在R中具有多个条件的.loc的等价物是什么?

Question

我想知道在R中是否有任何等效的。loc，我可以在for循环中有多个条件。

在python中，我使用.loc完成了这个任务，如下面的代码所示。然而，我无法在r中复制此

for column in df.columns[1:9]:
for i in range(4,len(df)):
col = 'AC' + str(column[-1])
df[col][i] = df['yw_lagged'].loc[(df['id'] == df[column][i]) & (df['yearweek'] == df['yearweek'][i])]

在R中，我认为这是可行的

df[i,col] <- df[df$id == df[column][i] & df$yearweek == df$yearweek[i], "yw_lagged"]

但是它似乎不像。loc那样过滤。

编辑:

structure(list(id = c(1, 2, 6, 7, 1, 2), v1 = c(2, 1, 1, 1, 2, 
1), v2 = c(6, 3, 2, 2, 6, 3), v3 = c(7, 6, 5, 3, 7, 6), v4 = 
c(NA, 7, 7, 6, NA, 7), v5 = c(NA, 8, 14, 8, NA, 8), v6 = c(NA, 
NA,15, 15, NA, NA), v7 = c(NA, NA, 16, 16, NA, NA), v8 = c(NA, 
NA,NA, 17, NA, NA), violent = c(1, 0, 1, 0, 0, 0), yw_lagged = 
c(NA, NA, NA, NA, 1, 0), yearweek = c(20161, 20161, 20161, 20161, 
20162, 20162), AC1 = c(NA, NA, NA, NA, NA, NA), AC2 = c(NA, NA, 
NA, NA, NA, NA), AC3 = c(NA, NA, NA, NA, NA, NA), AC4 = c(NA, NA, 
NA, NA, NA, NA), AC5 = c(NA, NA, NA, NA, NA, NA), AC6 = c(NA, 
NA, NA, NA, NA, NA), AC7 = c(NA, NA, NA, NA, NA, NA), AC8 = c(NA, 
NA, NA, NA, NA, NA)), row.names = c(NA, -6L), class = c("tbl_df", 
"tbl", "data.frame"))

期望输出的图片(尝试使用颜色添加一些连接)

Answer 1

示例数据缺少任何匹配，因此我将覆盖yw_lagged的几行:

df$yw_lagged[1:4] <- 1:4

从这里

dplyr

library(dplyr)
df %>%
group_by(yearweek) %>%
mutate(across(matches("^v[0-9]+"),
~ yw_lagged[match(., id)],
.names = "AC{sub('^v', '', .col)}")) %>%
ungroup()
# # A tibble: 6 × 20
#      id    v1    v2    v3    v4    v5    v6    v7    v8 violent yw_lagged yearweek   AC1   AC2   AC3   AC4   AC5   AC6   AC7   AC8
#   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>   <dbl>     <dbl>    <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1     1     2     6     7    NA    NA    NA    NA    NA       1         1    20161     2     3     4    NA    NA    NA    NA    NA
# 2     2     1     3     6     7     8    NA    NA    NA       0         2    20161     1    NA     3     4    NA    NA    NA    NA
# 3     6     1     2     5     7    14    15    16    NA       1         3    20161     1     2    NA     4    NA    NA    NA    NA
# 4     7     1     2     3     6     8    15    16    17       0         4    20161     1     2    NA     3    NA    NA    NA    NA
# 5     1     2     6     7    NA    NA    NA    NA    NA       0         1    20162     0    NA    NA    NA    NA    NA    NA    NA
# 6     2     1     3     6     7     8    NA    NA    NA       0         0    20162     1    NA    NA    NA    NA    NA    NA    NA

data.table

library(data.table)
cols <- grep("v[0-9]+", names(df), value = TRUE)
data.table(df)[, (sub("^v", "AC", cols)) :=
lapply(.SD, (z) yw_lagged[match(z, id)]),
.SDcols = cols][]
#       id    v1    v2    v3    v4    v5    v6    v7    v8 violent yw_lagged yearweek   AC1   AC2   AC3   AC4   AC5   AC6   AC7   AC8
#    <num> <num> <num> <num> <num> <num> <num> <num> <num>   <num>     <num>    <num> <num> <num> <num> <num> <num> <num> <num> <num>
# 1:     1     2     6     7    NA    NA    NA    NA    NA       1         1    20161     2     3     4    NA    NA    NA    NA    NA
# 2:     2     1     3     6     7     8    NA    NA    NA       0         2    20161     1    NA     3     4    NA    NA    NA    NA
# 3:     6     1     2     5     7    14    15    16    NA       1         3    20161     1     2    NA     4    NA    NA    NA    NA
# 4:     7     1     2     3     6     8    15    16    17       0         4    20161     1     2    NA     3    NA    NA    NA    NA
# 5:     1     2     6     7    NA    NA    NA    NA    NA       0         1    20162     2     3     4    NA    NA    NA    NA    NA
# 6:     2     1     3     6     7     8    NA    NA    NA       0         0    20162     1    NA     3     4    NA    NA    NA    NA