英文:
Add a column to data frame resulting from the merging of some values from other columns with R
问题
我有一个包含许多列的数据框，其中包括以下内容：
COLUMN_1 COLUMN_2 COLUMN_3red blue greennone blue nonered none greenred none nonenone none none
我想基于一个布尔值添加第四列COLORS，对于COLUMN_1的每个值为'red'，COLUMN_2的每个值为'blue'，或者COLUMN_3的每个值为'green'，都会给出'True'，如果COLUMN_1，COLUMN_2和COLUMN_3的值始终为'none'，则给出'False'。就像这样：
COLUMN_1 COLUMN_2 COLUMN_3 COLORSred blue green Truenone blue none Truered none green Truered none none Truenone none none False
一个不带布尔值的列也可以：
COLUMN_1 COLUMN_2 COLUMN_3 COLORSred blue green Colornone blue none Colorred none green Colorred none none Colornone none none No color
英文:
I have a data frame with many columns, among which I have the following:
COLUMN_1 COLUMN_2 COLUMN_3red blue greennone blue nonered none greenred none nonenone none none
I'd like to add a fourth column called COLORS based on a boolean which gives 'True' for every value 'red' of COLUMN_1, 'blue' of COLUMN_2, or 'green' of COLUMN_3, and gives 'False' if the value of COLUMN_1, COLUMN_2 and COLUMN_3 is always 'none'. Like this:
COLUMN_1 COLUMN_2 COLUMN_3 COLORSred blue green Truenone blue none Truered none green Truered none none Truenone none none False
A column without a boolean would also work:
COLUMN_1 COLUMN_2 COLUMN_3 COLORSred blue green Colornone blue none Colorred none green Colorred none none Colornone none none No color
答案1
得分: 2
Here is the translated code without the translation of code-related terms:
library(dplyr)df %>%mutate(COLORS = case_when(if_all(contains("COLUMN"), ~ .x == "none") ~ "无颜色",TRUE ~ "有颜色"))
And the table section:
# A tibble: 5 × 4COLUMN_1 COLUMN_2 COLUMN_3 COLORS<chr> <chr> <chr> <chr>1 red blue green 有颜色2 none blue none 有颜色3 red none green 有颜色4 red none none 有颜色5 none none none 无颜色
英文:
library(dplyr)df %>%mutate(COLORS = case_when(if_all(contains("COLUMN"), ~ .x == "none") ~ "No color",TRUE ~ "Color"))# A tibble: 5 × 4COLUMN_1 COLUMN_2 COLUMN_3 COLORS<chr> <chr> <chr> <chr>1 red blue green Color2 none blue none Color3 red none green Color4 red none none Color5 none none none No color
答案2
得分: 2
一种选择可能是：
df %>%rowwise() %>%mutate(COLORS = any(across(everything()) == c("red", "blue", "green")))COLUMN_1 COLUMN_2 COLUMN_3 COLORS<chr> <chr> <chr> <lgl>1 red blue green TRUE2 none blue none TRUE3 red none green TRUE4 red none none TRUE5 none none none FALSE
如果某个列的颜色与预期不同，它将返回 FALSE：
COLUMN_1 COLUMN_2 COLUMN_3 COLORS<chr> <chr> <chr> <lgl>1 red blue green TRUE2 none blue none TRUE3 red none green TRUE4 red none none TRUE5 blue none none FALSE
英文:
One option could be:
df %>%rowwise() %>%mutate(COLORS = any(across(everything()) == c("red", "blue", "green")))COLUMN_1 COLUMN_2 COLUMN_3 COLORS<chr> <chr> <chr> <lgl>1 red blue green TRUE2 none blue none TRUE3 red none green TRUE4 red none none TRUE5 none none none FALSE
If a given column would have a different color as expected, the it would return FALSE:
COLUMN_1 COLUMN_2 COLUMN_3 COLORS<chr> <chr> <chr> <lgl>1 red blue green TRUE2 none blue none TRUE3 red none green TRUE4 red none none TRUE5 blue none none FALSE
答案3
得分: 0
以下是您要翻译的内容：
"这并不是一个高效的答案，而是一个替代的答案
df2 <- df %>%mutate(across(c(COLUMN_1,COLUMN_2,COLUMN_3), ~ ifelse(.x=='none', 1, 0), .names = 'n_{.col}'),new=rowSums(across(c(n_COLUMN_1,n_COLUMN_2,n_COLUMN_3))),COLORS=ifelse(new==3,'no color', 'color')) %>%select(-contains('n_c'), -new)
<sup>创建于2023-06-08，使用 reprex v2.0.2</sup>
# A tibble: 5 × 4COLUMN_1 COLUMN_2 COLUMN_3 COLORS<chr> <chr> <chr> <chr>1 红色 蓝色 绿色 有颜色2 无 蓝色 无 有颜色3 红色 无 绿色 有颜色4 红色 无 无 有颜色5 无 无 无 无颜色
英文:
It is not an efficient answer but an alternate
df2 <- df %>%mutate(across(c(COLUMN_1,COLUMN_2,COLUMN_3), ~ ifelse(.x=='none', 1, 0), .names = 'n_{.col}'),new=rowSums(across(c(n_COLUMN_1,n_COLUMN_2,n_COLUMN_3))),COLORS=ifelse(new==3,'no color', 'color')) %>%select(-contains('n_c'), -new)
<sup>Created on 2023-06-08 with reprex v2.0.2</sup>
# A tibble: 5 × 4COLUMN_1 COLUMN_2 COLUMN_3 COLORS<chr> <chr> <chr> <chr>1 red blue green color2 none blue none color3 red none green color4 red none none color5 none none none no color
答案4
得分: 0
我们可以尝试使用 rowSums + col，如下所示：
transform(df,COLORS = rowSums(df == c("red", "blue", "green")[col(df)]) > 0)
这将得到以下结果：
COLUMN_1 COLUMN_2 COLUMN_3 COLORS1 red blue green TRUE2 none blue none TRUE3 red none green TRUE4 red none none TRUE5 none none none FALSE
英文:
We can try rowSums + col like below
transform(df,COLORS = rowSums(df == c("red", "blue", "green")[col(df)]) > 0)
which gives
COLUMN_1 COLUMN_2 COLUMN_3 COLORS1 red blue green TRUE2 none blue none TRUE3 red none green TRUE4 red none none TRUE5 none none none FALSE
答案5
得分: 0
library(dplyr)df |>mutate(COLORS = rowSums(pick(starts_with("COLUMN")) != "none") > 0)
注意：如果存在颜色“yellow”，则返回TRUE。
英文:
library(dplyr)df |>mutate(COLORS = rowSums(pick(starts_with("COLUMN")) != "none") > 0)
Note: this will return TRUE if there is a color "yellow".
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