根据组条件创建重复数据展开data.frame（3）

本文介绍了根据组条件创建重复数据展开data.frame（3）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 从此SO问题。 示例 data.frame ： df = read.table（text ='ID Day Count Count_group 18 1933 6 15 33 1933 6 15 37 1933 6 15 18 1933 6 15 16 1933 6 15 11 1933 6 15 111 1932 5 9 34 1932 5 9 60 1932 5 9 88 1932 5 9 18 1932 5 9 33 1931 3 4 13 1931 3 4 56 1931 3 4 23 1930 1 1 6 1800 6 12 37 1800 6 12 98 1800 6 12 52 1800 6 12 18 1800 6 12 76 1800 6 12 55 1799 4 6 6 1799 4 6 52 1799 4 6 133 1799 4 6 112 1798 2 2 677 1798 2 2 778 888 4 8 111 888 4 8 88 888 4 8 10 888 4 8 37 887 2 4 26 887 2 4 8 886 1 2 56 885 1 1 22 120 2 6 34 120 2 6 88 119 1 6 99 118 2 5 12 118 2 5 90 117 1 3 22 115 2 2 99 115 2 2'，header = TRUE） 计数 col显示总数 ID 每个的日期和 Count_group c ol显示每个 Day ， Day 1之前的 ID 值的总和， Day -2 ，第3天和 Day -4 。 例如1933 = Count_group 15，因为计数 6（1933）+ 计数 5（1932）+ 计数 3（1931）+ 计数 1（1930）+ 计数 0（1929）。 我需要做的是每个 Count_group 并将它们添加到它，以便每个 Count_group 显示其日，第1天，第-2天，第3天和 Day -4 。 eg Count_group = 15由计数值 Day 1933,1932,1931,1930（和1929不存在于 df ）。所以五天需要包含在 Count_group = 15中。下一个将是 Count_group = 9，组成到1932年，1931年，1930年，1929年和1928年;等等... 期望的输出： ID日数Count_group 18 1933 6 15 33 1933 6 15 37 1933 6 15 18 1933 6 15 16 1933 6 15 11 1933 6 15 111 1932 5 15 34 1932 5 15 60 1932 5 15 88 1932 5 15 18 1932 5 15 33 1931 3 15 13 1931 3 15 56 1931 3 15 23 1930 1 15 111 1932 5 9 34 1932 5 9 60 1932 5 9 88 1932 5 9 18 1932 5 9 33 1931 3 9 13 1931 3 9 56 1931 3 9 23 1930 1 9 33 1931 3 4 13 1931 3 4 56 1931 3 4 23 1930 1 4 23 1930 1 1 6 1800 6 12 37 1800 6 12 98 1800 6 12 52 1800 6 12 18 1800 6 12 76 1800 6 12 55 1799 4 12 6 1799 4 12 5 2 1799 4 12 133 1799 4 12 112 1798 2 12 677 1798 2 12 55 1799 4 6 6 1799 4 6 52 1799 4 6 133 1799 4 6 112 1798 2 6 677 1798 2 6 112 1798 2 2 677 1798 2 2 778 888 4 8 111 888 4 8 88 888 4 8 10 888 4 8 37 887 2 8 26 887 2 8 8 886 1 8 56 885 1 8 37 887 2 4 26 887 2 4 8 886 1 4 56 885 1 4 8 886 1 2 56 885 1 2 56 885 1 1 22 120 2 6 34 120 2 6 88 119 1 6 99 118 2 6 12 118 2 6 90 117 1 6 88 119 1 6 99 118 2 6 12 118 2 6 90 117 1 6 22 115 2 6 99 115 2 6 99 118 2 5 12 118 2 5 90 117 1 5 22 115 2 5 99 115 2 5 90 117 1 3 22 115 2 3 99 115 2 3 22 115 2 2 99 115 2 2 （注意不同的组5天，每一个都被一个空白行分隔，以使它们更清晰） 我有不同的数据框架，分为n天和因此，我想要适应每个人的代码（通过改变它）。 谢谢 #first add grouping变量天< - 5＃分组天数 df $ smalldaygroup< - c（0，cumsum（sapply（2：nrow（df），function（i）df $ Day [i] ！= df $ Day [i-1]）））#individual days df $ bigdaygroup< - c（0，cumsum（sapply（2：nrow（df）），function（i）df $ Day [i ]< df $ Day [i-1] -days + 1）））链接的日期的#block #duplicate da每个大组中的ys df2< - lapply（split（df，df $ bigdaygroup），function（x）{n dayvec< - （max（x $ Day）：min（x $ Day））可能的天数范围 daylog< - dayvec [dayvec％in ％x $ Day]＃范围内的实际天数模式< - data.frame（base = rep（dayvec，each = days））模式$ rep< - sapply（1：nrow ），function（i）pattern $ base [i] + 1-sum（pattern $ base [1：i] == pattern $ base [i]））#indices to repeat pattern $ offset< - match （pattern $ rep，daylog）-match（pattern $ base，daylog）#offsets（稍后使用） pattern< - pattern [（pattern $ base％in％x $ Day）& （pattern $ rep％in％x $ Day），] #remove无效元素 #store模式在列表中作为下一循环所需的偏移 return（list（df = split（x，x $ smalldaygroup） [match（pattern $ rep，daylog）]，pat = pattern））}） 将Count_group更改为添加条目中的上一个值 df2< - lapply df2，function（L）lapply（1：length（L $ df），function（i）{x offset< - L $ pat $ offset #pointer to day to copy Count_group from x $ Count_group< - L $ df [[i-offset [i]]] $ Count_group [1] return（x）}）） df2< - do.call（rbind，unlist（df2，recursive = FALSE））#bind返回 df2 [，5：6]< - NULL #remove分组变量 头（df2,30）#ignore rownames！ ID日数Count_group 01.1 18 1933 6 15 01.2 33 1933 6 15 01.3 37 1933 6 15 01.4 18 1933 6 15 01.5 16 1933 6 15 01.6 11 1933 6 15 02.7 111 1932 5 15 02.8 34 1932 5 15 02.9 60 1932 5 15 02.10 88 1932 5 15 02.11 18 1932 5 15 03.12 33 1931 3 15 03.13 13 1931 3 15 03.14 56 1931 3 15 04 23 1930 1 15 05.7 111 1932 5 9 05.8 34 1932 5 9 05.9 60 1932 5 9 05.10 88 1932 5 9 05.11 18 1932 5 9 06.12 33 1931 3 9 06.13 13 1931 3 9 06.14 56 1931 3 9 07 23 1930 1 9 08.12 33 1931 3 4 0 8.13 13 1931 3 4 08.14 56 1931 3 4 09 23 1930 1 4 010 23 1930 1 1 11.16 6 1800 6 12 Starting from this SO question.Example data.frame:df = read.table(text = 'ID Day Count Count_group 18 1933 6 15 33 1933 6 15 37 1933 6 15 18 1933 6 15 16 1933 6 15 11 1933 6 15 111 1932 5 9 34 1932 5 9 60 1932 5 9 88 1932 5 9 18 1932 5 9 33 1931 3 4 13 1931 3 4 56 1931 3 4 23 1930 1 1 6 1800 6 12 37 1800 6 12 98 1800 6 12 52 1800 6 12 18 1800 6 12 76 1800 6 12 55 1799 4 6 6 1799 4 6 52 1799 4 6 133 1799 4 6 112 1798 2 2 677 1798 2 2 778 888 4 8 111 888 4 8 88 888 4 8 10 888 4 8 37 887 2 4 26 887 2 4 8 886 1 2 56 885 1 1 22 120 2 6 34 120 2 6 88 119 1 6 99 118 2 5 12 118 2 5 90 117 1 3 22 115 2 2 99 115 2 2', header = TRUE)The Count col shows the total number of ID values per each Day and the Count_group col shows the sum of the ID values per each Day, Day - 1, Day -2, Day -3 and Day -4.e.g. 1933 = Count_group 15 because Count 6 (1933) + Count 5 (1932) + Count 3 (1931) + Count 1 (1930) + Count 0 (1929).What I need to do is to create duplicated observations per each Count_group and add them to it in order to show per each Count_group its Day, Day - 1, Day -2, Day -3 and Day -4.e.g. Count_group = 15 is composed by the Count values of Day 1933, 1932, 1931, 1930 (and 1929 not present in the df). So the five days needs to be included in the Count_group = 15. The next one will be Count_group = 9, composed by 1932, 1931, 1930, 1929 and 1928; etc...Desired output:ID Day Count Count_group18 1933 6 1533 1933 6 1537 1933 6 1518 1933 6 1516 1933 6 1511 1933 6 15111 1932 5 1534 1932 5 1560 1932 5 1588 1932 5 1518 1932 5 1533 1931 3 1513 1931 3 1556 1931 3 1523 1930 1 15111 1932 5 934 1932 5 960 1932 5 988 1932 5 918 1932 5 933 1931 3 913 1931 3 956 1931 3 923 1930 1 933 1931 3 413 1931 3 456 1931 3 423 1930 1 423 1930 1 16 1800 6 1237 1800 6 1298 1800 6 1252 1800 6 1218 1800 6 1276 1800 6 1255 1799 4 126 1799 4 1252 1799 4 12133 1799 4 12112 1798 2 12677 1798 2 1255 1799 4 66 1799 4 652 1799 4 6133 1799 4 6112 1798 2 6677 1798 2 6112 1798 2 2677 1798 2 2778 888 4 8111 888 4 888 888 4 810 888 4 837 887 2 826 887 2 88 886 1 856 885 1 837 887 2 426 887 2 48 886 1 456 885 1 48 886 1 256 885 1 256 885 1 122 120 2 634 120 2 688 119 1 699 118 2 612 118 2 690 117 1 688 119 1 699 118 2 612 118 2 690 117 1 622 115 2 699 115 2 699 118 2 512 118 2 590 117 1 522 115 2 599 115 2 590 117 1 322 115 2 399 115 2 322 115 2 299 115 2 2(note that different group of 5 days each one have been separated by a blank line in order to make them clearer)I have got different data.frames which are grouped by n days and therefore I would like to adapt the code (by changing it a little) specifically for each of them.Thanks 解决方案 A generalised version of my previous answer...#first add grouping variablesdays <- 5 #grouping no of daysdf$smalldaygroup <- c(0,cumsum(sapply(2:nrow(df),function(i) df$Day[i]!=df$Day[i-1]))) #individual daysdf$bigdaygroup <- c(0,cumsum(sapply(2:nrow(df),function(i) df$Day[i]<df$Day[i-1]-days+1))) #blocks of linked days#duplicate days in each big groupdf2 <- lapply(split(df,df$bigdaygroup),function(x) { n <- max(x$Day)-min(x$Day)+1 #number of consecutive days in big group dayvec <- (max(x$Day):min(x$Day)) #possible days in range daylog <- dayvec[dayvec %in% x$Day] #actual days in range pattern <- data.frame(base=rep(dayvec,each=days)) pattern$rep <- sapply(1:nrow(pattern),function(i) pattern$base[i]+1-sum(pattern$base[1:i]==pattern$base[i])) #indices to repeat pattern$offset <- match(pattern$rep,daylog)-match(pattern$base,daylog) #offsets (used later) pattern <- pattern[(pattern$base %in% x$Day) & (pattern$rep %in% x$Day),] #remove invalid elements #store pattern in list as offsets needed in next loop return(list(df=split(x,x$smalldaygroup)[match(pattern$rep,daylog)],pat=pattern))})#change the Count_group to previous value in added entriesdf2 <- lapply(df2,function(L) lapply(1:length(L$df),function(i) { x <- L$df[[i]] offset <- L$pat$offset #pointer to day to copy Count_group from x$Count_group <- L$df[[i-offset[i]]]$Count_group[1] return(x)}))df2 <- do.call(rbind,unlist(df2,recursive=FALSE)) #bind back togetherdf2[,5:6] <- NULL #remove grouping variableshead(df2,30) #ignore rownames! ID Day Count Count_group01.1 18 1933 6 1501.2 33 1933 6 1501.3 37 1933 6 1501.4 18 1933 6 1501.5 16 1933 6 1501.6 11 1933 6 1502.7 111 1932 5 1502.8 34 1932 5 1502.9 60 1932 5 1502.10 88 1932 5 1502.11 18 1932 5 1503.12 33 1931 3 1503.13 13 1931 3 1503.14 56 1931 3 1504 23 1930 1 1505.7 111 1932 5 905.8 34 1932 5 905.9 60 1932 5 905.10 88 1932 5 905.11 18 1932 5 906.12 33 1931 3 906.13 13 1931 3 906.14 56 1931 3 907 23 1930 1 908.12 33 1931 3 408.13 13 1931 3 408.14 56 1931 3 409 23 1930 1 4010 23 1930 1 111.16 6 1800 6 12 这篇关于根据组条件创建重复数据展开data.frame（3）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！