Question

我正在尝试为数据创建旅行号码。数据包含车辆ID和时间戳。对于相同ID，彼此接近的所有时间戳被认为是单次旅行的一部分。如果连续ID之间的时间差大于1小时，我希望它被视为新的旅行。

示例数据：

{
  "name": "rh-react",
  "version": "1.0.0",
  "description": "Pages accessible for outside world",
  "main": "index.js",
  "scripts": {
    "test": "echo \"Error: no test specified\" && exit 1",
    "dev": "webpack -d --watch",
    "build": "sh -ac '. .env; webpack -d'",
    "build:test": "sh -ac '. .env.test; webpack -p'",
    "build:prod": "sh -ac '. .env.prod; webpack -p'"
  },
  "author": "",
  "license": "ISC",
  "dependencies": {
    "ag-grid": "^13.3.1",
    "ag-grid-react": "^13.3.0",
    "axios": "^0.17.1",
    "babel-loader": "^7.1.2",
    "babel-plugin-add-module-exports": "^0.2.1",
    "babel-preset-es2015": "^6.24.1",
    "babel-preset-react": "^6.24.1",
    "bootstrap": "^3.3.7",
    "classnames": "^2.2.5",
    "dotenv": "^4.0.0",
    "fsevents": "^1.1.3",
    "jquery": "^3.2.1",
    "moment": "^2.19.1",
    "query-string": "^5.0.0",
    "react": "^16.2.0",
    "react-async-script-loader": "^0.3.0",
    "react-click-outside": "^3.0.0",
    "react-cookies": "^0.1.0",
    "react-datepicker": "^0.61.0",
    "react-datetime": "^2.11.1",
    "react-dom": "^16.2.0",
    "react-dom-factories": "^1.0.2",
    "react-dropzone": "^4.2.3",
    "react-dropzone-component": "^3.0.0",
    "react-moment": "^0.6.5",
    "react-notifications": "^1.4.3",
    "react-nvd3": "^0.5.7",
    "react-router": "^2.8.1",
    "react-router-dom": "^4.2.2",
    "react-scripts": "^1.0.17",
    "react-select": "^1.0.0-rc.10",
    "react-select2-wrapper": "^1.0.4-beta5",
    "reactjs-localstorage": "0.0.5",
    "webpack": "^3.9.1",
    "whatwg-fetch": "^2.0.3"
  },
  "devDependencies": {
    "babel-core": "^6.26.0",
    "babel-loader": "^7.1.2",
    "babel-preset-env": "^1.6.0",
    "babel-preset-stage-0": "^6.24.1",
    "compression-webpack-plugin": "^1.0.1",
    "css-loader": "^0.28.7",
    "eslint": "^4.11.0",
    "eslint-plugin-react": "^7.5.1",
    "style-loader": "^0.19.0",
    "uglifyjs-webpack-plugin": "^1.1.2",
    "webpack": "^3.6.0",
    "webpack-bundle-analyzer": "^2.9.1"
  }
}

预期产出：

data<-data.frame(ID = c('A', 'A', 'A', 'A','A', 'A', 'A', 'A','A','B','B','B','B','B','B','B','B','B'), date = c('2018-13-3 09:20:25','2018-13-3 09:23:42','2018-13-3 09:34:08','2018-13-3 11:25:25','2018-13-3 11:32:25','2018-13-3 11:33:42','2018-13-3 11:34:08','2018-13-3 11:36:25','2018-13-3 11:40:25','2018-13-3 11:23:42','2018-13-3 11:24:08','2018-13-3 11:25:25','2018-13-3 12:32:25','2018-13-3 12:33:42','2018-13-3 15:34:08','2018-13-3 15:36:25','2018-13-3 15:37:25','2018-13-3 15:38:25'))

我目前正在尝试使用未在R中优化的for循环。任何人都可以用更有效的方式帮助我吗？

Answer 1

我们将＆＃39;日期转换为＆＃39;到日期时间类，按＆＃39; ID＆＃39;分组，检查＆＃39; date＆＃39;的相邻元素之间的差异。大于1，得到逻辑向量的累积和

library(dplyr)
library(lubridate)
data %>% 
     mutate(date = ydm_hms(date)) %>%
     group_by(ID) %>% 
     mutate(Trip = cumsum(difftime(date, lag(date,default = date[1]), unit = 'hour')> 1) + 1)

# A tibble: 18 x 3
# Groups: ID [2]
#   ID     date                 Trip
#  <fctr> <dttm>              <dbl>
# 1 A      2018-03-13 09:20:25  1.00
# 2 A      2018-03-13 09:23:42  1.00
# 3 A      2018-03-13 09:34:08  1.00
# 4 A      2018-03-13 11:25:25  2.00
# 5 A      2018-03-13 11:32:25  2.00
# 6 A      2018-03-13 11:33:42  2.00
# 7 A      2018-03-13 11:34:08  2.00
# 8 A      2018-03-13 11:36:25  2.00
# 9 A      2018-03-13 11:40:25  2.00
#10 B      2018-03-13 11:23:42  1.00
#11 B      2018-03-13 11:24:08  1.00
#12 B      2018-03-13 11:25:25  1.00
#13 B      2018-03-13 12:32:25  2.00
#14 B      2018-03-13 12:33:42  2.00
#15 B      2018-03-13 15:34:08  3.00
#16 B      2018-03-13 15:36:25  3.00
#17 B      2018-03-13 15:37:25  3.00
#18 B      2018-03-13 15:38:25  3.00

基于组和列条件的唯一ID

1 个答案: