Question

我有一个数据框：

df<-structure(list(chrom = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 
                                       2L, 3L, 3L, 4L, 4L, 4L, 4L), .Label = c("1", "2", "3", "4"), class = "factor"), 
                     pos = c(10L, 200L, 134L, 400L, 600L, 1000L, 20L, 33L, 40L, 
                             45L, 50L, 55L, 100L, 123L)), .Names = c("chrom", "pos"), row.names = c(NA, -14L), class = "data.frame")

> head(df)
  chrom  pos
1     1   10
2     1  200
3     1  134
4     1  400
5     1  600
6     1 1000

我想计算样本染色体上的pos[i+1] - pos[i]（chrom）

通过在每个chrom级别上使用for循环，在每行上使用另一个循环，我得到了预期的结果：

 for (c in levels(df$chrom)){
    df_chrom<-filter(df, chrom == c)
    df_chrom<-arrange(df_chrom, df_chrom$pos)

    for (i in 1:nrow(df_chrom)){
      dist<-(df_chrom$pos[i+1] - df_chrom$pos[i])
      logdist<-log10(dist)
      cat(c, i, df_chrom$pos[i], dist, logdist, "\n")
    }
  }

但是，我想将其保存到数据框，并认为lapply或apply是正确的方法。我无法弄清楚如何进行pos[i+1] - pos[i]计算（看作lapply适用于每一行/每列。

任何指针都将不胜感激

以下是我的解决方案的输出：

chrom index pos dist log10dist
1 1 10 124 2.093422 
1 2 134 66 1.819544 
1 3 200 200 2.30103 
1 4 400 200 2.30103 
1 5 600 400 2.60206 
1 6 1000 NA NA 
2 1 20 13 1.113943 
2 2 33 NA NA 
3 1 40 5 0.69897 
3 2 45 NA NA 
4 1 50 5 0.69897 
4 2 55 45 1.653213 
4 3 100 23 1.361728 
4 4 123 NA NA

Answer 1

我们可以使用一组差异来做到这一点。转换＆＃39; data.frame＆＃39;到＆＃39; data.table＆＃39; （setDT(df)），按＆＃39; <＆＃39;，order＆＃39; pos＆＃39;分组，得到＆＃39; pos＆＃39; （diff）和差异的log

library(data.table)
setDT(df)[order(pos), {v1 <- diff(pos)
                   .(index = seq_len(.N), pos = pos, 
                      dist = c(v1, NA), logdiff = c(log10(v1), NA))}

               , by = chrom]
#    chrom index  pos dist  logdiff
# 1:     1     1   10  124 2.093422
# 2:     1     2  134   66 1.819544
# 3:     1     3  200  200 2.301030
# 4:     1     4  400  200 2.301030
# 5:     1     5  600  400 2.602060
# 6:     1     6 1000   NA       NA
# 7:     2     1   20   13 1.113943
# 8:     2     2   33   NA       NA
# 9:     3     1   40    5 0.698970
#10:     3     2   45   NA       NA
#11:     4     1   50    5 0.698970
#12:     4     2   55   45 1.653213
#13:     4     3  100   23 1.361728
#14:     4     4  123   NA       NA

运行OP代码后，打印的输出为

#1 1 10 124 2.093422 
#1 2 134 66 1.819544 
#1 3 200 200 2.30103 
#1 4 400 200 2.30103 
#1 5 600 400 2.60206 
#1 6 1000 NA NA 
#2 1 20 13 1.113943 
#2 2 33 NA NA 
#3 1 40 5 0.69897 
#3 2 45 NA NA 
#4 1 50 5 0.69897 
#4 2 55 45 1.653213 
#4 3 100 23 1.361728 
#4 4 123 NA NA

Answer 2

我们将df拆分为df$chrom（请注意，我们会在拆分前重新排序df和df$chrom）。然后我们使用a遍历每个子组（在此示例中，子组称为lapply）。在每个子组的pos列上，我们计算连续元素的差异（diff）并取log10。由于diff会将元素数量减少1，因此我们会在末尾添加NA。最后，我们使用rbind do.call将所有子组do.call(rbind, lapply(split(df[order(df$chrom, df$pos),], df$chrom[order(df$chrom, df$pos)]), function(a) data.frame(a, dist = c(log10(diff(a$pos)), NA)))) # chrom pos dist #1.1 1 10 2.093422 #1.3 1 134 1.819544 #1.2 1 200 2.301030 #1.4 1 400 2.301030 #1.5 1 600 2.602060 #1.6 1 1000 NA #2.7 2 20 1.113943 #2.8 2 33 NA #3.9 3 40 0.698970 #3.10 3 45 NA #4.11 4 50 0.698970 #4.12 4 55 1.653213 #4.13 4 100 1.361728 #4.14 4 123 NA组合在一起。

SELECT 
    bucket_index, user_type, medium, AVG(productivity) as avg_prod, 
    MAX(productivity) as max_prod, MIN(productivity) as min_prod
FROM
    (SELECT 
            floor(t1.total_docs_read/100) as bucket_index, 
            t2.user_type as user_type, t2.medium as medium, 
            t2.productivity as productivity
    FROM
        (SELECT 
            on_date::date as day, sum(docs_read) as total_docs_read 
        FROM work 
        GROUP BY day) as t1,
        (SELECT 
            on_date::date as day, user_type, medium, 
            sum(docs_read)/count(distinct(user)) as productivity
        FROM work
        GROUP BY day, user_type, medium) as t2
    WHERE t1.day=t2.day) as t3
GROUP BY bucket_index, user_type, medium

lapply替代for循环以附加到数据框

2 个答案: