data.frame

I have some time series (Monthly) that end in different months, for example, some end in March 2017, others in April 2017 and ... 3.1042 231 2017-03-01 1.05 9.85 3.1279 232 2017-04-01 0.79 3.1362 233 2017-05-01 0.08

I want to create a data frame as junction of 4 other data frames. I managed to do this using such commands: ZHO<-as.data ... o know if there is another way, because this way is laborious and does not work when I have a very large amount dates frames

Suppose I have the data frame iris, present in the memory of R: head(iris) Sepal.Length Sepal.Width Petal.Length Petal.Wi ... h has thousands of different species occurrences. It would be impossible to type in all the options that I have to work with.

I imported a table as a database to deal with in R. However, I need to do some calculations with only a few columns of that table. How do I select only these columns for the calculations?

In my case I have two date.frames: > head(Trecho) Xt Yt Zt 1 -75.56468 1.642710 0 2 -74.56469 1.639634 ... th their respective value closer to TrechoSim. I tried difference_inner_join but it returns values NA in a few lines. Grid,

I try to perform this excerpt below and it occurs: rain.df<-read.table("C:\\Users\\Marcia\\Desktop\\Sript R\\daily.dat", ... scan (file = file, what = what, sep = sep, quote = quote, dec = dec,: line 3572 had no 29 elements. Why ? grateful now

Suppose you want to create a blank data frame to receive the results of a adhesion test. x<-data.frame(Distribuicao=c(NUL ... XGumbel X0.9105 Gumbel 0.9105 How best to store these results or how can I work around the question of variable names?

It is common to read/hear comments saying that the data is in the format wideor long. What does it mean that a table is ... question , but it is restricted to one of the cases and only yours operationalization, without going into conceptual issues.

I need to calculate the fashion of inline values on a date.frame. Since in R there is no function already defined for fashion ... es in a single column. What should I do in this case? How can I calculate inline fashion with bimodal value on a date.frame?

Let's say I have a date.frame 6x5, example: print(Dados) Linha A B C D E L1 4 3 NA 2 4 L2 1 NA ... 0 2 3 4 L4 2 4 5 0 9 How can I do to replace variables with NA values by ZERO within my date.frame?

I have a date.frame and a value in R, needed this value to be included in the last row of certain column of this date.frame. ... function rbind but it did not work. x <- c(1:15) y <- c(1:15) z <- 4 df <- data.frame(x,y) df <- rbind(x, z)

Consider the following data.frame: df <- data.frame(x=c("a","b"), y=c(1,2)) How to include a new column, say, z = c(1,2)?

I need to create a vector with sequential dates between 01/01/2013 and 31/05/2018, but with repetitions, as an example below, ... 0:00 03/01/2013 1:00 03/01/2013 4:00 03/01/2013 8:00 03/01/2013 12:00 03/01/2013 18:00 How can I do this automatically?

If I have a date.frame: > dato<-as.data.frame(matrix(1:64,8,8,T))[-3,] > dato V1 V2 V3 V4 V5 V6 V7 V8 1 1 2 3 ... 59 60 61 62 63 64 How can I "call" the index vector (1, 2, 4, 5, 6, 7, 8)? Because I want to use it as an index in a for.

Assuming I have the following date.frame: print(DADOS) letra N1 N2 N3 N4 A 2 3 4 4 A 1 2 3 4 A 2 2 1 ... t sounds like a simple thing, but I've tried countless ways to no avail. How can I enumerate each row of my date.frame no R?

I have a database and my goal is to perform some class behavior analysis per row. Example: print(DADOS) Linha A B C ... andard, the famous "out of series". Knowing this, the question is: how can I calculate the median of a line in a date.frame?

Hello, I own a date.frame with thousands of rows and a vector with the rows I need from this date.frame, how do I select only the rows from the list in a separate object?

Why am I having the error File"", line 50 print ("this is the value of the sum of the products between the samples and ... álise da média ponderada", Nponderadai) print ("Esta é a média ponderada do conjunto de dados": PiXi/Nponderadai)

By downloading the database on daily rates and tickets for 2019 from the portal da Transparencia http://www.portaltransparen ... trim_ws = TRUE) Does anyone suggest any other function that can correctly read the database 2019_Viagem.csv?

When I'm going to "plot" a bar chart it modifies my Y-axis values proportionally , but I want them to stay with the actual va ... ns.barplot(data=dados_vendas_Activision_2012, x='Platform', y='Global_Sales', ci=None)