Question

我正在尝试计算tibble上的滚动相关性，在循环中迭代列名。不过，我似乎正在努力将变量传递给函数。这有效：

tbl <- tibble(date = seq(as.Date("1983-03-31"), by=7, length.out=100), 
    col1 = 1:100, col2 = sample(100, size = 100, replace=TRUE), col3 = col1 + col2)
tbl %>% 
    tq_mutate_xy(
        x = col1,
        y = col3,
        mutate_fun = runCor,
        n = 10,
        use = "pairwise.complete.obs",
        col_rename = "col1_col3_corr"
      )

但这并不是：

tbl <- tibble(date = seq(as.Date("1983-03-31"), by=7, length.out=100), 
    col1 = 1:100, col2 = sample(100, size = 100, replace=TRUE), col3 = col1 + col2)
c1 <- "col1"
c2 <- "col3"
tbl %>% 
    tq_mutate_xy(
        x = !!c1,
        y = !!c2,
        mutate_fun = runCor,
        n = 10,
        use = "pairwise.complete.obs",
        col_rename = paste0(c1, "_", c2, "_corr")
    )

错误是＆＃34; check_x_y_valid（data，x，y）出错：x =！（！c1）不是有效名称。＆＃34;

我做错了什么？

Answer 1

首先，我认为你想要tq_mutate_xy的非标准评估（NSE）版本 - 也就是tq_mutate_xy_。因此，当您使用这些函数的NSE时，您需要使用字符串 - 这意味着您的mutate_fun变量也应该是一个字符串。以下应该有效：

c1 <- "col1"
c2 <- "col3"
tbl %>% 
  tq_mutate_xy_(
    x = c1,
    y = c2,
    mutate_fun = "runCor",
    n = 10,
    use = "pairwise.complete.obs",
    col_rename = paste0(c1, "_", c2, "_corr")
  )

请务必查看帮助文档?tq_mutate_xy

中的示例5

R：滚动相关/非标准评估

1 个答案: