Question

我说过这个数据框：

gene0   1   2   3
gene1   0   0   5
gene2   6   8   0
gene3   5   5   5
        0   0   5
        1   2   3

我想将“未命名”列中的数字与基因相关联，以使其具有以下特征：

gene0   1   2   3
gene1   0   0   5
gene2   6   8   0
gene3   5   5   5
gene1   0   0   5
gene0   1   2   3

最好的方法是什么？我需要为此使用linux还是R？

Answer 1

一个dplyr和tidyr选项可以是：

df %>%
 group_by_at(-1) %>%
 fill(V1)

  V1       V2    V3    V4
  <chr> <int> <int> <int>
1 gene0     1     2     3
2 gene1     0     0     5
3 gene2     6     8     0
4 gene3     5     5     5
5 gene1     0     0     5
6 gene0     1     2     3

或者：

df %>%
 group_by(group = group_indices(., !!!select(., -1))) %>%
 fill(V1) %>%
 ungroup() %>%
 select(-group)

样本数据：

df <- read.table(text = "gene0   1   2   3
gene1   0   0   5
                 gene2   6   8   0
                 gene3   5   5   5
                 NA 0   0   5
                 NA 1   2   3",
                 header = FALSE,
                 na.strings = "NA",
                 stringsAsFactors = FALSE)

Answer 2

天真的解决方案

图书馆（tidyverse）

df <- tribble(~col1,~col2,~col3,
                     1,2,3,
                     0,0,5,
                     6,8,0,
                     5,5,5,
                     0,0,5,
                     1,2,3,
                     1,1,1)

df %>% 
  mutate(gene = case_when(col1 == 1 & col2 == 2 &col3 == 3 ~  "gene0",
                          col1 == 0 & col2 == 0 &col3 == 5 ~  "gene1",
                          col1 == 6 & col2 == 8 &col3 == 0 ~  "gene2",
                          col1 == 5 & col2 == 5 &col3 == 5 ~  "gene3",
                          TRUE ~ "unkown_gene"))

另一个更可扩展的选择是创建一个带有基因定义的表格（甚至可以从excel等导入）

df1 <- tribble(~gene,~col1,~col2,~col3,
              'gene0',1,2,3,
              'gene1',0,0,5,
              'gene2',6,8,0,
              'gene3',5,5,5)

并简单地加入新的观察

df %>% 
  left_join(df1)

Answer 3

我们可以使用match中的base R

a1 <- do.call(paste, df1[-1])    
df1$V1 <-  df1$V1[match(a1, unique(a1))]
df1$V1
#[1] "gene0" "gene1" "gene2" "gene3" "gene1" "gene0"

更新

使用OP的数据集

df1 <- read.csv("newest.csv", stringsAsFactors = FALSE)
df1$id[df1$id == ""] <- NA
a1 <- do.call(paste, df1[-1])    
df1$id <-  df1$id[match(a1, unique(a1))]
length(unique(df1$id))
#[1] 621    

head(df1$id, 20)
#[1] "pop13_110" "pop1_2"    "pop16_108" "pop2_10"   "pop2_2"    "pop2_3"    "pop2_4"    "pop2_5"    "pop2_6"    "pop2_7"    "pop2_8"   
#[12] "pop2_9"    "pop2_10"   "pop2_11"   "pop7_81"   "pop2_13"   "pop2_15"   "pop2_15"   "pop2_16"   "pop22_20" 
tail(df1$id, 20)
# [1] "pop22_2"   "pop22_3"   "pop22_4"   "pop22_5"   "pop22_8"   "pop22_9"   "pop13_60"  "pop16_131" "pop23_11"  "pop22_25"  "pop22"    
#[12] "pop22_14"  "pop22_15"  "pop22_32"  "pop22_28"  "pop16_56"  "pop22_18"  "pop9_9"    "pop22_21"  "pop22_22"

数据

df1 <- structure(list(V1 = c("gene0", "gene1", "gene2", "gene3", NA, 
NA), V2 = c(1L, 0L, 6L, 5L, 0L, 1L), V3 = c(2L, 0L, 8L, 5L, 0L, 
2L), V4 = c(3L, 5L, 0L, 5L, 5L, 3L)), class = "data.frame", 
row.names = c(NA, 
-6L))

Answer 4

这里是@akrun的解决方案之外的另一种解决方案，其中base R用于match()，用于从V2到{{1} }

V4

如此

df$V1[which(is.na(df$V1))] <- df$V1[match(data.frame(t(subset(df,is.na(df$V1))[-1])),
                                          data.frame(t(subset(df,is.na(df$V1))[-1])))]

如何将第一列中的名称与重复行相关联？

4 个答案:

更新

数据