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 99 100 146 147 148 149 150
print(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 with
df$var1[!grepl(exp, df$var2, fixed=T) & !is.na(df$var2)]
# [1] 146 147 148 149 150
## Incorrect results with
df$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 99 100 146 147 148 149 150
print(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5
df$var1
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
## 当我们使用少于15个TRUE/FALSE值的长度-15向量索引时会发生什么？较小的向量会被“recycled”，也就是重复，直到它的长度达到较大向量的长度。
## 这里有几个使用长度为3的索引的示例
df$var1[c(T, F, F)]
# [1] 1 4 7 10 13
df$var1[c(F, T, F)]
# [1] 2 5 8 11 14
df$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$var2)], fixed=T)]
# [1] 5 10 15
我认为使用这种子集方法很难得到正确的答案 - 你的第一种方法使用&是正确且合适的。
英文:
Your issue is due to R's "recycling". As usual, it's much easier to see with a smaller example - let's do 15 rows instead of 150, and we'll run each component separately to see what's going on:
## nice small sample
df <- data.frame(
var1 = 1:15,
var2 = c(rep(NA, 10), rep("ABC", 4), rep("CBA", 1))
)
## here's the non-missing var2 values, there are 5 of them
df$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$var2)], fixed=T))
# [1] FALSE FALSE FALSE FALSE TRUE
## but how long is var1? 15, not 5
df$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 index
df$var1[c(T, F, F)]
# [1] 1 4 7 10 13
df$var1[c(F, T, F)]
# [1] 2 5 8 11 14
df$var1[c(T, T, F)]
# [1] 1 2 4 5 7 8 10 11 13 14
## Remember that your grepl result is length 5
!grepl(exp, df$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：
``` r
df <- 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:
``` r
df <- 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.
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