtuous pagan beliefs—what is printed is all that there is If...
Subset multiples data frames in a list that match a certain condition
I created a sample data. When you use subset(), you need a data frame and a condition. When you use lapply(), you make your function anonymous. That is, you write function(x) and further write codes which you want R to loop through. In your case, you want to loop through...