Question

我的数据集如下：

            time type  amount
  1  2017/1/1 0:00   income  729.64
  2  2017/1/1 0:05   income 1465.15
  3  2017/1/1 0:10   outcome 1456.07
  4  2017/1/1 0:15   outcome 1764.28
        ...
  289  2017/1/2 0:00   income  719.64
  290  2017/1/2 0:05   income 165.15
  291  2017/1/2 0:10   income 1006.07
  292  2017/1/2 0:15   outcome 104.28

我想按日期计算净收入，如果您的收入超过结果，那么其他收益将为正数。结果应如下所示：

       date     netincome
  1  2017/1/1   -729.64
  2  2017/1/2   1465.15
  3  2017/1/3  1456.07
  4  2017/1/4   1764.28
    ...

我怎样才能有效地获得这个？

Answer 1

示例数据：

df <- data.frame(time=c("2017/1/1 0:00", "2017/1/1 0:05", "2017/1/1 0:10","2017/1/2 0:00", "2017/1/2 0:05", "2017/1/2 0:10"),
                 type=c("income", "income", "outcome", "income", "outcome", "outcome"),
                 amount=c(729.64, 1465.15, 1456.07, 729.64, 729.64, 1456.07))

将time转换为date，将outcome转换为负值：

df$date <- lubridate::date(df$time)
df$amount[df$type=="outcome"] <- df$amount[df$type=="outcome"]*-1

使用dplyr汇总数据（amount与date的总和）：

library(dplyr)

output <- df %>% group_by(date) %>% summarise(netincome=sum(amount))

结果：

output

# A tibble: 2 x 2
      date netincome
     <chr>     <dbl>
1 2017/1/1    738.72
2 2017/1/2  -1456.07

Answer 2

其他解决办法可能是：

library(tidyverse)
library(lubridate)

df %>% 
  spread(type, amount) %>% 
  group_by(date = date(time)) %>% 
  summarise(netincome = sum(income, na.rm = TRUE) - sum(outcome, na.rm = TRUE))

# # A tibble: 2 x 2
#   date       netincome
#   <date>         <dbl>
# 1 2017-01-01       739
# 2 2017-01-02     -1456

计算净收入

2 个答案: