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Warren Young wrote: > On Feb 15, 2019, at 1:14 AM, Turritopsis Dohrnii Teo En Ming > <tdteoenming at gmail.com> wrote: > >> >>>> Could you recommend affordable and reliable cloud storage for 50 TB >>>> of data? >>> >> My budget is around USD$50 per year. >> > > The cheapest RAID-friendly drives we?re buying these days are

On Feb 15, 2019, at 1:14 AM, Turritopsis Dohrnii Teo En Ming <tdteoenming at gmail.com> wrote: > >>> Could you recommend affordable and reliable cloud storage for 50 TB of data? >> > My budget is around USD$50 per year. The cheapest RAID-friendly drives we?re buying these days are about US $37/TB in low quantities. A big data warehouser will be getting a substantial

Hi! I am doing a research about the prices of SIP telephones. If someone can tell me which one are the cheapest and have an acceptable quality... it will be very kind. Best Regards, Mireia

On Fri, Feb 15, 2019 at 4:10 PM Walter H. <Walter.H at mathemainzel.info> wrote: > > On 15.02.2019 06:29, Turritopsis Dohrnii Teo En Ming wrote: > > Hi, > > > > Could you recommend affordable and reliable cloud storage for 50 TB of data? > whats your budget? > > and 50 TB = 50 000 GB is a big amount which isn't this cheap ... Hi Walter H, My budget is

Dear group, Here is my data frame: position100415 <- structure(list(DESCRIPTION = structure(1:9, .Label = c(" SUGAR NO.11 Jul/10 ", " SUGAR NO.11 May/10 ", "CORN May/10 ", "COTTON NO.2 Jul/10 ", "CRUDE OIL miNY May/10 ", "ROBUSTA COFFEE (10) Jul/10 ", "SILVER May/10 ", "SOYBEANS Jul/10 ", "WHEAT May/10

Can someone verify for me if the for loop below is really calculating the nonzero min for each row of a matrix? I have a bug somewhere in the is section of code. My first guess is how I am find the the nonzero min of each row of my matrix. The overall idea is to make sure I am investing all of my money, i.e. new.set is a set of indicator variables for each stock for a particular portfolio, i.e.

I have continous oom&panic situation unresolved. I am not sure system fills up all the ram (36GB). Why this system triggered this oom situation? Is it about some other memory? highmem? lowmem? stack size? Best Regards, Kernel 3.10.24 Dec 27 09:19:05 2013 kernel: : [277622.359064] squid invoked oom-killer: gfp_mask=0x42d0, order=3, oom_score_adj=0 Dec 27 09:19:05 2013 kernel: :

Hello, I''m running into some trouble with what appear on the surface to be OOM issues in Dom0, but I''m not seeing any other evidence. This typically happens during periods of high I/O, and has occurred during RAID initial sync, and mkfs.ext4ing (as a test, no intention to keep ext4 on this array). I''ve found some older posts citing very similar circumstances, however

Dear Everybody, I am telling you the whole story, perhaps you can give me a better idea. Almost 3 years ago I purchased Arima HDAMA rev. G motherboard with integrated Silicon Image Sil3114 Serial ATA 4 Ports SATA I adapter. ( http://www.arimacorp.com/ViewProduct.asp?View=86 ) . BIOS version is 2.13. My intention is to make a file server, using linux software raid, so I bought 2 new shiny 1TB

Dear group, I know this issue has been already covered, and before you reply I must say I have read the R-FAQ and search the mailing list archive. I still can't manage to change my factor to numeric as I couldn't find any clear answer. Here is my df : Pose1 <- structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L, 8L), .Label = c(" SUGAR NO.11 May/10 ", "COTTON

Hello All, Anyone have seen this before? Did samba_spnupdate really caused the crash??? [ 49.753564] block drbd1: drbd_sync_handshake: [ 49.753571] block drbd1: self BB16E125AF60AEDC:0000000000000000:30D97136FB1DA7A3:30D87136FB1DA7A3 bits:0 flags:0 [ 49.753576] block drbd1: peer 6365B5AFF049F16D:BB16E125AF60AEDD:30D97136FB1DA7A2:30D87136FB1DA7A3 bits:1 flags:0 [ 49.753580] block drbd1:

My mistake sir. I was literally engrossed in my stupid logic, and while doing so, overlooked the simple and very effective solution you had offered. Sorry once again sir and will certainly try to be very careful in future. Thanks again and have a great weekend sir. Regards Amelia --- On Fri, 7/1/11, Petr PIKAL <petr.pikal@precheza.cz> wrote: From: Petr PIKAL

Found this solution. It is maybe not the most elegant way, but it does the job. > a=as.data.frame(substr(lme$DESCRIPTION,1,14)) > colnames(a)=c("DESCRIPTION") > lme=as.data.frame(c(a,lme[,2:3])) > lme DESCRIPTION CLOSING.PRICE POSITION 1 PRIMARY NICKEL 25,755.7100 0 2 PRIMARY NICKEL 25,760.8600 0 3 PRM HGH GD ALU 2,415.9000 0

Not sure if it''s anything interesting - I had the following entry in dmesg a few days ago, on a server with 32 GB RAM. The system is still working fine. [1878432.675210] btrfs-qgroup-re: page allocation failure: order:5, mode:0x104050 [1878432.675319] CPU: 5 PID: 22251 Comm: btrfs-qgroup-re Not tainted 3.11.0-rc7 #2 [1878432.675417] Hardware name: System manufacturer System Product

Dear R forum, (Pl note this is not a finance problem) I have two data.frames as currency_df = data.frame(current_date = c("3/4/2013", "3/4/2013", "3/4/2013", "3/4/2013"), issue_date = c("27/11/2012", "9/12/2012", "14/01/2013", "28/02/2013"), maturity_date = c("27/04/2013", "3/5/2013",

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 sir, I am extremely sorry for messing up the logic asking for help w.r.t. my earlier mails   I have tried to explain below what I am looking for.     I have a database (say, currency_rates) storing datewise currency exchange rates with some base currency XYZ.   currency_rates <- data.frame(date = c("12/31/2010", "12/30/2010", "12/29/2010",

Dear R forum helpers, I am extremely sorry for the receipt of my incomplete mail yesterday. There was connectivity problem at my end and so I chose to send the mail through my cell, only to realize today about the way mail has been transmitted. I am again sending my complete mail through regular channel and sincerely apologize for the inconvenience caused. ## Here is my actual mail Dear R

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

Hi R users, I have a dataframe in the below format xyz 01/03/2007 15.25 USD xyz 01/04/2007 15.32 USD xyz 01/02/2008 23.22 USD abc 01/03/2007 45.2 EUR abc 01/04/2007 45.00 EUR