在 dplyr 中筛选字符串列上的多个值

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最佳答案

You need %in% instead of ==:

library(dplyr)
target <- c("Tom", "Lynn")
filter(dat, name %in% target)  # equivalently, dat %>% filter(name %in% target)

Produces

  days name
1   88 Lynn
2   11  Tom
3    1  Tom
4  222 Lynn
5    2 Lynn

To understand why, consider what happens here:

dat$name == target
# [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE

Basically, we're recycling the two length target vector four times to match the length of dat$name. In other words, we are doing:

 Lynn == Tom
Tom == Lynn
Chris == Tom
Lisa == Lynn
... continue repeating Tom and Lynn until end of data frame

In this case we don't get an error because I suspect your data frame actually has a different number of rows that don't allow recycling, but the sample you provide does (8 rows). If the sample had had an odd number of rows I would have gotten the same error as you. But even when recycling works, this is clearly not what you want. Basically, the statement dat$name == target is equivalent to saying:

return TRUE for every odd value that is equal to "Tom" or every even value that is equal to "Lynn".

It so happens that the last value in your sample data frame is even and equal to "Lynn", hence the one TRUE above.

To contrast, dat$name %in% target says:

for each value in dat$name, check that it exists in target.

Very different. Here is the result:

[1]  TRUE  TRUE FALSE FALSE FALSE  TRUE  TRUE  TRUE

Note your problem has nothing to do with dplyr, just the mis-use of ==.

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Using the base package:

df <- data.frame(days = c(88, 11, 2, 5, 22, 1, 222, 2), name = c("Lynn", "Tom", "Chris", "Lisa", "Kyla", "Tom", "Lynn", "Lynn"))


# Three lines
target <- c("Tom", "Lynn")
index <- df$name %in% target
df[index, ]


# One line
df[df$name %in% c("Tom", "Lynn"), ]

Output:

  days name
1   88 Lynn
2   11  Tom
6    1  Tom
7  222 Lynn
8    2 Lynn

Using sqldf:

library(sqldf)
# Two alternatives:
sqldf('SELECT *
FROM df
WHERE name = "Tom" OR name = "Lynn"')
sqldf('SELECT *
FROM df
WHERE name IN ("Tom", "Lynn")')

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This can be achieved using dplyr package, which is available in CRAN. The simple way to achieve this:

Install dplyr package.
Run the below code

library(dplyr)


df<- select(filter(dat,name=='tom'| name=='Lynn'), c('days','name))

Explanation:

So, once we’ve downloaded dplyr, we create a new data frame by using two different functions from this package:

filter: the first argument is the data frame; the second argument is the condition by which we want it subsetted. The result is the entire data frame with only the rows we wanted. select: the first argument is the data frame; the second argument is the names of the columns we want selected from it. We don’t have to use the names() function, and we don’t even have to use quotation marks. We simply list the column names as objects.

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 by_type_year_tag_filtered <- by_type_year_tag %>%
dplyr:: filter(tag_name %in% c("dplyr", "ggplot2"))

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In case you have long strings as values in your string columns you can use this powerful method with the stringr package. A method that filter( %in% ) and base R can't do.

library(dplyr)
library(stringr)


sentences_tb = as_tibble(sentences) %>%
mutate(row_number())
sentences_tb
# A tibble: 720 x 2
value                                       `row_number()`
<chr>                                                <int>
1 The birch canoe slid on the smooth planks.               1
2 Glue the sheet to the dark blue background.              2
3 Its easy to tell the depth of a well.                   3
4 These days a chicken leg is a rare dish.                 4
5 Rice is often served in round bowls.                     5
6 The juice of lemons makes fine punch.                    6
7 The box was thrown beside the parked truck.              7
8 The hogs were fed chopped corn and garbage.              8
9 Four hours of steady work faced us.                      9
10 Large size in stockings is hard to sell.                10
# ... with 710 more rows


matching_letters <- c(
"canoe","dark","often","juice","hogs","hours","size"
)
matching_letters <- str_c(matching_letters, collapse = "|")
matching_letters
[1] "canoe|dark|often|juice|hogs|hours|size"


letters_found <- str_subset(sentences_tb$value,matching_letters)
letters_found_tb = as_tibble(letters_found)
inner_join(sentences_tb,letters_found_tb)


# A tibble: 16 x 2
value                                          `row_number()`
<chr>                                                   <int>
1 The birch canoe slid on the smooth planks.                  1
2 Glue the sheet to the dark blue background.                 2
3 Rice is often served in round bowls.                        5
4 The juice of lemons makes fine punch.                       6
5 The hogs were fed chopped corn and garbage.                 8
6 Four hours of steady work faced us.                         9
7 Large size in stockings is hard to sell.                   10
8 Note closely the size of the gas tank.                     33
9 The bark of the pine tree was shiny and dark.             111
10 Both brothers wear the same size.                         253
11 The dark pot hung in the front closet.                    261
12 Grape juice and water mix well.                           383
13 The wall phone rang loud and often.                       454
14 The bright lanterns were gay on the dark lawn.            476
15 The pleasant hours fly by much too soon.                  516
16 A six comes up more often than a ten.                     609

It's a bit verbose, but it's very handy and powerful if you have long strings and want to filter in what row is located a specific word.

Comparing with the accepted answers:

> target <- c("canoe","dark","often","juice","hogs","hours","size")
> filter(sentences_tb, value %in% target)
# A tibble: 0 x 2
# ... with 2 variables: value <chr>, row_number() <int>


> df<- select(filter(sentences_tb,value=='canoe'| value=='dark'), c('value','row_number()'))
> df
# A tibble: 0 x 2
# ... with 2 variables: value <chr>, row_number() <int>


> target <- c("canoe","dark","often","juice","hogs","hours","size")
> index <- sentences_tb$value %in% target
> sentences_tb[index, ]
# A tibble: 0 x 2
# ... with 2 variables: value <chr>, row_number() <int>

You need to write all the sentences to get the desired result.

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Write that. Example:

library (dplyr)


target <- YourData%>% filter (YourColum %in% c("variable1","variable2"))

Example with your data

target <- df%>% filter (names %in% c("Tom","Lynn"))