how to select specific row by a column - r

I have a data, as an example I show below
a = rep(1:5, each=3)
b = rep(c("a","b","c","a","c"), each = 3)
df = data.frame(a,b)
I want to select all the rows that have the "a"
I tried to do it with
df[df$a %in% a,]
Can someone give me an idea how to get them out?
df2<- structure(list(V1 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), V2 = structure(c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L,
5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L), .Label = c("B02", "B03",
"B04", "B05", "B06", "B07", "C02", "C03", "C04", "C05", "C06",
"C07"), class = "factor")), .Names = c("V1", "V2"), class = "data.frame", row.names = c(NA,
-24L))
I want to select specific rows that start with B but not all of them and just 02, 03, 04, 05
1 B02
1 B03
1 B04
1 B05
2 B02
2 B03
2 B04
2 B05
I also want to have the original data without them too


We need to check the 'b' column
df[df$b %in% 'a',]
For the updated question with 'df2', we can use paste to create the strings 'B02' to 'B05' and use %in% to subset
df2[df2$V2 %in% paste0("B0", 2:5),]
Or another option is grep
df2[grep("^B0[2-5]$", df2$V2),]


> df
a b
1 1 a
2 1 a
3 1 a
4 2 b
5 2 b
6 2 b
7 3 c
8 3 c
9 3 c
10 4 a
11 4 a
12 4 a
13 5 c
14 5 c
15 5 c
This basically says:
For all columns in df choose rows that have value equal to a
> rows_with_a<-df[df$b=='a', ]
> rows_with_a
a b
1 1 a
2 1 a
3 1 a
10 4 a
11 4 a
12 4 a

Related

How to select every 5 other observations in R? [duplicate]

This question already has answers here:
Subset dataframe by multiple logical conditions of rows to remove
(8 answers)
Closed 6 years ago.
I have a dataset that contains 10 "houses" with energy production for every minute of the day. Like so:
HouseID Time KwH
1 1 X
2 1 X
3 1 X
4 1 X
5 1 X
6 1 X
7 1 X
8 1 X
9 1 X
10 1 X
1 2 X
2 2 X
3 2 X
4 2 X
5 2 X
6 2 X
7 2 X
8 2 X
9 2 X
10 2 X
I would like to delete the rows with houseIDs 6 until 10 so that I would be left with only the observations of houseID 1,2,3,4 and 5.

You can try
newdf <- df1[!df1$HouseID %in% 6:10,]
# HouseID Time KwH
#1 1 1 X
#2 2 1 X
#3 3 1 X
#4 4 1 X
#5 5 1 X
#11 1 2 X
#12 2 2 X
#13 3 2 X
#14 4 2 X
#15 5 2 X
data
df1 <- structure(list(HouseID = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L,
10L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L), Time = c(1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L), KwH = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "X",
class = "factor")), .Names = c("HouseID", "Time", "KwH"),
class = "data.frame", row.names = c(NA, -20L))

Assuming df is the name of your data frame then just use the following:
df2 <- subset(df, df$HouseID==1:5)

Get sum of unique rows in table function in R

Suppose I have data which looks like this
Id Name Price sales Profit Month Category Mode Supplier
1 A 2 5 8 1 X K John
1 A 2 6 9 2 X K John
1 A 2 5 8 3 X K John
2 B 2 4 6 1 X L Sam
2 B 2 3 4 2 X L Sam
2 B 2 5 7 3 X L Sam
3 C 2 5 11 1 X M John
3 C 2 5 11 2 X L John
3 C 2 5 11 3 X K John
4 D 2 8 10 1 Y M John
4 D 2 8 10 2 Y K John
4 D 2 5 7 3 Y K John
5 E 2 5 9 1 Y M Sam
5 E 2 5 9 2 Y L Sam
5 E 2 5 9 3 Y M Sam
6 F 2 4 7 1 Z M Kyle
6 F 2 5 8 2 Z L Kyle
6 F 2 5 8 3 Z M Kyle
if I apply table function, it will just combines are the rows and result will be
K L M
X 4 4 1
Y 2 1 3
Z 0 1 2
Now what if I want not the sum of all rows but only sum of those rows with Unique Id
so it looks like
K L M
X 2 2 1
Y 1 1 2
Z 0 1 1
Thanks

If df is your data.frame:
# Subset original data.frame to keep columns of interest
df1 <- df[,c("Id", "Category", "Mode")]
# Remove duplicated rows
df1 <- df1[!duplicated(df1),]
# Create table
with(df1, table(Category, Mode))
# Mode
# Category K L M
# X 2 2 1
# Y 1 1 2
# Z 0 1 1
Or in one line using unique
table(unique(df[c("Id", "Category", "Mode")])[-1])
df <- structure(list(Id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L,
4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L), Name = structure(c(1L, 1L, 1L,
2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L), .Label = c("A",
"B", "C", "D", "E", "F"), class = "factor"), Price = c(2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
), sales = c(5L, 6L, 5L, 4L, 3L, 5L, 5L, 5L, 5L, 8L, 8L, 5L,
5L, 5L, 5L, 4L, 5L, 5L), Profit = c(8L, 9L, 8L, 6L, 4L, 7L, 11L,
11L, 11L, 10L, 10L, 7L, 9L, 9L, 9L, 7L, 8L, 8L), Month = c(1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L), Category = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L), .Label = c("X", "Y", "Z"
), class = "factor"), Mode = structure(c(1L, 1L, 1L, 2L, 2L,
2L, 3L, 2L, 1L, 3L, 1L, 1L, 3L, 2L, 3L, 3L, 2L, 3L), .Label = c("K",
"L", "M"), class = "factor"), Supplier = structure(c(1L, 1L,
1L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 2L, 2L, 2L
), .Label = c("John", "Kyle", "Sam"), class = "factor")), .Names = c("Id",
"Name", "Price", "sales", "Profit", "Month", "Category", "Mode",
"Supplier"), class = "data.frame", row.names = c(NA, -18L))

We can try
library(data.table)
dcast(unique(setDT(df1[c('Category', 'Mode', 'Id')])),
Category~Mode, value.var='Id', length)
# Category K L M
#1: X 2 2 1
#2: Y 1 1 2
#3: Z 0 1 1
Or with dplyr
library(dplyr)
df1 %>%
distinct(Id, Category, Mode) %>%
group_by(Category, Mode) %>%
tally() %>%
spread(Mode, n, fill=0)
# Category K L M
# (chr) (dbl) (dbl) (dbl)
#1 X 2 2 1
#2 Y 1 1 2
#3 Z 0 1 1
Or as #David Arenburg suggested, a variant of the above is
df1 %>%
distinct(Id, Category, Mode) %>%
select(Category, Mode) %>%
table()

Rescaling by group across data frames

I have two data frames
df1 <- structure(list(g1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A", "B"), class = "factor"), g2 = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L), .Label = c("a", "b", "c"), class = "factor"), val1 = 1:20, val2 = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 4L, 1L, 2L, 3L)), .Names = c("g1", "g2", "val1", "val2"), row.names = c(NA, -20L), class = "data.frame")
df2 <- structure(list(g1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A", "B"), class = "factor"), g2 = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L), .Label = c("a", "b", "c"), class = "factor"), val3 = c(5L, 6L, 7L, 3L, 4L, 5L, 2L, 3L, 4L, 8L, 9L, 10L, 4L, 5L, 6L, 5L, 6L)), .Names = c("g1", "g2", "val3"), row.names = c(NA, -17L), class = "data.frame")
> df1
g1 g2 val1 val2
1 A a 1 1
2 A a 2 2
3 A a 3 3
4 A a 4 4
5 A b 5 1
6 A b 6 2
7 A b 7 3
8 A c 8 1
9 A c 9 2
10 A c 10 3
11 B a 11 1
12 B a 12 2
13 B a 13 3
14 B b 14 1
15 B b 15 2
16 B b 16 3
17 B b 17 4
18 B c 18 1
19 B c 19 2
20 B c 20 3
> df2
g1 g2 val3
1 A a 5
2 A a 6
3 A a 7
4 A b 3
5 A b 4
6 A b 5
7 A c 2
8 A c 3
9 B c 4
10 B a 8
11 B a 9
12 B a 10
13 B b 4
14 B b 5
15 B b 6
16 B c 5
17 B c 6
My aim is to rescale df1$val2 to take values between the min and max values of df2$val3 within the respective groups.
I tried this:
library(dplyr)
df1 <- df1 %.% group_by(g1, g2) %.% mutate(rescaled=(max(df2$val3)-min(df2$val3))*(val2-min(val2))/(max(val2)-min(val2))+min(df2$val3))
But the output is different from what I expect. The problem is that I can neither cbind nor merge the two data frames due to their different lengths. Any hints?

Does this work?
library(plyr)
df3 <- ddply(df2, .(g1, g2), summarize, max.val=max(val3), min.val=min(val3))
merged.df <- merge(df1, df3, by=c("g1", "g2"), all.x=TRUE)
## Now rescale merged.df$val2 as desired

Replace NA values in dataframe variable with values from other dataframe by "ID"

I would like to know if there is a more concise way to replace NA values for a variable in a dataframe than what I did below. The code below seems to be longer than what I think might be possible in R. For example, I am unaware of some package/tool that might do this more succinctly.
Is there a way to replace, or merge values only if they are NA? After merging two dataframes using all.x = T I have some NA values, I'd like to replace those with information from another dataframe using a common variable to link the replacement.
# get dataframes
breaks <- structure(list(Break = 1:11, Value = c(2L, 13L, 7L, 9L, 40L,
21L, 10L, 37L, 7L, 26L, 42L)), .Names = c("Break", "Value"), class = "data.frame", row.names = c(NA,
-11L))
fsites <- structure(list(Site = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L,
3L, 3L, 3L, 3L), Plot = c(0L, 1L, 2L, 3L, 4L, 0L, 1L, 2L, 0L,
1L, 2L, 3L, 4L, 5L), Break = c(1L, 5L, 7L, 8L, 11L, 1L, 6L, 11L,
1L, 4L, 6L, 8L, 9L, 11L)), .Names = c("Site", "Plot", "Break"
), class = "data.frame", row.names = c(NA, -14L))
bps <- structure(list(Site = c(1L, 1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 3L,
3L), Plot = c(0L, 1L, 2L, 3L, 1L, 2L, 0L, 1L, 2L, 3L, 4L), Value = c(0.393309653,
0.12465733, 0.27380161, 0.027288989, 0.439712533, 0.289724079,
0.036429062, 0.577460008, 0.820375917, 0.323217357, 0.28637503
)), .Names = c("Site", "Plot", "Value"), class = "data.frame", row.names = c(NA,
-11L))
# merge fsites and bps
df1 <- merge(fsites, bps, by=c("Site", "Plot"), all.x=T)
# merge df1 and breaks to get values to eventually replace the NA values in
# df1$Values.x, here "Break" is the ID by which to replace the NA values
df2 <- merge(df1, breaks, by=c("Break"))
# Create a new column 'Value' that uses Value.x, unless NA, then Value.y
df3 <- df2
df3$Value <- df2$Value.x
df2.na <- is.na(df2$Value.x)
df3$Value[df2.na] <- df2$Value.y[df2.na]
# get rid of unnecessary columns
cols <- c(1:3,6)
df4 <- df3[,cols]

At the stage where there is only (breaks, fsites, bps and) df1 around:
df1$Value <- ifelse(is.na(df1$Value),
breaks$Value[match(df1$Break, breaks$Break)], df1$Value)
#> df1
# Site Plot Break Value
#1 1 0 1 0.39330965
#2 1 1 5 0.12465733
#3 1 2 7 0.27380161
#4 1 3 8 0.02728899
#5 1 4 11 42.00000000
#6 2 0 1 2.00000000
#7 2 1 6 0.43971253
#8 2 2 11 0.28972408
#9 3 0 1 0.03642906
#10 3 1 4 0.57746001
#11 3 2 6 0.82037592
#12 3 3 8 0.32321736
#13 3 4 9 0.28637503
#14 3 5 11 42.00000000
#just to test with your `df4`
> sort(df1$Value) == sort(df4$Value)
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE

ddply summarise proportional count

I am having some trouble using the ddply function from the plyr package. I am trying to summarise the following data with counts and proportions within each group. Here's my data:
structure(list(X5employf = structure(c(1L, 3L, 1L, 1L, 1L, 3L,
1L, 1L, 1L, 3L, 1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 1L, 2L, 2L, 2L,
2L, 2L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 1L, 1L, 3L, 1L,
3L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L,
3L, 3L, 1L), .Label = c("increase", "decrease", "same"), class = "factor"),
X5employff = structure(c(2L, 6L, NA, 2L, 4L, 6L, 5L, 2L,
2L, 8L, 2L, 2L, 2L, 7L, 7L, 8L, 11L, 7L, 2L, 8L, 8L, 11L,
7L, 6L, 2L, 5L, 2L, 8L, 7L, 7L, 7L, 8L, 6L, 7L, 5L, 5L, 7L,
2L, 6L, 7L, 2L, 2L, 2L, 2L, 2L, 5L, 5L, 5L, 2L, 5L, 2L, 2L,
2L, 5L, 12L, 2L, 2L, 2L, 2L, 5L, 5L, 5L, 5L, 2L, 5L, 2L,
13L, 9L, 9L, 9L, 7L, 8L, 5L), .Label = c("", "1", "1 and 8",
"2", "3", "4", "5", "6", "6 and 7", "6 and 7 ", "7", "8",
"1 and 8"), class = "factor")), .Names = c("X5employf", "X5employff"
), row.names = c(NA, 73L), class = "data.frame")
And here's my call using ddply:
ddply(kano_final, .(X5employf, X5employff), summarise, n=length(X5employff), prop=(n/sum(n))*100)
This gives me the counts of each instance of X5employff correctly, but but seems as though the proportion is being calculated across each row and not within each level of the factor X5employf as follows:
X5employf X5employff n prop
1 increase 1 26 100
2 increase 2 1 100
3 increase 3 15 100
4 increase 1 and 8 1 100
5 increase <NA> 1 100
6 decrease 4 1 100
7 decrease 5 5 100
8 decrease 6 2 100
9 decrease 7 1 100
10 decrease 8 1 100
11 same 4 4 100
12 same 5 6 100
13 same 6 5 100
14 same 6 and 7 3 100
15 same 7 1 100
When manually calculating the proportions within each group I get this:
X5employf X5employff n prop
1 increase 1 26 59.09
2 increase 2 1 2.27
3 increase 3 15 34.09
4 increase 1 and 8 1 2.27
5 increase <NA> 1 2.27
6 decrease 4 1 10.00
7 decrease 5 5 50.00
8 decrease 6 2 20.00
9 decrease 7 1 10.00
10 decrease 8 1 10.00
11 same 4 4 21.05
12 same 5 6 31.57
13 same 6 5 26.31
14 same 6 and 7 3 15.78
15 same 7 1 5.26
As you can see the sum of proportions in each level of factor X5employf equals 100.
I know this is probably ridiculously simple, but I can't seem to get my head around it despite reading all sorts of similar posts. Can anyone help with this and my understanding of how the summarise function works?!
Many, many thanks
Marty

You cannot do it in one ddply call because what gets passed to each summarize call is a subset of your data for a specific combination of your group variables. At this lowest level, you do not have access to that intermediate level sum(n). Instead, do it in two steps:
kano_final <- ddply(kano_final, .(X5employf), transform,
sum.n = length(X5employf))
ddply(kano_final, .(X5employf, X5employff), summarise,
n = length(X5employff), prop = n / sum.n[1] * 100)
Edit: using a single ddply call and using table as you hinted towards:
ddply(kano_final, .(X5employf), summarise,
n = Filter(function(x) x > 0, table(X5employff, useNA = "ifany")),
prop = 100* prop.table(n),
X5employff = names(n))

I'd add here an example with dplyr which makes it quite easily in one step, with a short-code and easy-to-read syntax.
d is your data.frame
library(dplyr)
d%.%
dplyr:::group_by(X5employf, X5employff) %.%
dplyr:::summarise(n = length(X5employff)) %.%
dplyr:::mutate(ngr = sum(n)) %.%
dplyr:::mutate(prop = n/ngr*100)
will result in
Source: local data frame [15 x 5]
Groups: X5employf
X5employf X5employff n ngr prop
1 increase 1 26 44 59.090909
2 increase 2 1 44 2.272727
3 increase 3 15 44 34.090909
4 increase 1 and 8 1 44 2.272727
5 increase NA 1 44 2.272727
6 decrease 4 1 10 10.000000
7 decrease 5 5 10 50.000000
8 decrease 6 2 10 20.000000
9 decrease 7 1 10 10.000000
10 decrease 8 1 10 10.000000
11 same 4 4 19 21.052632
12 same 5 6 19 31.578947
13 same 6 5 19 26.315789
14 same 6 and 7 3 19 15.789474
15 same 7 1 19 5.263158

What you apparently want to do is to find out the proportions of X5employff for every value of X5employf. However, you don't tell ddply that X5employf and X5employff are different; to ddply, these two variables are just two variables to split up the data. Also, since there is one observation per line, i.e. count = 1 for every line of the data, the length of each (X5employf, X5employff) combination equals the sum of each (X5employf, X5employff) combination.
The simplest "plyr way" to solve your problem that I can think of is the following:
result <- ddply(kano_final, .(X5employf, X5employff), summarise, n=length(X5employff), drop=FALSE)
n <- result$n
n2 <- ddply(kano_final, .(X5employf), summarise, n=length(X5employff))$n
result <- data.frame(result, prop=n/rep(n2, each=13)*100)
You can also use good old xtabs:
a <- xtabs(~X5employf + X5employff, kano_final)
b <- xtabs(~X5employf, kano_final)
a/matrix(b, nrow=3, ncol=ncol(a))

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