英文:
Where to place is.na() when using grepl to find non-missing non-matches?
问题
我想找到那些在var2中没有包含已定义表达式的非缺失（！）条目，并检索它们相应的var1值。默认情况下，grepl会返回缺失的条目，我想避免这种情况。我想到了两种方法，其中一种提供了错误的结果。我想了解为什么它提供了错误的结果？请查找下面的代码和带有输出的代码。谢谢！## 正确结果df$var1[!grepl(exp, df$var2, fixed=T) & !is.na(df$var2)]# [1] 146 147 148 149 150## 不正确的结果df$var1[!grepl(exp, df$var2[!is.na(df$var2)], fixed=T)]print(df[c(46, 47, 48, 49, 50, 96, 97, 98, 99, 100, 146, 147, 148, 149, 150),7])# [1] 46 47 48 49 50 96 97 98 99 100 146 147 148 149 150print(df[c(46, 47, 48, 49, 50, 96, 97, 98, 99, 100, 146, 147, 148, 149, 150),7])# [1] NA NA NA NA NA NA NA NA NA NA "CBA" "CBA" "CBA" "CBA" "CBA"
英文:
I want to find those non-missing (!) entries that do not contain a defined expression in var2 and retrieve their respective value of var1. By default, grepl will return also missing entries which I want to avoid. I came up with two approaches and one of them delivers wrong results. I would like to understand why it delivers wrong results? Find both the code and the code with output below, please. Thank you!
df <- data.frame(var1 = 1:150,var2 = c(rep(NA, 100), rep("ABC", 45), rep("CBA", 5)))exp <- "BC"## Correct results withdf$var1[!grepl(exp, df$var2, fixed=T) & !is.na(df$var2)]# [1] 146 147 148 149 150## Incorrect results withdf$var1[!grepl(exp, df$var2[!is.na(df$var2)], fixed=T)]print(df[c(46, 47, 48, 49, 50, 96, 97, 98, 99, 100, 146, 147, 148, 149, 150),7])# [1] 46 47 48 49 50 96 97 98 99 100 146 147 148 149 150print(df[c(46, 47, 48, 49, 50, 96, 97, 98, 99, 100, 146, 147, 148, 149, 150),7])# [1] NA NA NA NA NA NA NA NA NA NA "CBA" "CBA" "CBA" "CBA" "CBA"
答案1
得分: 4
你的问题是由于R的“recycling”机制引起的。通常情况下，通过一个较小的示例更容易看出问题 - 我们将使用15行而不是150行，并分别运行每个组件来查看发生了什么：
## 不错的小样本df <- data.frame(var1 = 1:15,var2 = c(rep(NA, 10), rep("ABC", 4), rep("CBA", 1)))## 这里是非缺失的var2值，共有5个df$var2[!is.na(df$var2)]# [1] "ABC" "ABC" "ABC" "ABC" "CBA"## 当我们使用grep时，会得到5个TRUE/FALSE值(!grepl(exp, df$var2[!is.na(df$var2)], fixed=T))# [1] FALSE FALSE FALSE FALSE TRUE## 但var1的长度是多少？是15，而不是5df$var1# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15## 当我们使用少于15个TRUE/FALSE值的长度-15向量索引时会发生什么？较小的向量会被“recycled”，也就是重复，直到它的长度达到较大向量的长度。## 这里有几个使用长度为3的索引的示例df$var1[c(T, F, F)]# [1] 1 4 7 10 13df$var1[c(F, T, F)]# [1] 2 5 8 11 14df$var1[c(T, T, F)]# [1] 1 2 4 5 7 8 10 11 13 14## 请记住，grepl的结果长度为5!grepl(exp, df$var2[!is.na(df$var2)], fixed=T)# [1] FALSE FALSE FALSE FALSE TRUE## 因此，它将被重复3次，直到达到长度15，## 希望现在这个结果有意义了！df$var1[!grepl(exp, df$var2[!is.na(df$var2)], fixed=T)]# [1] 5 10 15
我认为使用这种子集方法很难得到正确的答案 - 你的第一种方法使用&是正确且合适的。
英文:
Your issue is due to R's "recycling". As usual, it's much easier to see with a smaller example - let's do 15 rows instead of 150, and we'll run each component separately to see what's going on:
## nice small sampledf <- data.frame(var1 = 1:15,var2 = c(rep(NA, 10), rep("ABC", 4), rep("CBA", 1)))## here's the non-missing var2 values, there are 5 of themdf$var2[!is.na(df$var2)]# [1] "ABC" "ABC" "ABC" "ABC" "CBA"## and when we grep them, we get 5 TRUE/FALSE values(!grepl(exp, df$var2[!is.na(df$var2)], fixed=T))# [1] FALSE FALSE FALSE FALSE TRUE## but how long is var1? 15, not 5df$var1# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15## what happens when we index a length-15 vector by less than## 15 TRUE/FALSE values? The smaller vector is "recycled", that## is, repeated until it gets to the length of the larger vector.## Here's a couple examples of that with a length-3 indexdf$var1[c(T, F, F)]# [1] 1 4 7 10 13df$var1[c(F, T, F)]# [1] 2 5 8 11 14df$var1[c(T, T, F)]# [1] 1 2 4 5 7 8 10 11 13 14## Remember that your grepl result is length 5!grepl(exp, df$var2[!is.na(df$var2)], fixed=T)# [1] FALSE FALSE FALSE FALSE TRUE## So it will be recycled 3 times up to length 15,## and hopefully now this result makes sense!df$var1[!grepl(exp, df$var2[!is.na(df$var2)], fixed=T)]# [1] 5 10 15
I don't think there's a great way to get the right answer with this subset approach - your first approach with & is correct and proper.
答案2
得分: 1
regexpr() 在提供 `NA` 时返回 `NA`。如果你想要那些**不匹配**表达式的元素，你可以这样做：``` rdf <- data.frame(var1 = 1:150,var2 = c(rep(NA, 100), rep("ABC", 45), rep("CBA", 5)))exp <- "BC"df[regexpr(exp, df$var2, fixed = TRUE) %in% -1, ]#> var1 var2#> 146 146 CBA#> 147 147 CBA#> 148 148 CBA#> 149 149 CBA#> 150 150 CBA
regexpr() 在无法匹配表达式时返回 -1，如果提供 NA 则返回 NA，如果找到匹配的表达式则返回位置。
如果你想要匹配的元素，可以使用：
!(regexpr(exp, df$var2, fixed = TRUE) %in% c(NA, -1))
会返回所有匹配且非缺失的元素。
<details><summary>英文:</summary>`regexpr()` returns `NA`s if it is supplied `NA`s.If you want elements that **don't match** your expression you can do:``` rdf <- data.frame(var1 = 1:150,var2 = c(rep(NA, 100), rep("ABC", 45), rep("CBA", 5)))exp <- "BC"df[regexpr(exp, df$var2, fixed = TRUE) %in% -1, ]#> var1 var2#> 146 146 CBA#> 147 147 CBA#> 148 148 CBA#> 149 149 CBA#> 150 150 CBA
<sup>Created on 2023-06-30 with reprex v2.0.2</sup>
regexpr() returns -1 when it can't match the expression, NA if given NA or the position at which it finds the matching expression.
If you wanted the elements that do match
!(regexpr(exp, df$var2, fixed = TRUE) %in% c(NA, -1))
would give all matched and non-missing elements.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论