如何逐行比较两个数据帧？答案

【问题标题】：How to compare two data frames line by line?如何逐行比较两个数据帧？
【发布时间】：2021-07-13 03:40:52
【问题描述】：

大约一个月前，我发布了原始问题，我需要逐行比较两个数据帧并标记 df2（第二个文件）中与 df1（第一个文件）不匹配的行。解决方案是使用反连接。这很有效，直到我添加了一个带有文本字符串的附加列。我还需要在比较中包含该列并检测文本字符串的哪些记录不匹配。

附件是示例数据框。我需要将 df2 与 df1 进行比较，并显示 df2 中的哪些行与 df1 不匹配。我可以使用 R 中的反连接来显示哪些行不匹配，但是当我在行中有文本字符串时它不起作用。

df1

Product basecode    A   B   C   D   E   F
Tractor A810     382    512 363 553 530 A dog ran fast
Tractor A773     222    155 650 278 215 A dog ran fast
Tractor A203     382    512 363 553 530 A dog ran fast
Tractor A329     332    459 251 341 475 A dog ran fast
Combine B244     244    714 467 122 340 A dog ran fast
Combine B302     257    758 230 704 715 A dog ran fast
Combine B681     670    626 572 795 323 A dog ran fast
Combine B514     768    510 546 542 582 A dog ran fast
Sprayer C850     553    624 557 660 337 A dog ran fast
Sprayer C202     561    733 443 107 526 A dog ran fast
Sprayer C619     256    226 257 770 633 A dog ran fast
Sprayer C292     256    226 257 770 633 A dog ran fast
SPFH    D126     323    597 647 159 317 A dog ran fast
SPFH    D307     711    535 323 793 769 A dog ran fast
SPFH    D355     155    744 772 689 509 A dog ran fast
SPFH    D893     155    744 772 689 509 A dog ran fast

df2

Product basecode    A   B   C   D   E   F
Tractor A810     382    512 363 553 530 A dog ran fast
Tractor A773     222    155 650 278 215 A dog ran fast
Tractor A203     382    512 363 553 530 A dog ran fast
Tractor A329     332    459 251 341 475 A dog ran fast
Combine B 244    244    714 467 122 340 A dog ran fast
Combine B302     257    758 230 704 715 A dog ran fast
Combine B681     670    626 572 795 323 A dog ran fast
Combine B514     768    510 546 542 582 A dog ran fast
Sprayer C850     553    624 557 660 337 A dog ran fast
Sprayer C202     561    733 443 107 526 A dog ran fast
Sprayer C619     256    226 257 770 633 A dog ran fast
Sprayer C292     1  1   1   1   1   A dog ran fast
SPFH    D126     323    597 647 159 317 A dog ran fast
SPFH    D307     711    535 323 793 769 A dog ran fast
SPFH    D355     155    744 772 689 509 A dog ran fast
SPFH    D893     1  1   1   1   1   A dog ran fast
Tractor A810     491    765 457 249 641 A dog ran fast
Tractor A773     222    155 650 278 215 A dog ran fast
Tractor A203     382    512 363 553 530 A dog ran fast
Tractor A329     332    459 251 341 475 A dog ran fast
Combine B 244    244    714 467 122 340 A dog ran fast
Combine B302     257    758 230 704 715 A cat ran slow
Combine B681     670    626 572 795 323 cat
Combine B514     768    510 546 542 582 A dog ran fast
Sprayer C850     553    624 557 660 337 A dog ran fast
Sprayer C202     561    733 443 107 526 A dog ran fast
Sprayer C619     256    226 257 770 633 A dog ran fast

代码

# add id to identify which rows are not matching
df2 <- df2 %>% mutate(id = basecode)

df_unmatch <- anti_join(df2, df1)

# list of non-match are the ids of df_unmatch
df_unmatch$id

数据

#structure(list(Product = c("Tractor", "Tractor", "Tractor", "Tractor", 
"Combine", "Combine", "Combine", "Combine", "Sprayer", "Sprayer", 
"Sprayer", "Sprayer", "SPFH", "SPFH", "SPFH", "SPFH", "Tractor", 
"Tractor", "Tractor", "Tractor", "Combine", "Combine", "Combine", 
"Combine", "Sprayer", "Sprayer", "Sprayer"), basecode = c("A810", 
"A773", "A203", "A329", "B 244", "B302", "B681", "B514", "C850", 
"C202", "C619", "C292", "D126", "D307", "D355", "D893", "A810", 
"A773", "A203", "A329", "B 244", "B302", "B681", "B514", "C850", 
"C202", "C619"), A = c(382, 222, 382, 332, 244, 257, 670, 768, 
553, 561, 256, 1, 323, 711, 155, 1, 491, 222, 382, 332, 244, 
257, 670, 768, 553, 561, 256), B = c(512, 155, 512, 459, 714, 
758, 626, 510, 624, 733, 226, 1, 597, 535, 744, 1, 765, 155, 
512, 459, 714, 758, 626, 510, 624, 733, 226), C = c(363, 650, 
363, 251, 467, 230, 572, 546, 557, 443, 257, 1, 647, 323, 772, 
1, 457, 650, 363, 251, 467, 230, 572, 546, 557, 443, 257), D = c(553, 
278, 553, 341, 122, 704, 795, 542, 660, 107, 770, 1, 159, 793, 
689, 1, 249, 278, 553, 341, 122, 704, 795, 542, 660, 107, 770
), E = c(530, 215, 530, 475, 340, 715, 323, 582, 337, 526, 633, 
1, 317, 769, 509, 1, 641, 215, 530, 475, 340, 715, 323, 582, 
337, 526, 633), F = c("A dog ran fast", "A dog ran fast", "A dog ran fast", 
"A dog ran fast", "A dog ran fast", "A dog ran fast", "A dog ran fast", 
"A dog ran fast", "A dog ran fast", "A dog ran fast", "A dog ran fast", 
"A dog ran fast", "A dog ran fast", "A dog ran fast", "A dog ran fast", 
"A dog ran fast", "A dog ran fast", "A dog ran fast", "A dog ran fast", 
"A dog ran fast", "A dog ran fast", "A dog ran fast", "cat", 
"A dog ran fast", "A dog ran fast", "A dog ran fast", "A dog ran fast"
)), row.names = c(NA, -27L), class = c("tbl_df", "tbl", "data.frame"
))

【问题讨论】：

您能否提供可重现格式的数据并显示共享数据的预期输出？
您好！我很想......但由于某种原因无法弄清楚如何去做。你能提供说明吗？
您可以在控制台中运行dput(df1) 和dput(df2)，复制输出并编辑您的帖子以将其包含在此处。如果数据框很大，请使用 dput(head(df1, 10)) 提供前 10 行。这个帖子有更多信息stackoverflow.com/questions/5963269
请显示您常用的反连接例程的代码，如本例中所尝试的那样
# 添加 id 以识别哪些行不匹配 df2 % mutate(id = basecode) df_unmatch

标签： r string-comparison

【解决方案1】：

它确实有效，除非你有一些特别的期望（见 Limey 的评论）。您提供的两个文件实际上是相同的（请参阅 MonJeanJean 的评论），所以让我们从创建不匹配的行开始：

df1$F <- "A dog ran faster" ## df2 has "cat" somewhere
df2$A[16] <- 155
anti_join(df2, df1)

# A tibble: 2 x 8
  Product basecode     A     B     C     D     E F             
  <chr>   <chr>    <dbl> <dbl> <dbl> <dbl> <dbl> <chr>         
1 SPFH    D893         1     1     1     1     1 A dog ran fast
2 Combine B681       670   626   572   795   323 cat

你期待什么结果？

【讨论】：

嗯....这正是我所期望的，所以谢谢！但我无法让它发挥作用，所以我发布了这个问题。经过一番调查，我确定有些列是数字的，有些是字符的。我将两个数据帧都转换为字符，这很有效。非常感谢您的帮助！

【解决方案2】：

不是最易读的解决方案，但如果你有很多行可能会很有用.. 你可以得到数学和不匹配的线..

library(data.table)

match <- merge(as.data.table(df1)[, c(.SD, .(source = "df1", id1 = 1:nrow(df1)))], 
      as.data.table(df2)[, c(.SD, .(source = "df2", id2 = 1:nrow(df1)))], 
      by = c("Product",  "basecode", "A",        "B",        "C",        "D",        "E",        "F" ), 
      all = TRUE)[!is.na(source.x) & !is.na(source.y)]

unmatch <- merge(as.data.table(df1)[, c(.SD, .(source = "df1", id1 = 1:nrow(df1)))], 
                 as.data.table(df2)[, c(.SD, .(source = "df2", id2 = 1:nrow(df1)))], 
               by = c("Product",  "basecode", "A",        "B",        "C",        "D",        "E",        "F" ), 
               all = TRUE)[is.na(source.x) | is.na(source.y)]

【讨论】：