在“ apply”中使用“ with”时，“ character”类型的“ environment”参数无效

Question

I'm applying two conditions to each row in the dataframe trades1 (basically comparing each row with each other row within one column). conditions vector is supposed to consist of 1 when both conditions are satisfied and 0 otherwise.

Data example:

  order     date         time           ms price dir amount    hour index i
1 FUT-3 14.02.06 10:00:00.567 1.950535e+16 66.97 BUY      1 5418154     1 1
2 FUT-3 14.02.06 10:00:00.574 1.950535e+16 66.97 BUY      1 5418154     2 2
3 FUT-3 14.02.06 10:00:00.577 1.950535e+16 66.97 BUY      1 5418154     3 3
4 FUT-3 14.02.06 10:00:00.585 1.950535e+16 66.97 BUY      1 5418154     4 4
5 FUT-3 14.02.06 10:00:00.587 1.950535e+16 66.97 BUY      1 5418154     5 5
6 FUT-3 14.02.06 10:00:00.594 1.950535e+16 66.97 BUY      1 5418154     6 6

Code:

conditions <- apply(trades1, 1, function(x) with(x, as.integer(ms - trades$ms == 1e+6 & price/trades1$price >= 1)))

I've checked that trades1 is a dataframe and that columns are numeric. The error I recieve:

Error in eval(substitute(expr), data, enclos = parent.frame()) : 
  invalid 'envir' argument of type 'character'

In case the problem is not with the data argument, but with placing with inside of apply , I'd appreciate suggestions on how to solve this the other way.

Answer 1

The first thing that apply does is convert its argument into a matrix. Once that's done, with doesn't work any more.

If you want to loop over rows (are you sure this is the best solution?), use a plain old for loop, or lapply over a vector of row numbers:

lst <- lapply(seq_len(nrow(trades1)), function(row) { with(trades1[row, ], ...) } )
do.call(rbind, lst)

Answer 2

As mentioned by @Hong Ooi apply converts the dataframe to matrix and hence all your numbers are converted to characters. You can correct the apply loop by doing

apply(df, 1, function(x) as.integer(any(as.numeric(x["ms"]) - df$ms == 1e+6 & 
                                    as.numeric(x["price"])/df$price >= 1)))

However, I think a better approach here would be to use mapply since you want to check the conditions for price and ms .

as.integer(mapply(function(x, y) 
          any(x - df$ms == 1e+6 & y/df$price >= 1),df$ms, df$price))

---

Similar approach using data.table

library(data.table)
setDT(df)[, ans := as.integer(any(ms - df$ms == 1e+6 & 
                                  price/df$price >= 1)), by = seq_len(nrow(df))]

and tidyverse

library(dplyr)
library(purrr)

df %>%
   mutate(ans = map2(ms, price, 
               ~as.integer(any(.x - df$ms == 1e+6 & .y/df$price >= 1))))