Question

我有两个df，一个是df1

df1<- structure(list(V1 = structure(c(1L, 2L, 3L, 7L, 5L, 6L, 4L, 9L, 
8L), .Label = c("A0A061ACH4;Q95Q10;Q9U1W6", "A0A061ACL3;Q965I6;O76618", 
"A0A061ACR1;Q2XN02;F5GUA3;Q22498", "A0A061AJJ3;A0A061AEA8", "A0A061AL01", 
"C1P641", "H2FLH3;H2FLH2;A0A061ACT3;A0A061AE24;Q23551-2;Q23551;Q23551-4;Q23551-3;Q23551-5", 
"Q22501;A0A061AE05", "Q86CZ7"), class = "factor")), .Names = "V1", class = "data.frame", row.names = c(NA, 
-9L))

另一个是df2

df2 <- structure(list(V1 = structure(c(1L, 2L, 3L, 6L, 5L, 4L, 8L, 9L, 
7L), .Label = c("A0A061ACH4;Q95Q10;Q9U1W6", "A0A061ACL3;Q965I6;O76618", 
"A0A061ACR1;Q2XN02;F5GUA3;Q22498", "A0A061AJJ3;A0A061AEA8", "A0A061AL01", 
"H2FLH3;H2FLH2;A0A061ACT3;A0A061AE24;Q23551-2;Q23551;Q23551-4;Q23551-3;Q23551-5", 
"Q22501;A0A061AE05", "Q27GQ4", "Q86CZ7"), class = "factor")), .Names = "V1", class = "data.frame", row.names = c(NA, 
-9L))

我想逐行比较这两者。哪条线从df1到df2相似，反之亦然

然后使用df1和df2中的所有唯一行创建一个输出（表示来自这两个df的所有行在一个新的df中）

然后在df1中我们没有的那些行的前面，但是我们在df2中，我们提到零和df2相同。

预期输出可以如下

output<- structure(list(V1 = structure(c(1L, 2L, 3L, 4L, 8L, 6L, 7L, 5L, 
10L, 11L, 9L), .Label = c("", "A0A061ACH4;Q95Q10;Q9U1W6", "A0A061ACL3;Q965I6;O76618", 
"A0A061ACR1;Q2XN02;F5GUA3;Q22498", "A0A061AJJ3;A0A061AEA8", "A0A061AL01", 
"C1P641", "H2FLH3;H2FLH2;A0A061ACT3;A0A061AE24;Q23551-2;Q23551;Q23551-4;Q23551-3;Q23551-5", 
"Q22501;A0A061AE05", "Q27GQ4", "Q86CZ7"), class = "factor"), 
    V2 = structure(c(3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 
    1L), .Label = c("", "0", "df1"), class = "factor"), V3 = structure(c(3L, 
    1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L), .Label = c("", "0", 
    "df2"), class = "factor")), .Names = c("V1", "V2", "V3"), class = "data.frame", row.names = c(NA, 
-11L))

此处Q27GQ4在df1中不存在但存在于df2中，因此在输出中的df1列中我们提到零此处C1P641存在于df1中但不存在于df2中，因此在输出中的df2列中我们提到零。

我很感激任何帮助，因为我是R的新手，我无法弄清楚如何做到这一点

Answer 1

这是一种方法：

strServername

Answer 2

试试这个：

op <- merge(df1,df2, 
      all.x = TRUE,
      all.y = TRUE) 

op$df1 <- 1*(op$V1 %in% df1$V1)

op$df2 <- 1*(op$V1 %in% df2$V1)

> op
                                                                               V1 df1 df2
1                                                        A0A061ACH4;Q95Q10;Q9U1W6   1   1
2                                                        A0A061ACL3;Q965I6;O76618   1   1
3                                                 A0A061ACR1;Q2XN02;F5GUA3;Q22498   1   1
4                                                           A0A061AJJ3;A0A061AEA8   1   1
5                                                                      A0A061AL01   1   1
6                                                                          C1P641   1   0
7  H2FLH3;H2FLH2;A0A061ACT3;A0A061AE24;Q23551-2;Q23551;Q23551-4;Q23551-3;Q23551-5   1   1
8                                                               Q22501;A0A061AE05   1   1
9                                                                          Q86CZ7   1   1
10                                                                         Q27GQ4   0   1

OR

library(dplyr)

op <- merge(df1,df2, 
             all.x = TRUE,
             all.y = TRUE) %>% 
        mutate(df1=1*(V1 %in% df1$V1),
               df2=1*(V1 %in% df2$V1))

以下是您额外问题的答案：

- 知道df1和df2有多少行相似？

sum(df1$V1 %in% df2$V1)

- df1中存在哪些df1并不存在？

sum(!(df1$V1 %in% df2$V1))

- df2中存在多少df2并不存在？

sum(!(df2$V1 %in% df1$V1))

如何将两列字符串进行比较

2 个答案: