Question

我有这样的数据

    df1 <- structure(list(V1 = structure(c(1L, 2L, 3L, 5L, 4L), .Label = c("A0A061ACH4;Q95Q10;Q9U1W6", 
"A0A061ACL3;Q965I6;O76618", "A0A061ACR1;Q2XN02;F5GUA3;Q22498", 
"A0A061AL01", "H2FLH3;H2FLH2;A0A061ACT3;A0A061AE24;Q23551-2;Q23551;Q23551-4;Q23551-3;Q23551-5"
), class = "factor"), V2 = c(1L, 5L, 100L, 645L, 11L), V3 = c(67L, 
10L, 33L, 99L, 10L), V4 = c(7L, 16L, 0L, 1L, 5L)), .Names = c("V1", 
"V2", "V3", "V4"), class = "data.frame", row.names = c(NA, -5L
))

我想要的是将所有字符串保持在彼此之下并复制每行前面的值，在该行的每个字符串前面
预期的输出应该是这样的

output <- structure(list(V1 = structure(c(1L, 18L, 20L, 2L, 19L, 19L, 10L, 
3L, 17L, 7L, 11L, 9L, 8L, 4L, 5L, 13L, 12L, 15L, 14L, 16L, 6L
), .Label = c("A0A061ACH4", "A0A061ACL3", "A0A061ACR1", "A0A061ACT3", 
"A0A061AE24", "A0A061AL01", "F5GUA3", "H2FLH2", "H2FLH3", "O76618", 
"Q22498", "Q23551", "Q23551-2", "Q23551-3", "Q23551-4", "Q23551-5", 
"Q2XN02", "Q95Q10", "Q965I6", "Q9U1W6"), class = "factor"), V2 = c(1L, 
1L, 1L, 5L, 5L, 5L, 5L, 100L, 100L, 100L, 100L, 645L, 645L, 645L, 
645L, 645L, 645L, 645L, 645L, 645L, 11L), V3 = c(67L, 67L, 67L, 
10L, 10L, 10L, 10L, 33L, 33L, 33L, 33L, 99L, 99L, 99L, 99L, 99L, 
99L, 99L, 99L, 99L, 10L), V4 = c(7L, 7L, 7L, 16L, 16L, 16L, 16L, 
0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 5L)), .Names = c("V1", 
"V2", "V3", "V4"), class = "data.frame", row.names = c(NA, -21L
))

例如，如果我们查看df1第一行，它看起来像这样

A0A061ACH4;Q95Q10;Q9U1W6    1   67  7

然后我分开琴弦;并将它们放在彼此之下并复制它们前面的值。所以对于df1的第一行，我会有

  A0A061ACH4    1   67  7
    Q95Q10      1   67  7
    Q9U1W6      1   67  7

Answer 1

可以使用separate_rows

中的tidyr来完成此操作

library(tidyr)
separate_rows(df1, V1, sep=";")

或者使用cSplit选项<{1}}

long

或使用其他library(splitstackshape) cSplit(df1, 'V1', ';', 'long')选项

base R

如果我们只需要第一个lst <- strsplit(as.character(df1$V1), ";") cbind(V1= unlist(lst), df1[rep(1:nrow(df1), lengths(lst)),-1])之前的子字符串，请使用;匹配模式sub后跟字符，直到字符串结尾（;）和将其替换为空白（.*）。

""

Answer 2

这是＆＃34;基地＆＃34;方式。

# for each line...
out <- do.call(rbind, apply(df1, MARGIN = 1, FUN = function(x) {
  # split by ; and...
  do.call(rbind, sapply(unlist(strsplit(x[1], ";")), FUN = function(y, y2) {
    # ... append the rest of the columns to individual element
    c(y, y2)
  }, simplify = FALSE, y2 = x[2:4]))
  }))

rownames(out) <- NULL

out <- as.data.frame(out)

out

           V1  V2 V3 V4
1  A0A061ACH4   1 67  7
2      Q95Q10   1 67  7
3      Q9U1W6   1 67  7
4  A0A061ACL3   5 10 16
5      Q965I6   5 10 16
6      O76618   5 10 16
7  A0A061ACR1 100 33  0
8      Q2XN02 100 33  0
9      F5GUA3 100 33  0
10     Q22498 100 33  0
11     H2FLH3 645 99  1
12     H2FLH2 645 99  1
13 A0A061ACT3 645 99  1
14 A0A061AE24 645 99  1
15   Q23551-2 645 99  1
16     Q23551 645 99  1
17   Q23551-4 645 99  1
18   Q23551-3 645 99  1
19   Q23551-5 645 99  1
20 A0A061AL01  11 10  5

如何保持每行的第一个字符串

2 个答案: