0

我目前正在以下列形式将多个 csv 文件加载到 R 中：

read.csv("Cashflows2.csv", header = F, )
           V1      V2
1        Date Payments
2  18/08/2017  -20495*
3  18/04/2018  639.76*
4  18/05/2018  639.76
5  18/06/2018  639.76
6  18/07/2018  639.76
7  18/08/2018  639.76
8  18/09/2018  639.76
9  18/10/2018  639.76
10 18/11/2018  639.76*
11 18/05/2019  639.76*
12 18/06/2019  639.76
13 18/07/2019  639.76
14 18/08/2019  639.76
15 18/09/2019  639.76
16 18/10/2019  639.76
17 18/11/2019  639.76
18 18/12/2019  639.76
19 18/01/2020  639.76
20 18/02/2020  639.76
21 18/03/2020  639.76
22 18/04/2020  639.76
23 18/05/2020  639.76
24 18/06/2020  639.76
25 18/07/2020  639.76
26 18/08/2020  639.76
27 18/09/2020  639.76
28 18/10/2020  639.76
29 18/11/2020  639.76
30 18/12/2020  639.76
31 18/01/2021  639.76
32 18/02/2021  639.76
33 18/03/2021  639.76
34 18/04/2021  639.76
35 18/05/2021  639.76
36 18/06/2021  639.76
37 18/07/2021  734.76

但是，如星号（未出现在 csv 文件中）所示，有两个时期没有付款。是否有一个函数可以将此 csv 文件转换为 R 中的以下形式：

read.csv("Cashflows2.csv", header = F, )
           V1      V2
1        Date Payment
2  18/08/2017  -20495
3  18/09/2017       0
4  18/10/2017       0
5  18/11/2017       0
6  18/12/2017       0
7  18/01/2018       0
8  18/02/2018       0
9  18/03/2018       0
10 18/04/2018  639.76
11 18/05/2018  639.76
12 18/06/2018  639.76
13 18/07/2018  639.76
14 18/08/2018  639.76
15 18/09/2018  639.76
16 18/10/2018  639.76
17 18/11/2018  639.76
18 18/12/2018       0
19 18/01/2019       0
20 18/02/2019       0
21 18/03/2019       0
22 18/04/2019       0
23 18/05/2019  639.76
24 18/06/2019  639.76
25 18/07/2019  639.76
26 18/08/2019  639.76
27 18/09/2019  639.76
28 18/10/2019  639.76
29 18/11/2019  639.76
30 18/12/2019  639.76
31 18/01/2020  639.76
32 18/02/2020  639.76
33 18/03/2020  639.76
34 18/04/2020  639.76
35 18/05/2020  639.76
36 18/06/2020  639.76
37 18/07/2020  639.76
38 18/08/2020  639.76
39 18/09/2020  639.76
40 18/10/2020  639.76
41 18/11/2020  639.76
42 18/12/2020  639.76
43 18/01/2021  639.76
44 18/02/2021  639.76
45 18/03/2021  639.76
46 18/04/2021  639.76
47 18/05/2021  639.76
48 18/06/2021  639.76
49 18/07/2021  734.76

并非所有 csv 文件都有相同的问题，因此理想情况下，该功能将适用于多个类似的 csv 文件，其中并非所有文件都经历 0 付款期。

任何帮助将不胜感激。

 dput(df)
structure(list(V1 = structure(c(37L, 22L, 7L, 10L, 14L, 18L, 
23L, 26L, 29L, 32L, 11L, 15L, 19L, 24L, 27L, 30L, 33L, 35L, 1L, 
3L, 5L, 8L, 12L, 16L, 20L, 25L, 28L, 31L, 34L, 36L, 2L, 4L, 6L, 
9L, 13L, 17L, 21L), .Label = c("18/01/2020", "18/01/2021", "18/02/2020", 
"18/02/2021", "18/03/2020", "18/03/2021", "18/04/2018", "18/04/2020", 
"18/04/2021", "18/05/2018", "18/05/2019", "18/05/2020", "18/05/2021", 
"18/06/2018", "18/06/2019", "18/06/2020", "18/06/2021", "18/07/2018", 
"18/07/2019", "18/07/2020", "18/07/2021", "18/08/2017", "18/08/2018", 
"18/08/2019", "18/08/2020", "18/09/2018", "18/09/2019", "18/09/2020", 
"18/10/2018", "18/10/2019", "18/10/2020", "18/11/2018", "18/11/2019", 
"18/11/2020", "18/12/2019", "18/12/2020", "Date"), class = "factor"), 
    V2 = structure(c(4L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L), .Label = c("-20495", 
    "639.76", "734.76", "Payment"), class = "factor")), class = "data.frame", row.names = c(NA, 
-37L))
4

2 回答 2

1

我们可以tidyr::complete在读取数据后使用header = TRUE，将date列转换为实际的 Date 对象。

df <- read.csv("Cashflows2.csv", header = TRUE)

library(dplyr)

df %>%
  mutate(Date = as.Date(Date, "%d/%m/%Y")) %>%
  tidyr::complete(Date = seq(min(Date), max(Date), by = "1 month"), 
          fill = list(Payments = 0))


# A tibble: 48 x 2
#   Date       Payments
#   <date>        <dbl>
# 1 2017-08-18  -20495 
# 2 2017-09-18       0 
# 3 2017-10-18       0 
# 4 2017-11-18       0 
# 5 2017-12-18       0 
# 6 2018-01-18       0 
# 7 2018-02-18       0 
# 8 2018-03-18       0 
# 9 2018-04-18     640.
#10 2018-05-18     640.
# … with 38 more rows

max在基础 R 中，您可以使用和来创建一个新的数据框，min将它们替换为 0并将s 替换为 0。DatemergeDateNA

df$Date <- as.Date(df$Date, "%d/%m/%Y")
compare_df <- data.frame(Date = seq(min(df$Date), max(df$Date), by = "1 month"))
df1 <- merge(compare_df, df, by = "Date", all.x = TRUE)
df1$Payments[is.na(df1$Payments)] <- 0

要将其应用于多个 csv 文件，我们可以将其更改为一个函数并使用lapply

read_fun  <- function(df) {
   df$Date <- as.Date(df$Date, "%d/%m/%Y")
   compare_df <- data.frame(Date = seq(min(df$Date), max(df$Date), by = "1 month"))
   df1 <- merge(compare_df, df, by = "Date", all.x = TRUE)
   df1$Payments[is.na(df1$Payments)] <- 0
   df1
 }

list_df <- lapply(list_df, read_fun)
于 2019-08-15T09:27:55.920 回答
0

header = TRUE您应该使用in读取数据，read.csv因为您有列名。

my_data <- read.csv("Cashflows2.csv", header = TRUE)

然后您可以将 Date 列转换为“正确”的日期列

my_data$Date <- as.Date(my_data$Date, format = "%d/%m/%Y")

然后，我认为解决您的任务的简单方法如下。但是，这需要您安装tidyr-package：（使用install.packages("tidyr")）

tidyr::complete(my_data, Date = seq.Date(min(Date), max(Date), by = "month"), 
                fill = list(Payments = 0)) 

# A tibble: 48 x 2
#    Date       Payments
#    <date>        <dbl>
#  1 2017-08-18  -20495 
#  2 2017-09-18       0 
#  3 2017-10-18       0 
#  4 2017-11-18       0 
#  5 2017-12-18       0 
#  6 2018-01-18       0 
#  7 2018-02-18       0 
#  8 2018-03-18       0 
#  9 2018-04-18     640.
# 10 2018-05-18     640.
# ... with 38 more rows

此函数获取您的数据，并扩展日期序列，我们从数据中的最小日期开始，到最大日期，步长为一个月。此外，我们希望在缺少的月​​份中用零填充 Payments 列。

您可以保存更新的数据

write.csv(my_data, "Cashflows2_updated.csv")

如果要恢复以前的日期格式，可以使用

my_data$Date <- format(my_data$Date, format = "%d/%m/%Y")

在保存文件之前。

于 2019-08-15T09:28:02.220 回答