Recently when I tried to plot in R I keep getting this error. Can anyone tell me why I can't seem to do a scatter plot? I've pasted the terminal screen below.
tcmg2o4 <-read.table("~/Documents/research/metal.oxides/TcMg2O4.inverse/energydata.txt")
tcmg2o4
V1 V2
1 Lattice_constant Total_energy
2 8.0 -371.63306746
3 8.1 -375.035492
4 8.2 -378.8669067
5 8.3 -380.34136459
6 8.4 -382.3921237
7 8.5 -383.60394736
8 8.6 -384.09517631
9 8.7 -383.77668067
10 8.8 -382.43806866
11 8.9 -381.42213458
12 9.0 -379.63327976
attach(tcmg2o4)
plot(Lattice_constant, Total_energy)
Error in plot(Lattice_constant, Total_energy) :
object 'Lattice_constant' not found
plot(V1,V2)
Your problem is that you are not reading the column names as column names. to do this use
header = T
tcmg2o4 <-read.table("~/Documents/research/metal.oxides/TcMg2O4.inverse/energydata.txt", header = T)
In your case, the read.table call has created column names V1 and V2 and these columns will both be factor variables.
You can check the structure of your read in object by
str(tcmg2o4)
## 'data.frame': 11 obs. of 2 variables:
## $ Lattice_constant: num 8 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 ...
## $ Total_energy : num -372 -375 -379 -380 -382 ...
I would also avoid using attach
instead use with or
with(tcmg2o4, plot(Lattice_constant, Total_energy))
or the fact that it is a 2 column data.frame
plot(tcmg2o4)
or use a formula to specify your x and y axis (y~x)
plot(Total_energy ~ Lattice_constant, data = tcmg2o4)
which will all give the same result and be much clearer as to where the data is stored
Related
I have a dataframe df with various columns.
In column df$xyz I have about 20 character variables.
I want to retain 3 variables ("HL%", "HH$", "LL$") and all other variables ("truncated", "kk$", "hhb"...) should be replaced with "other".
Thats my data frame:
xz xyz
2.5 HL%
4.4 HH$
9.3 kk$
2.4 kk$
4.5 LL$
5.6 truncated
I need:
xz xyz
2.5 HL%
4.4 HH$
9.3 other
2.4 other
4.5 LL$
5.6 other
I tried:
df$xyz[df$xyz!="HL%"|
df$xyz!="HH$"|
df$xyz!="LL$"] <- "other"
That doesn't seem to do the trick.
As #nya already stated in comments your df$xyz is probably a factor variable, check with str(df).
str(df)
# 'data.frame': 6 obs. of 2 variables:
# $ xz : num 2.5 4.4 9.3 2.4 4.5 5.6
# $ xyz: Factor w/ 6 levels "HH$","HL%","kk$",..: 2 1 6 6 4 6
In this case first update your factor levels with the new level "other" you introduce. Otherwise skip this step.
levels(df$xyz) <- c(levels(df$xyz), "other")
After that just do.
df$xyz[-which(df$xyz %in% c("HL%", "HH$", "LL$"))] <- "other"
Your approach will also work, but you need to replace the | with &.
I have a data.frame as following:
Lot Wafer Voltage Slope Voltage_irradiated Slope_irradiated m_dist_lot
1 8 810 356.119 6.08423 356.427 6.13945 NA
2 8 818 355.249 6.01046 354.124 6.20855 NA
3 9 917 346.921 6.21474 346.847 6.33904 NA
4 (...)
120 9 914 353.335 6.15060 352.540 6.19277 NA
121 7 721 358.647 6.10592 357.797 6.17244 NA
122 (...)
My goal is simple but also a bit difficult. Definitely it is doable to solve it in several ways:
I want to apply a function "func" to each row according to a factor, e.g. the factor "Lot". This is done via
m_dist_lot<- by(data.frame, data.frame$Lot,func)
This actually works but the result is a by-list:
data.frame$Lot: 7
354 355 363 367 378 419 426 427 428 431 460 477 836
3.5231249 9.4229589 1.4996504 7.2984485 7.6883170 1.2354754 1.8547674 3.1129814 4.4303001 1.9634573 3.7281868 3.6182559 6.4718306
data.frame$Lot: 8
1 2 11 15 17 18 19 20 21 22 24 25
2.1415352 4.6459868 1.3485551 38.8218984 3.9988686 2.2473563 6.7186047 2.6433790 0.5869746 0.5832567 4.5321623 1.8567318
The first row seems to be the row of the initial data.frame where the data is taken from. The second row are the calculated values.
My problem now is: How can I store these values properly into the origin data.frame according to the correct rows?
For example in case of one certain calculation/row of the data frame:
m_dist_lot<- by(data.frame, data.frame$Lot,func)
results for the second row of the data.frame in
data.frame$Lot: 8
2
4.6459868
I want to store the value 4.6459868 in data.frame$m_dist_lot according to the correct row "2":
Lot Wafer Voltage Slope Voltage_irradiated Slope_irradiated m_dist_lot
1 8 810 356.119 6.08423 356.427 6.13945 NA
2 8 818 355.249 6.01046 354.124 6.20855 4.6459868
3 9 917 346.921 6.21474 346.847 6.33904 NA
4 (...)
120 9 914 353.335 6.15060 352.540 6.19277 NA
121 7 721 358.647 6.10592 357.797 6.17244 NA
122 (...)
but I don't know how. My best try actually is to use "unlist".
un<- unlist(m_dist_lot) results in
un[1]
6.354
3.523125
un[2]
6.355
9.422959
un[3]
(..)
But I still don't know how I can "separate" the information of "factor.row" and "calculcated" value in such a way that the information is stored correctly in the data frame.
At least when using un<- unlist(m_dist_lot, use.names = FALSE) the factors are not present:
un[1]
3.523125
un[2]
9.422959
un[3]
1.49965
(..)
But now I lack the information of how to assign these values properly into the data.frame.
Using un<- do.call(rbind, lapply(m_dist_lot, data.frame, stringsAsFactors=FALSE)) results in
(...)
7.922 0.94130936
7.976 4.89560441
8.1 2.14153516
8.2 4.64598677
8.11 1.34855514
(...)
Here I still lack a proper assignment of calculated values <> data.frame.
I'm sure there must be a doable way. Do you know a good method?
Without reproducible data or an example of what you want func to do, I am guessing a bit here. However, I think that dplyr is going to be the answer for you.
First, I am going to use the pipe (%>%) from dplyr (exported from magrittr) to pass the builtin iris data through a series of functions. If what you are trying to calculate requires the full data.frame (and not just a column or two), you could modify this approach to do what you want (just write your function to take a data.frame, add the column(s) of interest, then return the full data.frame).
Here, I first split the iris data by Species (this creates a list, with a separate data.frame for each species). Next, I use lapply to run the function head on each element of the list. This returns a list of data.frames that now each only have three rows. (You could replace head with your function of interest here, as long as it returns a full data.frame.) Finally, I stitch each element of the list back together with bind_rows.
topIris <-
iris %>%
split(.$Species) %>%
lapply(head, n = 3) %>%
bind_rows()
This returns:
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 5.1 3.5 1.4 0.2 setosa
2 4.9 3.0 1.4 0.2 setosa
3 4.7 3.2 1.3 0.2 setosa
4 7.0 3.2 4.7 1.4 versicolor
5 6.4 3.2 4.5 1.5 versicolor
6 6.9 3.1 4.9 1.5 versicolor
7 6.3 3.3 6.0 2.5 virginica
8 5.8 2.7 5.1 1.9 virginica
9 7.1 3.0 5.9 2.1 virginica
Which I am going to use to illustrate the approach that I think will actually address your underlying problem.
The group_by function from dplyr allows a similar approach, but without having to split the data.frame. When a data.frame is grouped, any functions applied to it are applied separately by group. Here is an example in action, which ranks the sepal lengths within each species. This is obviously not terribly useful directly, but you could write a custom function which took any number of columns as arguments (which are then passed in as vectors) and returned a vector of the same length (to create a new column or update an existing one). The select function at the end is only there to make it easier to see what I did
topIris %>%
group_by(Species) %>%
mutate(rank_Sepal_Length = rank(Sepal.Length)) %>%
select(Species, rank_Sepal_Length, Sepal.Length)
Returns:
Species rank_Sepal_Length Sepal.Length
<fctr> <dbl> <dbl>
1 setosa 3 5.1
2 setosa 2 4.9
3 setosa 1 4.7
4 versicolor 3 7.0
5 versicolor 1 6.4
6 versicolor 2 6.9
7 virginica 2 6.3
8 virginica 1 5.8
9 virginica 3 7.1
I got a workaround with the help of Force gsub to keep trailing zeros :
un<- do.call(rbind, lapply(list, data.frame, stringsAsFactors=FALSE))
un<- gsub(".*.","", un)
un<- regmatches(un, gregexpr("(?<=.).*", un, perl=TRUE))
rows<- data.frame(matrix(ncol = 1, nrow = lengths(un)))
colnames(rows)<- c("row_number")
rows["row_number"]<- sprintf("%s", rownames(un))
rows["row_number"]<- as.numeric(un[,1])
rows["row_number"]<- sub("^[^.]*[.]", "", format(rows[,1], width = max(nchar(rows[,1]))))
I need some help with R coding here.
The data set Glass consists of 214 rows of data in which each row corresponds to a glass sample. Each row consists of 10 columns. When viewed as a classification problem, column 10
(Type) specifies the class of each observation/instance. The remaining columns are attributes that might beused to infer column 10. Here is an example of the first row
RI Na Mg Al Si K Ca Ba Fe Type
1 1.52101 13.64 4.49 1.10 71.78 0.06 8.75 0.0 0.0 1
First, I casted column 10 so that it is interpreted by R as a factor instead of an integer value.
Now I need to create a vector with indices for all observations (must have values 1-214). This needs to be done to creating training data for Naive Bayes. I know how to create a vector with 214 values, but not one that has specific indices for observations from a data frame.
If it helps this is being done to set up training data for Naive Bayes, thanks
I'm not totally sure that I get what you're trying to do... So please forgive me if my solution isn't helpful. If your df's name is 'df', just use the dplyr package for reordering your columns and write
library(dplyr)
df['index'] <- 1:214
df <- df %>% select(index,everything())
Here's an example. So that I can post full dataframes, my dataframes will only have 10 rows...
Let's say my dataframe is:
df <- data.frame(col1 = c(2.3,6.3,9.2,1.7,5.0,8.5,7.9,3.5,2.2,11.5),
col2 = c(1.5,2.8,1.7,3.5,6.0,9.0,12.0,18.0,20.0,25.0))
So it looks like
col1 col2
1 2.3 1.5
2 6.3 2.8
3 9.2 1.7
4 1.7 3.5
5 5.0 6.0
6 8.5 9.0
7 7.9 12.0
8 3.5 18.0
9 2.2 20.0
10 11.5 25.0
If I want to add another column that just is 1,2,3,4,5,6,7,8,9,10... and I'll call it 'index' ...I could do this:
library(dplyr)
df['index'] <- 1:10
df <- df %>% select(index, everything())
That will give me
index col1 col2
1 1 2.3 1.5
2 2 6.3 2.8
3 3 9.2 1.7
4 4 1.7 3.5
5 5 5.0 6.0
6 6 8.5 9.0
7 7 7.9 12.0
8 8 3.5 18.0
9 9 2.2 20.0
10 10 11.5 25.0
Hope this will help
df$ind <- seq.int(nrow(df))
This question already has answers here:
Remove rows with all or some NAs (missing values) in data.frame
(18 answers)
Closed 7 years ago.
I apologize if this is elementary or has been answered before, but I haven't found an answer to my question despite extensive searching. I'm also very new to programming so please bear with me here.
I have a bunch of 25 by 2 matrices of data, however some of the cells have NA values. I'm looking to extract a subset of the matrix consisting of only the complete paired values (so no NA values).
So say I have:
3.6 4.2
9.2 8.4
4.8 NA
1.1 8.2
NA 11.6
NA NA
2.7 3.5
I want:
3.6 4.2
9.2 8.4
1.1 8.2
2.7 3.5
Is there some function that would do this easily?
Thanks!
Try this
df <- read.table(text = "3.6 4.2
9.2 8.4
4.8 NA
1.1 8.2
NA 11.6
NA NA
2.7 3.5")
df[complete.cases(df), ]
# V1 V2
# 1 3.6 4.2
# 2 9.2 8.4
# 4 1.1 8.2
# 7 2.7 3.5
df[ apply(!is.na(df), 1, all) , ]
df <- data.frame(V1 = c(3.6,9.2,4.8,1.1,NA,NA,2.7),
V2 = c(4.2,8.4,NA,8.2,11.6,NA,3.5))
EDIT: I forgot na.omit or complete.cases Doh.
I would like to compute a moving average (ma) over some time series data but I would like the ma to consider the order n starting from the rightmost of my series so my last ma value corresponds to the ma of the last n values of my series. The desired function rightmost_ma would produce this output:
data <- seq(1,10)
> data
[1] 1 2 3 4 5 6 7 8 9 10
rightmost_ma(data, n=2)
NA 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
I was reviewing the different ma possibilities e.g. package forecast and could not find how to cover this use case. Note that the critical requirement for me is to have valid non NA ma values for the last elements of the series or in other words I want my ma to produce valid results without "looking into the future".
Take a look at rollmean function from zoo package
> library(zoo)
> rollmean(zoo(1:10), 2, align ="right", fill=NA)
1 2 3 4 5 6 7 8 9 10
NA 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
you can also use rollapply
> rollapply(zoo(1:10), width=2, FUN=mean, align = "right", fill=NA)
1 2 3 4 5 6 7 8 9 10
NA 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
I think using stats::filter is less complicated, and might have better performance (though zoo is well written).
This:
filter(1:10, c(1,1)/2, sides=1)
gives:
Time Series:
Start = 1
End = 10
Frequency = 1
[1] NA 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
If you don't want the result to be a ts object, use as.vector on the result.