search for: zinc

...RD LEAD USD 23/06/10 ", "STANDARD LEAD USD 23/06/10 ", "STANDARD LEAD USD 23/06/10 ", "STANDARD LEAD USD 25/06/10 ", "STANDARD LEAD USD 25/06/10 ", "STANDARD LEAD USD 29/06/10 ", "STANDARD LEAD USD 29/06/10 ", "SPCL HIGH GRADE ZINC USD 06/07/10 ", "SPCL HIGH GRADE ZINC USD 06/07/10 ", "SPCL HIGH GRADE ZINC USD 06/07/10 ", "SPCL HIGH GRADE ZINC USD 06/07/10 ", "SPCL HIGH GRADE ZINC USD 06/07/10 ", "SPCL HIGH GRADE ZINC USD 06/07/10 ", "SPCL HIGH GRADE ZINC USD 06/0...

...ta.frame, "lme": > lme DESCRIPTION CLOSING.PRICE POSITION 4 PRIMARY NICKEL USD 04/06/10 25,755.7100 0 5 PRIMARY NICKEL USD 10/06/10 25,760.8600 0 6 PRM HGH GD ALUMINIUM USD 09/07/10 2,415.9000 0 8 SPCL HIGH GRADE ZINC USD 06/07/10 2,420.1000 -1 9 SPCL HIGH GRADE ZINC USD 07/07/10 2,420.4100 -1 10 SPCL HIGH GRADE ZINC USD 08/07/10 2,420.7300 1 11 SPCL HIGH GRADE ZINC USD 09/07/10 2,421.0500 1 12 SPCL HIGH GRADE ZINC USD 13/04/10 2,388.4300 0 13 SPCL HIGH GRADE...

...e(c(1L, 2L, 3L, 4L, 5L, 8L), .Label = c(" SUGAR NO.11 May/10 ", "COTTON NO.2 May/10 ", "PLATINUM Jul/10 ", "ROBUSTA COFFEE (10) May/10 ", "WHEAT May/10 ", "PRIMARY NICKEL USD", "PRM HGH GD ALUMINIUM USD", "SPCL HIGH GRADE ZINC USD", "STANDARD LEAD USD"), class = "factor"), POSITION = c(5, 3, -1, 15, 4, 2), SETTLEMENT = structure(c(3L, 5L, 2L, 1L, 4L, 8L), .Label = c("1,353.0000", "1,739.4000", "16.5400", "467.7500", "78.1300", "25,760.8600...

...ass about analysis of variance using R, > I was doing a one-way analysis for the two data-sets given below in the > R-class. I got a typical error in "emmeans" package, please help: > > Data-set-1: > -------------- > Medley and Clements (1998) investigated the impact of zinc contamination > (and other heavy metals) on the diversity of diatom species in the USA > Rocky Mountains. The diversity of diatoms (number of species) and degree of > zinc contamination (categorized as either of high, medium, low or natural > background level) were recorded from between...

...too. Dr. A. K. Singh On 09-Jan-2018 8:18 PM, "Sal Mangiafico" <salvatore.s.mangiafico at gmail.com> wrote: One way to avoid this error is to create the aov without using the with function, but instead use the data= option in the aov function. That is, medley2 = aov(diversity ~ zinc, data=medley.clementis) emmeans::emmeans(medley2, "zinc") You can see the difference in the calls: medley2$call medley.clementis.aov$call This works for the other data set as well, e.g. keough2 = aov(serpulid.ln ~ biofilm, data=keough.raimondi.ln) ~ Sal Mangiafico On 1/8/2018 4:44 P...

Dear all can you help me? i have this problem: i have a dataset in a text file in a matrix of 3 columns: x, y, z where x and y are the coordinates and Z are the mesurements. How can i do a variogram with R?

...r, India. While teaching in class about analysis of variance using R, I was doing a one-way analysis for the two data-sets given below in the R-class. I got a typical error in "emmeans" package, please help: Data-set-1: -------------- Medley and Clements (1998) investigated the impact of zinc contamination (and other heavy metals) on the diversity of diatom species in the USA Rocky Mountains. The diversity of diatoms (number of species) and degree of zinc contamination (categorized as either of high, medium, low or natural background level) were recorded from between four and six sampli...

Dear Manager, Wish you have a nice day! We are the manufacture and exporter of Zinc ingots. Should this item be of interest to you, please feel free to contact me. Thanks, Sunshine Hebei Xingtai chemical Technology Co. ,Ltd Web:http://www.hbxthg.com -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.alioth.debian.org/pipermail/pk...

Hi; I cannot find in the R html documentation a way to evaluate cross-correlation in 2D data sets. I would like to evaluate cross-correlation in a series of moving windows between two maps. i,e, specify several windows inside the complete 2D spatial matrixes and for each one ofthese windows evaluate the 2D cross-correlation (commonly conducted in the spectral domain). Thanks in advance and best

...e 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 2 5 SPCL HIGH GRADE ZINC USD Jul/10 -2464.0 -1 6 STANDARD LEAD USD Aug/10 -118.0 0 For each df$DESCRIPTION element containing "USD" I want to remove the last part of it(i.e Jul/10, or Aug/10...). I was thinking of something like that: >Df$DESCRIPTION<-gsub("USD","new n...

...;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) Jul/10", "SILVER May/10", "SOYBEANS Jul/10", "SPCL HIGH GRADE ZINC USD", "STANDARD LEAD USD", "SUGAR NO.11 Jul/10", "SUGAR NO.11 May/10", "WHEAT Jul/10", "WHEAT May/10"), class = "factor"), pl = c(4.75000000000003, -14, -1.5, 7.75, 3.74999999999999, 5.22500000000005, 21.6999999999998, -0.0...

...uot;, "CORN", "CORN", "CORN", "CORN", "SOYBEANS", "SOYBEANS", "SOYBEANS", "SOYBEANS", "SOYBEANS", "SOYBEANS", "STANDARD LEAD USD", "STANDARD LEAD USD", "SPCL HIGH GRADE ZINC USD", "SPCL HIGH GRADE ZINC USD", "SPCL HIGH GRADE ZINC USD", "ROBUSTA COFFEE (10)", "ROBUSTA COFFEE (10)", "ROBUSTA COFFEE (10)", "ROBUSTA COFFEE (10)", "ROBUSTA COFFEE (10)", "ROBUSTA COFFEE (10)", "RO...

...TION = 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 NO.11 Oct/10", "WHEAT Sep/10"), prix = c(-168.3, -1.60000000000002, -773.75, -78.75, -1168.3, -0.0919999999999996, -470.75, 1758.5, -975.25, 1964, -19.09, -605.75), pos = c(-1, 0, -2, -1, -1, 0, -1, 1, -1, 1, -1, -1), PL = c(-12.03, -31.68, -43.2, -70.49, -11.8...

...of brute force looping approach I have drawn a blank. Would somebody please suggest something? Thanks Example. mylist <- list(aa=c("cat","peach" ), bb=c("dog", "apple", "iron"), cc = c("rabbit", "orange", "zinc", "silk")) myfun <- function(dff) dff[1] vect1 <- unlist(lapply(mylist, myfun)) Looking for the perfect gift? Give the gift of Flickr!

...Yes, I have tried testparam. All is OK . Note that I can write > anything for unix charset and testparm does not complain. > > Load smb config files from /etc/samba/smb.conf > rlimit_max: increasing rlimit_max (1024) to minimum Windows limit > (16384) Processing section "[zinc1]" > Loaded services file OK. > Server role: ROLE_DOMAIN_MEMBER > > > The systeme locale : > > [root at zinc samba]# locale > LANG=en_CA.iso88591 > LC_CTYPE="en_CA.iso88591" > LC_NUMERIC="en_CA.iso88591" > LC_TIME="en_CA.iso88591&quo...

...t. Description of the experiment: After the calcination at different temperatures my calcinated sludge are submeted to the lixiviation test. In my tables i show the concentration of the some elements in the extract phase. The results: Method A (calcination at 1100 C) Chromium Nickel Aluminum Zinc 2.0 3.0 2.7 2.3 2.1 3.1 2.8 2.2 1.9 3.2 2.6 2.3 2.2 3.1 2.8 2.2 Method B (calcination at 1500 C) Chromium Nickel Aluminum Zinc 2.1 3.1 2.6 2.2 2.3 2.9 3.0 2.5 1.9 3.0 2.6 2.2 2.2 3.0 2.9 2.1 2.4 3.1 2.7 2.2 2.3 3.0 2.6 2.1 1.8 2.8 2.7 2.0 Assumptions: lambda = |W|/(|W + B|) W is the pooled c...

Hello, I am learning how to use functions, but I'm running into a roadblock. I would like my function to do two things: 1) convert an object to a dataframe, 2) and then subset the dataframe. Both of these commands work fine outside the function, but I would like to wrap them in a function so I can apply the code iteratively to many such objects. Here's what I wrote, but it doesn't

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, -25)), .Names = c("DESCRIPTION", "prix", "quantity", "fees"), row.names = c(NA, -...

...L 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", "PRIMARY NICKEL USD", "PRM HGH GD ALUMINIUM USD", "SPCL HIGH GRADE ZINC USD", "STANDARD LEAD USD"), class = "factor"), POSITION = c(2, -3, -2, -2, -18, -3, -5, 1, -5, 1), prix = c(702.9, -244.77, -166.48, -2273.8, -24876, -2985, -80.9, 15.95, -2512.5, 2423)), .Names = c("DESCRIPTION", "POSITION", "prix&qu...

Hi, I would know how to plot two variograms on a same graph. I can plot one by one but I would draw both on the same one. Is it possible? Do i need any special package? Thanks! Cordialement Damien Landais