There may well be neater ways to do this, but if you have only a limited number of zeros in any run, this is probably as quick as any. Suppose your data frame is 'dat':
> fixCol <- function(x) { y <- x n <- length(x) while(any(zx <- x == 0)) { y <- c(NA, y[-n]) x[zx] <- y[zx] } x } > dat <- transform(dat, x = fixCol(x), y = fixCol(y), z = fixCol(z)) If you have a great number of columns to fix up like this, a more succinct way would be something like > dat[, 3:5] <- lapply(dat[, 3:5], fixCol) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of saikat sarkar Sent: Sunday, 6 April 2008 11:05 PM To: r-help@r-project.org Subject: [R] how to manupute data frame with conditions fill cell with previous value if next cell is zero Dear R Experts, This is the 2nd time in the chat room. Its a great place to get help from R experts. I have a data frame problem, it contains thousands of data. part of it, I am giving for explaining the problem date day x y z 82 1989-04-28 Fri 2118.0 2418.80 33713 83 1989-05-01 Mon 0.0 2414.96 33793 84 1989-05-02 Tue 2103.1 2402.86 33955 85 1989-05-03 Wed 2105.7 2393.70 0 86 1989-05-04 Thu 2119.0 2384.90 0 87 1989-05-05 Fri 2132.8 2381.96 0 103 1989-05-29 Mon 0.0 0.00 34161 127 1989-06-30 Fri 2151.0 2440.06 32949 128 1989-07-03 Mon 2165.6 2452.77 33236 129 1989-07-04 Tue 2174.4 0 35376 130 1989-07-05 Wed 2162.9 2456.56 33310 167 1989-08-25 Fri 2397.4 2732.36 34740 168 1989-08-28 Mon 0.0 2743.36 34607 169 1989-08-29 Tue 2380.8 2726.63 34688 170 1989-08-30 Wed 2381.3 2728.15 34472 171 1989-08-31 Thu 2387.9 2737.27 34431 172 1989-09-01 Fri 2407.5 2752.09 34348 173 1989-09-04 Mon 2419.2 0.00 34484 174 1989-09-05 Tue 2426.0 2744.68 34442 175 1989-09-06 Wed 2390.8 2719.79 34271 176 1989-09-07 Thu 2415.9 2706.88 34153 177 1989-09-08 Fri 2423.9 2709.54 34116 178 1989-09-11 Mon 2400.6 2704.41 34114 179 1989-09-12 Tue 2397.6 2707.26 34333 180 1989-09-13 Wed 2401.5 2679.52 34287 181 1989-09-14 Thu 2382.0 2664.89 34402 182 1989-09-15 Fri 2366.5 2674.58 0 183 1989-09-18 Mon 2373.8 2687.50 34473 195 1989-10-04 Wed 2312.1 2771.09 35383 196 1989-10-05 Thu 2281.6 2773.56 35523 197 1989-10-06 Fri 2277.5 2785.52 35209 198 1989-10-09 Mon 2247.0 2791.41 35376 199 1989-10-10 Tue 2218.8 2785.33 0 200 1989-10-11 Wed 2218.8 2773.36 35240 In this data frame, I need to replace cells which is zero with previous value. For example 82 1989-04-28 Fri 2118.0 2418.80 33713 83 1989-05-01 Mon 0.0 2414.96 33793 in 2nd line 0.0 should be replaced by 2118.0 Another Example 84 1989-05-02 Tue 2103.1 2402.86 33955 85 1989-05-03 Wed 2105.7 2393.70 0 86 1989-05-04 Thu 2119.0 2384.90 0 87 1989-05-05 Fri 2132.8 2381.96 0 3 lines filled by 33955 value. Another example 198 1989-10-09 Mon 2247.0 2791.41 35376 199 1989-10-10 Tue 2218.8 2785.33 0 200 1989-10-11 Wed 2218.8 2773.36 35240 in 2nd line filled by zero should be replaced by value 35376 Please help me. Thanking you saikat -- View this message in context: http://www.nabble.com/how-to-manupute-data-frame-with-conditions-fill-ce ll-with-previous-value-if-next-cell-is-zero-tp16524107p16524107.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.