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conditional removal from data frame


conditional removal from data frame

By : Raqib Ahmed Asif
Date : November 22 2020, 02:59 PM
wish of those help I have a data frame , Use this:
code :
r[r$a!=0 | (r$a==0 & r$b=="a"),]
r[r$a!=0 | r$b=="a",]


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Data frame selection and NA removal

Data frame selection and NA removal


By : vidhyag
Date : March 29 2020, 07:55 AM
I hope this helps . I am working on the data frame called "juul" that can be found in library(ISwR). How can I re-write the following expression: selection= juul[juul$sex==2 & juul$tanner>2,] so that NA is excluded in the criteria? , You can do that with:
code :
selection <- na.omit(juul[juul$sex==2 & juul$tanner>2,])
selection <- juul[juul$sex==2 & juul$tanner>2,]
selection <- na.omit(juul[juul$sex==2 & juul$tanner>2,c(1:5)])
selection <- na.omit(juul[juul$sex==2 & juul$tanner>2,-6])
Conditional row removal in an R data frame

Conditional row removal in an R data frame


By : user3714549
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , How about just using which.max since that would select just one value (i.e. the index of the first occurrence of the maximum value):
code :
as.data.frame(lapply(df, function(x) x[-which.max(x)]))
#   t1 t2 t3
# 1  1  7  1
# 2  2  3  1
# 3  3  1  1
apply function in data table for conditional removal of row

apply function in data table for conditional removal of row


By : apenman
Date : March 29 2020, 07:55 AM
this one helps. I have a data table, dt: , If I understand you correctly I would do:
code :
dt[!ew, on = c(V2 = "V1")][nchar(V2) > 5]
       V1                      V2             V3 PubMedCounts
1: 100005                100-00-5     CAS_Number            6
2: 100005 1-Chloro-4-nitrobenzene DescriptorName           12
dt[nchar(V2) > 5][!ew, on = c(V2 = "V1")]
dt[nchar(V2) > 5 & !( V2 %chin% ew$V1 )]
dt <- structure(list(V1 = c(100005L, 100005L, 100005L, 100005L), V2 = c("100-00-5", 
"1-Chloro-4-nitrobenzene", "aahs", "PNCB"), V3 = c("CAS_Number", 
"DescriptorName", "DescriptorName", "Synonym"), PubMedCounts = c(6L, 
12L, 111L, 35L)), .Names = c("V1", "V2", "V3", "PubMedCounts"
), row.names = c(NA, -4L), class = c("data.table", "data.frame"))

ew <- structure(list(V1 = c("aah", "aahed", "aahing", "aahs", "aardvark")), .Names = "V1", row.names = c(NA, -5L), class = c("data.table", "data.frame"))
Conditional removal of rows for grouped data in R

Conditional removal of rows for grouped data in R


By : Shakif Fahim
Date : March 29 2020, 07:55 AM
it helps some times Raw data: Remove the default factoring of variables while creating data frames.
code :
df<-data.frame(id = id,year = year, month = month, day = day, value = value, stringsAsFactors = FALSE)
library("data.table")
setDT(df)[, .(value, .N), by = .(id, year, month)][N > 15, .(avg_rainfall = mean(value)), by = .(id, year)]

#           id year avg_rainfall
# 1: Station_1 1950     4.852840
# 2: Station_1 1951     5.138069
# 3: Station_1 1952     4.934006
# 4: Station_2 1950     4.870335
# 5: Station_2 1951     5.179425
# 6: Station_2 1952     5.055026
# 7: Station_3 1950     4.959524
# 8: Station_3 1951     5.049996
# 9: Station_3 1952     4.927548
How to apply conditional duplicate removal for list of data.frame?

How to apply conditional duplicate removal for list of data.frame?


By : Hussein M.
Date : March 29 2020, 07:55 AM
it fixes the issue We can do this Map to subset the rows of the list elements based on a logical index created with the corresponding number specified in the vector (1:3). Convert the data.frame elements in list to data.table (setDT(x)) , grouped by the columns ('start.pos', 'end.pos', 'pos.score'), we get the number of rows (.N), create a logical index with if/else and get the sequence of rows that satisfies the condition specified in the OP's post, use .I to get the row index, extract that index column ($V1) and use that to subset the dataset.
code :
library(data.table)
res <- Map(function(x,y) setDT(x)[x[,  .I[if(.N > y) seq_len(pmax(y-1, 1)) 
        else seq_len(.N)]  , .(start.pos, end.pos, pos.score)]$V1], myList, 1:3)
sapply(res, nrow)
#bar cat foo 
#  7  12   9 

sapply(expectedList, nrow) 
#bar cat foo 
#7  12   9 
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