dplyr滤波函数与agrep相结合

Question

我试图只从表中筛选标题列中有“狗”一词的行，但我无法让它工作。

下面是一个数据示例：

    ID NozamaItemID                                                    NozamaTitle 
1 4557  12000017544 Starbucks Double Shot Espresso Light (4 Count, 6.5 Fl Oz Each) 
2 4558  12000021992                                        Pepsi, 8Ct, 12Oz Bottle 
3 4559  12000024542                     Zuke'S Natural Hip Action dog Treats, 3 Oz 
4 4560  12000030680                  Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans 
5 4561  12000030680                  Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans 
6 4562  12000030680                  Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans 

下列代码应该有效，但不起作用：

amzp <- select(amz, ID, NozamaItemID, NozamaTitle, NozamaCustomerID)

searchTerm="cat|dog"
amzp.a <- mutate(amzp, animalFood = ifelse(grepl(searchTerm, amzp$NozamaTitle, ignore.case = TRUE) == TRUE, TRUE, FALSE))

我希望在第3行能看到一个真实的结果。任何帮助都是非常感谢的。谢谢

Answer 1

你离我很近，你只需要摆脱ifelse

amzp.a <- mutate(amzp, animalFood = grepl(searchTerm, 
                         NozamaTitle, ignore.case = TRUE))

这意味着：

> amzp.a
    ID NozamaItemID                                                     NozamaTitle animalFood
1 4557  12000017544  Starbucks Double Shot Espresso Light (4 Count, 6.5 Fl Oz Each)      FALSE
2 4558  12000021992                                         Pepsi, 8Ct, 12Oz Bottle      FALSE
3 4559  12000024542                      Zuke'S Natural Hip Action dog Treats, 3 Oz       TRUE
4 4560  12000030680                   Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans      FALSE
5 4561  12000030680                   Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans      FALSE
6 4562  12000030680                   Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans      FALSE

使用的数据：

amzp <- structure(list(ID = 4557:4562,
                       NozamaItemID = c(12000017544, 12000021992, 12000024542, 12000030680, 12000030680, 12000030680),
                       NozamaTitle = structure(c(4L, 1L, 2L, 3L, 3L, 3L), .Label = c("Pepsi, 8Ct, 12Oz Bottle","Zuke'S Natural Hip Action dog Treats, 3 Oz","Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans","Starbucks Double Shot Espresso Light (4 Count, 6.5 Fl Oz Each)"), class = "factor")),
                  .Names = c("ID", "NozamaItemID", "NozamaTitle"), class = "data.frame", row.names = c(NA, -6L))

编辑:您的原始代码：

amzp.a <- mutate(amzp, animalFood = ifelse(grepl(searchTerm, amzp$NozamaTitle, ignore.case = TRUE) == TRUE, TRUE, FALSE))

真的很管用。虽然它包含一些不需要的组件( ifelse-statement和在标准dplyr函数中使用data$column )，但它提供了所需的结果：

> amzp.a
    ID NozamaItemID                                                     NozamaTitle animalFood
1 4557  12000017544  Starbucks Double Shot Espresso Light (4 Count, 6.5 Fl Oz Each)      FALSE
2 4558  12000021992                                         Pepsi, 8Ct, 12Oz Bottle      FALSE
3 4559  12000024542                      Zuke'S Natural Hip Action dog Treats, 3 Oz       TRUE
4 4560  12000030680                   Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans      FALSE
5 4561  12000030680                   Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans      FALSE
6 4562  12000030680                   Pepsi Made With Real Sugar, 12 Ct, 12 Oz Cans      FALSE

因此，您可能需要更详细地描述“不工作”语句。

Answer 2

我不太清楚您想要达到什么目的，但是如果您的目标是只留下“NozamaTitle”一词出现在dplyr::filter列中的行，则只需使用dplyr::filter。使用chickwts作为一个示例来代替一个最小可复制的示例：

levels(chickwts$feed)
# [1] "casein"    "horsebean" "linseed"   "meatmeal"  "soybean"  
# [6] "sunflower"

df <- filter(chickwts, grepl("bean", feed))
df
#    weight      feed
# 1     179 horsebean
# 2     160 horsebean
# 3     136 horsebean
# ...
# 11    243   soybean
# 12    230   soybean
# 13    248   soybean
# ...

这就是你想要的吗？