similar to: Asterisk, she no hang uppa the phone!

Bryan... NOBODY CARES how many times Chris is wrong. NOBODY CARES what you prove on this point. Whether Chris is right, wrong or stoopid is NOT proper fodder for this (CENTOS) list. Chris being wrong is *strictly* a "Bryan Fixation". Part of YOU growing up is YOU realizing that whether Chris is or is not wrong is UNIMPORTANT to ANYTHING that you or anybody important to you think is

Sharing my stoopid mistake in the hopes of saving someone else the same grief: I had a boolean toggle that was not performing as expected. Long story short: I had put quotes around the word "false" class { ''foo'' : boolFlag => "false" } was coming up TRUE To fix it, lose the quotes class { ''foo'' : boolFlag => false } “Sometimes I

On Friday 20 February 2004 00:09, Renaud Waldura shaped the electrons to shout: > 1- this is such a stoopid question I should be ashamed of myself -- > RTFM > > 2- this is a well-known issue that everybody's aware of, it doesn't > need further discussion thank you, as it is actively being worked on > > 3- it is meant to be that way and will not change. No, if you

I am interested in testing two similar nls models to determine if the lines are statistically different when fitted with two different data sets; one corn, another soybean. I know I can do this in linear models by testing for interactions. See Introductory Statistics with R by Dallgaard p212-218 for an example. I have two different data sets I am comparing to lai. ci.re should have very

Hello, I have a simple question but I don't know which method is best to use for my problem. I have the following strings: str1 <- "My_name_is_peter" str2 <- "what_is_your_surname_peter" I would like to apply predefined abbreviations for peter=p and name=n to both strings so that the new strings look like the followings: str1: "My_n_is_p" str2:

Dear group, Here is my df, trades1 : trades1 <- structure(list(Instrument.Long.Name = c("CORN", "CORN", "CORN", "CORN", "CORN", "SOYBEANS", "SOYBEANS", "SOYBEANS", "SOYBEANS", "SOYBEANS", "SOYBEANS", "STANDARD LEAD USD", "STANDARD LEAD USD", "SPCL

Dear group, I need to modify some characters in a data frame. I want to use gsub and the regex functionalities to do this. Here is the data frame (df): DESCRIPTION prix quantity 1 CORN Jul/10 -1.5 0 2 CORN May/10 -1082.0 -3 3 ROBUSTA COFFEE (10) Jul/10 11084.0 8 4 SOYBEANS Jul/10 1983.5

Dear group, Here is my data frame: futures <- structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10", "LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11

Dear group, Here is my df : trades <- structure(list(Trade.Status = c("DEL", "INS", "INS"), Instrument.Long.Name = c("SUGAR NO.11", "CORN", "CORN"), Delivery.Prompt.Date = c("Jul/10", "Jul/10", "Jul/10"), Buy.Sell..Cleared. = c("Sell", "Buy", "Buy"), Volume = c(1L, 2L, 1L),

Dear group, I have 3 data frames I would like to merge. Here they are: pose16 <- structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 12L), .Label = c("COPPER May/10", "COTTON NO.2 Jul/10", "CRUDE OIL miNY May/10", "GOLD Jun/10", "ROBUSTA COFFEE (10) Jul/10", "SOYBEANS Jul/10", "SUGAR NO.11

Dear group, Here is my df: avprix <- structure(list(DESCRIPTION = c("CORN Jul/10", "CORN May/10", "ROBUSTA COFFEE (10) Jul/10", "SOYBEANS Jul/10", "SPCL HIGH GRADE ZINC USD", "STANDARD LEAD USD"), prix = c(-1.5, -1082, 11084, 1983.5, -2464, -118), quantity = c(0, -3, 8, 2, -1, 0), fees = c(-8.64, -30.24, -26.4, -25.92, -37.5,

Dear group, I have these 2 following data frame: allcon <- structure(list(DESCRIPTION = structure(1:17, .Label = c("COFFEE C Jul/10", "COPPER May/10", "CORN Jul/10", "CORN May/10", "COTTON NO.2 Jul/10", "CRUDE OIL miNY May/10", "GOLD Jun/10", "HENRY HUB NATURAL GAS May/10", "ROBUSTA COFFEE (10)

Dear group, Here is the kind of data.frame I obtain every day with my function : futures <- structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10", "LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR

Dear group, Here is my data frame: > dput(u) structure(list(DESCRIPTION = structure(c(2L, 5L, 6L, 7L, 9L, 11L, 12L, 15L, 14L, 16L, 1L, 10L, 3L, 4L, 13L, 8L, 17L), .Label = c("COFFEE C Jul/10", "COPPER May/10", "CORN Jul/10", "CORN May/10", "COTTON NO.2 Jul/10", "CRUDE OIL miNY May/10", "GOLD Jun/10", "HENRY HUB

Dear R-list, Is there any way I can extract part or whole column name from two variables and combine them? Not values, but column names. For example, I have data set with 70 variables. The column names are "Alfalfa", "Corn","Soybean",..."Sunflower". After I combine the two variable (say Alfalfa and Corn), I want to rename it as "AlfCor" or

Dear group, Here is my df : df3 <- structure(list(DESCRIPTION = c("COPPER May/10", "COTTON NO.2 Jul/10", "CRUDE OIL miNY May/10", "GOLD Jun/10", "ROBUSTA COFFEE (10) Jul/10", "SOYBEANS Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 May/10", "WHEAT Jul/10", "SPCL HIGH GRADE ZINC USD",

Dear list, here are my two data frames: av <- structure(list(DESCRIPTION = c("COFFEE C Sep/10", "COPPER Sep/10", "CORN Dec/10", "CRUDE OIL miNY Sep/10", "GOLD Aug/10", "HENRY HUB NATURAL GAS Sep/10", "PALLADIUM Sep/10", "SILVER Sep/10", "SOYBEANS Nov/10", "SPCL HIGH GRADE ZINC USD", "SUGAR

Hi List l am interested in developing price model. I have found a research paper related to price model of corn in US market where it has taken demand & supply forces into consideration. Following are the equation: Supply equation: St= a0+a1Pt-1+a2Rt-1+a3St-1+a5D1+a6D2+a7D3+U1 -(1) Where D1,D2,D3=Quarterly Dummy Variables(Since quarterly data are considered) Here, Supply

Did it stay stable this time? It looks like the restart of the service restored communication. If it reoccurs: >>> Bus 001 Device 011: ID 09ae:3024 Tripp Lite I wonder if this is the same device number after? Something to keep an eye on Thank you, David Zomaya ________________________________ This message is for the addressee's use only. It may contain confidential information. If

Muchas gracias Carlos, Me da error al hacerlo. Mi variable donde quiero que localice las palabras de la lista tiene más de una palabra, no se si puede ser por eso. Gracias El Mar, 22 de Mayo de 2018, 20:15, Carlos Ortega escribió: > Hola, > > Aquí tienes un ejemplo (reproducible)... > > #----------------------- >> # Generar nombres de mujer >> library(randNames)