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sum date different frames based on column names

Y
yajahira last edited by
I really need your help to finish an analysis. I have the following tables (date frames):
```
df1<-data.frame(c(0,1,1,0),c(1,0,0,1),c(1,1,0,0))
colnames(df1)<-c("0","0.5","1")
df2<-data.frame(c(1,1,1,0),c(0,1,0,0))
colnames(df2)<-c("0","0.5")
df3<-data.frame(c(1,1,1,1),c(0,0,0,1))
colnames(df3)<-c("0.5","1")
```
understand that for the analysis work I need the three tables to have the columns of name 0, 0.5 and 1. I thought about adding all the tables with a mold table containing the three columns and values all equal to zero, but still couldn't solve this. You think that's the solution? How can I do that? Thank you guys!!
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1 Reply Last reply

Anderson

Sorry friends, it was very easy

molde<-data.frame(c(0,0,0,0),c(0,0,0,0),c(0,0,0,0))
colnames(molde)<-c("0","0.5","1")
melted <- cbind(df2,molde)

aggr <- as.data.frame(do.call(cbind, by(t(melted),INDICES=names(melted),FUN=colSums)))

Thank you anyway

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How to pass my data set from wide format to long cm multiple variables in R
Software Programming • database data.frame • • Kadyn

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Hello, to do this task it is necessary to primarily combine the functions to melt and dcast package reshape2 along with the function sub.require(reshape2) longo <- melt(df, id.vars=c("id","grupo")) # Passo as variáveis do formato wide para longo longo$tempo <- factor(with(longo, sub(".*_","",variable))) #Crio a variável tempo longo$variavel <- factor(with(longo, sub("_.*","",variable))) #Crio a variável em questão longo <- longo[,-3] longo <- dcast(longo, id + grupo + tempo ~ variavel) # Volto as variáveis que ficaram no formato "wide" para longo. After that, the dataframe will look like this: id grupo tempo fc pa 1 1 A pos 70 120 2 1 A pre 90 130 3 2 A pos 77 110 4 2 A pre 98 140 5 3 A pos 77 140 6 3 A pre 77 160 Note that for this code to work, it is necessary that the columns in the original dataframe (df) obey some pattern so that the sub() function is able to properly separate the variable name, from the time it was measured.
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How to find the accuracy between two columns of a frame date?
Software Programming • python spanking • • terrea

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Considering a dateframe in the following format, as your example:>>> df ColA ColB 0 Bola Sola 1 Rua Rua And you want to count how many rows have the equal A and B column values, you could use the https://numpy.org/doc/stable/reference/generated/numpy.sum.html?highlight=sum#numpy.sum :import numpy total_iguais = numpy.sum(df['ColA'] == df['ColB']) total = len(df) acuracia = total_iguais / total print(f'Acuracia: {acuracia}') Now if you want to know how close the two words are (Bola and Sola are different by just one letter), you would have to add some logic of comparing each character or then use some similarity algorithm, such as https://docs.python.org/3/library/difflib.html#difflib.SequenceMatcher :import difflib s = difflib.SequenceMatcher(None, 'Bola', 'Sola') print(s.ratio()) # 0.75
File import Excel to R
Software Programming • data.frame • • morde

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If your problem is the path specification, try to bring it with the following function:file.choose() It will open the folder menu of your operating system. Give two clicks on the file that you want to import. The path of this file will be issued to the R Console. Something like this:"/home/user/importR/my1/MeusDados.xlsx" Copy this path ("/home/user/importR/my1/MeusDados.xlsx") without the file name. In this way:"/home/user/importR/my1" Use it in the function below. So:setwd("/home/user/importR/my1") Try to carry out the import with the name complete of the archive ("/home/user/importR/my1/MeusDados.xlsx"library(readxl) meus_dados <- read_excel("/home/user/importR/my1/MeusDados.xlsx")
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About merge when the label name is different
Software Programming • python pandas python3 • • gionna

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"About the merge when the label name is different" => align the label name! I think it is not answeredIf you want results, you can:Change index and column name to merge with concat.import pandas as pd dateA = pd.DatetimeIndex(['2000-01-05','2000-01-06','2000-01-07'],name='日付') dateB = pd.DatetimeIndex(['2000-01-05','2000-01-06','2000-01-08'],name='Date') A = pd.DataFrame({'価格':[100,110,120]},index=dateA) B = pd.DataFrame({'price':[50,60,70]},index=dateB) tmp = B.rename_axis('日付') #indexの名前を変える tmp = tmp.rename(index=str, columns={'price': '価格'}) #columnの名前を変える tmp = pd.concat([A, tmp], ignore_index=False) # concatでマージ AB = tmp The result is included in AB, but the date is the same.tmp['日付'] = tmp.index # AとBで日付が重複しているので違うなにかの価格とするとindexにいると扱いにくいので tmp.index = range(len(tmp.index)) #日付をcolumnに追い出す tmp = tmp[['日付','価格']] # columnの順番入れ替え AB = tmp.sort_values(by=['日付'], ascending=True) #日付でソート Maybe I don’t want to write this code, so I asked how to do it.If you have a smart answer, you can merge.
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How to change the class of a variable within a frame/tibble table/date?
Software Programming • data.frame dplyr • • nishika

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first would be good if you exposed your data or a part of them to make the problem playable. But here taking a generic case.library(dplyr) tibble(uf=c("AM", "RJ", "SC"), regiao=c("norte", "sudeste", "sul"), ano=c(1989, 1987, 1986), pop=runif(3)) %>% print() %>% {.} -> dt uf regiao ano pop <chr> <chr> <dbl> <dbl> #1 AM norte 1989 0.05092618 #2 RJ sudeste 1987 0.54035176 #3 SC sul 1986 0.54615493 To change the typing of the variables is only to apply the function mutate package dplyr. On the part of the date should be established a day and a month to use the type date, in the example below I used the first of January, which will be applied to all lines.dt %>% mutate(regiao=as.factor(regiao)) %>% mutate(ano=as.Date(ISOdate(ano, 1, 1))) uf regiao ano pop <chr> <fctr> <date> <dbl> #1 AM norte 1989-01-01 0.05092618 #2 RJ sudeste 1987-01-01 0.54035176 #3 SC sul 1986-01-01 0.54615493 What you could do different here would be to use a single mutate to change both columns, such as:dt %>% mutate(regiao=as.factor(regiao), ano=as.Date(ISOdate(ano, 1, 1)))
How to transform the values of the variable of my date. frame in columns in R?
Software Programming • data.frame • • inna

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You can use the tidyverse to do what you want. First, I create a reproducible example:set.seed(123) df_1 <- data.frame( cod = 1, qtde = sample(x = 1:4, size = 11, replace = TRUE), mes = c(1, 1, 2, 3, 3, 3, 4, 4, 5, 6, 6) ) Now the analysis:library(tidyverse) df_1 %>% group_by(mes) %>% mutate(n = row_number()) %>% spread(key = mes, value = qtde) A tibble: 3 x 8 cod n 1 2 3 4 5 6 <dbl> <int> <int> <int> <int> <int> <int> <int> #1 1 1 3 3 3 2 4 4 #2 1 2 4 NA 3 1 NA 3 #3 1 3 NA NA 3 NA NA NA You can use the argument fill to fill in NAs with the digit you want. For example, replace NAs by 0:df_1 %>% group_by(mes) %>% mutate(n = row_number()) %>% spread(key = mes, value = qtde, fill = 0) A tibble: 3 x 8 cod n 1 2 3 4 5 6 <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> #1 1 1 3 3 3 2 4 4 #2 1 2 4 0 3 1 0 3 #3 1 3 0 0 3 0 0 0
EWMA Volatility on a date. frame
Software Programming • data.frame • • Marcee

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Follow the code with the answer The second function variance.ewma receives a matrix with the returns of the actions and the decay factor value of the EWMA, for each column of the matrix it calls the function ewma.func and pass the decay factor by applying the ewma formula to each column, when you finish applying the formula in each column it returns a matrix with the results.ewma.func <- function(rets, lambda) { rets[is.infinite(rets)] = 0 rets[is.nan(rets)] = 0 ewma.ant <- rets[1]^2 ewma <- vapply(rets, function(r) ewma.ant <<- ewma.ant *lambda + (r^2)*(1 - lambda), 0) return(ewma) } variance.ewma <- function(bancodados,decayfactor){ EWMA = bancodados for (i in 1:ncol(EWMA)){ EWMA[, i] = ewma.func(EWMA[, i], decayfactor) } return(EWMA)}
Filter data from a Dataframe pandas by a specified column and the last four dates of a set of dates
Software Programming • python pandas dataframe numpy • • jeanid

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ATUALIZED TO PARTNERSI hope I understand the problem.Let's go straight to the point:Creating A Dataframe>>> import pandas as pd >>> dfA = pd.DataFrame({"sinal": ["SP1", "BH", "NAC", "SP1", "SP1", "BH", "BH", "BH", "BH", "BH"], "cod_programa": ["XXXXX", "XXXXX", "XXXXX", "XXXXX", "XXXXX", "ZZZZZ", "ZZZZZ", "ZZZZZ", "ZZZZZ", "ZZZZZ"], "data": ["23/05/2020", "01/05/2020", "02/05/2020", "01/05/2020", "02/05/2020", "02/05/2020", "03/05/2020", "01/05/2020", "02/05/2020", "03/05/2020"], "tempo_disponivel": [10, 300, 118, 106, 6, 218, 66, 44, 190, 202]}) >>> dfA sinal cod_programa data tempo_disponivel 0 SP1 XXXXX 23/05/2020 10 1 BH XXXXX 01/05/2020 300 2 NAC XXXXX 02/05/2020 118 3 SP1 XXXXX 01/05/2020 106 4 SP1 XXXXX 02/05/2020 6 5 BH ZZZZZ 02/05/2020 218 6 BH ZZZZZ 03/05/2020 66 7 BH ZZZZZ 01/05/2020 44 8 BH ZZZZZ 02/05/2020 190 9 BH ZZZZZ 03/05/2020 202 Transforming date (string) for date>>> dfA["data"] = pd.to_datetime(dfA['data'], format="%d/%m/%Y") >>> dfA sinal cod_programa data tempo_disponivel 0 SP1 XXXXX 2020-05-23 10 1 BH XXXXX 2020-05-01 300 2 NAC XXXXX 2020-05-02 118 3 SP1 XXXXX 2020-05-01 106 4 SP1 XXXXX 2020-05-02 6 5 BH ZZZZZ 2020-05-02 218 6 BH ZZZZZ 2020-05-03 66 7 BH ZZZZZ 2020-05-01 44 8 BH ZZZZZ 2020-05-02 190 9 BH ZZZZZ 2020-05-03 202 Creating DataFrame B>>> dfB = pd.DataFrame({"sinal": ["SP1", "SP1", "BH", "SP1", "NAC", "SP1", "SP1", "BH", "NAC", "SP1", "DF", "RJ"], ... "cod_programa": ["XXXX", "XXXX", "XXXX", "XXXX", "XXXX", "XXXX", "XXXX", "ZZZZ", "ZZZZ", "ZZZZ", "ZZZZ", "ZZZZ"], ... "data_exibicao": ["23/05/2020", "09/05/2020", "02/05/2020", "16/05/2020", "16/05/2020", "30/05/2020", "02/05/2020", "01/05/2020", "01/05/2020", "01/05/2020", "01/05/2020", "01/05/2020"], ... "data_inicio_programa": ["02/05/2020 19:01:00", "02/05/2020 19:01:00", "02/05/2020 19:01:00", "02/05/2020 19:05:00", "02/05/2020 19:05:00", "02/05/2020 19:05:00", "02/05/2020 19:05:00", "01/05/2020 11:30:00", "01/05/2020 11:30:00", "01/05/2020 11:30:00", "01/05/2020 11:30:00", "01/05/2020 11:30:00"], ... "media_audiencia": [325104.42, 255104.42, 455104.42, 118375.44, 1053999.80, 1299650.50, 14091402.82, 150998.68, 7846891.62, 841810.51, 123071.38, 581696.70]}) Transforming Dates>>> dfB["data_exibicao"] = pd.to_datetime(dfB['data_exibicao'], format="%d/%m/%Y") >>> dfB["data_inicio_programa"] = pd.to_datetime(dfB['data_inicio_programa'], format="%d/%m/%Y %H:%M:%S") >>> dfB sinal cod_programa data_exibicao data_inicio_programa media_audiencia 0 SP1 XXXX 23/05/2020 02/05/2020 19:01:00 325104.42 1 SP1 XXXX 09/05/2020 02/05/2020 19:01:00 255104.42 2 BH XXXX 02/05/2020 02/05/2020 19:01:00 455104.42 3 SP1 XXXX 16/05/2020 02/05/2020 19:05:00 118375.44 4 NAC XXXX 16/05/2020 02/05/2020 19:05:00 1053999.80 5 SP1 XXXX 30/05/2020 02/05/2020 19:05:00 1299650.50 6 SP1 XXXX 02/05/2020 02/05/2020 19:05:00 14091402.82 7 BH ZZZZ 01/05/2020 01/05/2020 11:30:00 150998.68 8 NAC ZZZZ 01/05/2020 01/05/2020 11:30:00 7846891.62 9 SP1 ZZZZ 01/05/2020 01/05/2020 11:30:00 841810.51 10 DF ZZZZ 01/05/2020 01/05/2020 11:30:00 123071.38 11 RJ ZZZZ 01/05/2020 01/05/2020 11:30:00 581696.70 Catching the Existing Signals>>> sinais = list(set(dfB["sinal"])) >>> sinais ['NAC', 'BH', 'DF', 'RJ', 'SP1'] Catching the biggest dates by sign>>> datas = [] >>> for sinal in sinais: ... dts = dfA[(dfA["sinal"] == sinal)].nlargest(4, "data") ... tmp = [row["data"] for _, row in dts.iterrows()] ... datas.append((sinal, tmp)) ... >>> datas [('NAC', [Timestamp('2020-05-02 00:00:00')]), ('BH', [Timestamp('2020-05-03 00:00:00'), Timestamp('2020-05-03 00:00:00'), Timestamp('2020-05-02 00:00:00'), Timestamp('2020-05-02 00:00:00')]), ('DF', []), ('RJ', []), ('SP1', [Timestamp('2020-05-23 00:00:00'), Timestamp('2020-05-02 00:00:00'), Timestamp('2020-05-01 00:00:00')])] Note: O numeric_only=True is already predicting future versions of pandas. The default of this parameter is None what in the future will bring the median of dates (datetime64 and datetime64tz).Catching medians by signal and date>>> for item in datas: ... sinal = item[0] ... dts = item[1] ... for dt in dts: ... d = f"{dt.year}-{dt.month}-{dt.day}" ... mediana = dfB[(dfB["sinal"] == sinal) & (dfB["data_exibicao"] == d)].median(numeric_only=True) ... print(sinal, d, mediana["media_audiencia"]) BH 2020-5-3 nan BH 2020-5-3 nan BH 2020-5-2 455104.42 BH 2020-5-2 455104.42 NAC 2020-5-2 nan SP1 2020-5-23 325104.42 SP1 2020-5-2 14091402.82 SP1 2020-5-1 841810.51 Note: As I created the dataframes in hand, some dates did not hit, but reference server.I hope I helped
Build Data Frame with "get" Function
Software Programming • orientation to objec lang data.frame • • Alberto

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The function get() is not suitable for what you want to do. She takes only one object at a time. To get more than one object you would have to use the function mget, but still doesn't make much sense in this case, since you can select columns by name in one data.frame.One way to solve your problem would be the following:mylist<-c("A","B","C") colunas <- lapply(mylist, function(x) paste(c("P", "i", "S"), x, sep = ".")) lista_de_dfs <- list() for(i in seq_along(colunas)) lista_de_dfs[[mylist[[i]]]] <- data[c(colunas[[i]], "P.D", "i.D")] And the object lista_de_dfs is a list with the data.frames that you want, for example:lista_de_dfs[["A"]] P.A i.A S.A P.D i.D 1 16.86 59.67768 12.3125 28.41693 12.400 2 16.69 59.90459 12.2500 28.73513 12.980 3 16.62 60.28277 12.0625 28.90646 13.220 4 16.90 60.43405 12.3125 29.17570 495.300 5 17.05 60.73660 12.0000 29.59179 39.125 If you want to play these data.frames for the global environment (which I would not recommend, it is much easier to work with them on the list), just run list2env(lista_de_dfs, globalenv()).Just to illustrate, if you want to do using the mget() -- what do not recommend, since besides much more complicated is much slower --, the logic would be quite similar:mylist<-c("A","B","C") colunas <- lapply(mylist, function(x) paste(c("P", "i", "S"), x, sep = ".")) lista_de_dfs <- list() for(i in seq_along(colunas)) lista_de_dfs[[mylist[[i]]]] <- data.frame(mget(c(colunas[[i]], "P.D", "i.D"), as.environment(data))) There is no reason to do this way instead of working with the natural way of selecting data.frames.
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Calculate difference between two dates to validate date fields
Software Programming • javascript data • • tahishae

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Good morning. Try this:var date1 = new Date("7/11/2010"); var date2 = new Date("12/12/2010"); var timeDiff = Math.abs(date2.getTime() - date1.getTime()); var diffDays = Math.ceil(timeDiff / (1000 * 3600 * 24)); alert(diffDays); http://jsfiddle.net/JS69L/1/ Source: https://stackoverflow.com/questions/3224834/get-difference-between-2-dates-in-javascript
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Sum values of a column in the database through date filter
Software Programming • mysql c sharp winforms • • kalyn

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Why not add a MySqlParameter the MySqlCommand passing the value of the date?string query = "SELECT IFNULL(SUM(quantidadeVagas), 0) AS Vagas FROM vagas WHERE data = @Data"; using(MySqlConnection conexao = new MySqlConnection(DadosDaConexao.StringDeConexao)) { conexao.Open(); using(MySqlCommand cmdDataBase = new MySqlCommand(query, conexao)) { cmdDataBase.Parameters.AddWithValue("@Data", mTxtDataVagas.Text); using(MySqlDataReader myReader = cmdDataBase.ExecuteReader()) { int quantidade = 0; if(myReader.HasRows) { myReader.Read(); quantidade = myReader.GetInt32("Vagas"); } } } } The way I was passing the date parameter almost certainly would give problems.
Add a year column to a date column in scala spark
Software Programming • dataframe date sc languageala spark • • Alberto

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I already found the solution, which was to create an UDF, which adds the year to date.import org.apache.spark.sql.functions._ import org.joda.time.DateTime import java.util.Calendar import scala.language.postfixOps import java.text.SimpleDateFormat // Método que realiza la suma del año a una fecha def addDaysPolTrad(fechainicial:String, anio:String) :String = { try { val formato = new SimpleDateFormat("yyyy-MM-dd") val date = formato.parse(fechainicial) val calendario = Calendar.getInstance() val anioASumar = anio.toInt calendario.setTime(date) calendario.add(Calendar.YEAR, anioASumar) val nuewYeardate = calendario.getTime var fechaSumada = formato.format(nuewYeardate) var fechafinal = fechaSumada return fechafinal } catch { case e: Exception => "" } } // Variable que almacena el resultado que devvuelve el método val addDaysPolTradUDF = udf((fechainicial: String,anio: String) => (addDaysPolTrad(fechainicial,anio))) //Para este ejemplo cree la fecha, pero puede tomarse de un campo de tu df val dt: DateTime = new DateTime("2011-02-28") val dateSub: String = dt.toString.substring(0, 10) import spark.implicits._ //DataFrame ejemplo val tabla = Seq(5).toDF("anio"). withColumn("fecha",lit(dateSub)) // se agrega como una nueva columna a tu df, //indicando tus campos que tomara como los parámetros de entra val tablaFinal = tabla. select( col("fecha"), col("anio")). withColumn("fecha_add_anio",addDaysPolTradUDF(col("fecha"),col("anio"))) when performing the show to your table, the result is the followingtablaFinal.show() +----------+----+--------------+ | fecha|anio|fecha_add_anio| +----------+----+--------------+ |2011-02-28| 5| 2016-02-28| +----------+----+--------------+ I'm waiting for you, I've been trying days and I'm not leaving, but with help... you could.Greetings!
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Replacing values NA of a column by the top row value of the same column of a dataframe
Software Programming • data.frame • • Duquan

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The function na.locf package zoo does exactly what is requested:library(zoo) #> #> Attaching package: 'zoo' #> The following objects are masked from 'package:base': #> #> as.Date, as.Date.numeric na.locf(DADOS) #> a b #> 1 1 5 #> 2 2 6 #> 3 2 7 #> 4 3 7 #> 5 4 7 Created on 2021-06-02 by the https://reprex.tidyverse.org (v2.0.0)
How to delete rows of a R Data Frame based on the values of one of the columns?
Software Programming • data.frame • • briley

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To research alphanumeric patterns, the best is to use grep or grepl.set.seed(6323) # Torna os resultados reprodutíveis n <- 100 DF <- data.frame(A = sample(LETTERS, n, TRUE), X = rnorm(n)) inx <- grepl("[AEIOU]", toupper(DF[[1]])) DF2 <- DF[!inx, ] # Usando um indíce lógico DF3 <- subset(DF, !inx) # Usando avaliação não-standard identical(DF2, DF3) #[1] TRUE head(DF2) A X #2 F -0.54113708 #3 D -0.72646708 #4 V 0.02213349 #6 X -0.64141533 #7 F -1.06416864 #8 Y -0.90681239 Another equivalent way will be with %in%.inx2 <- toupper(DF[[1]]) %in% c("A", "E", "I", "O", "U") DF4 <- DF[!inx2, ] # Usando um indíce lógico mais uma vez identical(DF2, DF4) #[1] TRUE
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How to select all variables from the date. once frame for a regression?
Software Programming • data.frame regression lm • • amarantha

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The operator . in this context formula function lm) means "all other columns that are not in the formula".Thus y regression against all other columns of data.frame can be obtained as follows:modelo <- lm(y~., data=dados) Reference: ?formula

sum date different frames based on column names

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