我有一些数据，其中包含日期和名称，以及一个我想订购的任务，可以计算出人们执行的任务的顺序和任务的流程。 因此，非常简单的是一些示例数据。
Name Date Food
Fred 01/01/2018 Peanuts
Jim 03/02/2018 Banana
Barney 02/02/2018 Rice
Fred 06/03/2018 Rice
Barry 12/02/2018 Peanuts
John 04/04/2018 Rice
Jim 03/03/2018 Rice
Fred 20/04/2018 Rice
Den 12/02/2018 Banana
Barney 04/05/2018 Banana
Jim 05/06/2018 Rice
John 06/07/2018 Peanuts
Jim 30/06/2018 Banana
Fred 05/05/2018 Rice
这给了我每个指定人士吃指定食物的日期。我想为每个人提供的是他们所吃食物的完整清单以及他们进餐的顺序。
我在R中使用了order函数，并使用从1到nrow的顺序创建了一个seq以获取顺序，但是我不知道如何为每个人获取该序列。
第二步，我想创建一个流表并记录每个流的记录次数，因此最终结果将是一个这样的表。
Flow count
Peanuts to rice 1
Peanuts to banana 0
Peanuts to peanuts 0
Rice to peanuts 1
Rice to banana 2
Rice to rice 3
Banana to rice 1
Banana to peanuts 0
Banana to banana 0
谢谢
更新：
与这些事情一样，我越深入研究某件事，我就越希望对数据进行更改！
因此，下面提供的答案已经给了我我想要的流程图-谢谢。 现在，我想做的是能够编辑原始数据框，以删除我不感兴趣或不想分析的流实例。
例如，我如何从数据帧中删除所有从水稻流向花生或从香蕉流向水稻的流（不论人流）？
答案 0 :(得分：3)
让您的数据框为dat，并假定：
Date列的升序排序（或Date在Name内排序，如您目前所拥有的那样）； Name和Food是因子列。
## split by person; not to be messed up by "between person" flow
x <- split(levels(dat$Food)[dat$Food], dat$Name)
#$Barney
#[1] "Rice" "Banana"
#
#$Barry
#[1] "Peanuts"
#
#$Den
#[1] "Banana"
#
#$Fred
#[1] "Peanuts" "Rice" "Rice" "Rice"
#
#$Jim
#[1] "Banana" "Rice" "Rice" "Banana"
#
#$John
#[1] "Rice" "Peanuts"
方法1
getFlow1 <- function (u) {
if (length(u) == 1L) NULL
else paste(u[-length(u)], u[-1], sep = " to ")
}
Flow1 <- unlist(lapply(x, getFlow1), use.names = FALSE)
#[1] "Rice to Banana" "Peanuts to Rice" "Rice to Rice" "Rice to Rice"
#[5] "Banana to Rice" "Rice to Rice" "Rice to Banana" "Rice to Peanuts"
## maybe you can control the order of factor levels here
All_Flow <- outer(levels(dat$Food), levels(dat$Food), paste, sep = " to ")
Flow1 <- table("Flow" = factor(Flow1, levels = All_Flow))
#Flow
# Banana to Banana Peanuts to Banana Rice to Banana Banana to Peanuts
# 0 0 2 0
#Peanuts to Peanuts Rice to Peanuts Banana to Rice Peanuts to Rice
# 0 1 1 1
# Rice to Rice
# 3
as.data.frame(Flow1)
# Flow Freq
#1 Banana to Banana 0
#2 Peanuts to Banana 0
#3 Rice to Banana 2
#4 Banana to Peanuts 0
#5 Peanuts to Peanuts 0
#6 Rice to Peanuts 1
#7 Banana to Rice 1
#8 Peanuts to Rice 1
#9 Rice to Rice 3
方法2（我更喜欢这种方法）
getFlow2 <- function (u) {
if (length(u) == 1L) NULL
else cbind(u[-length(u)], u[-1])
}
Flow2 <- do.call("rbind", lapply(x, getFlow2))
# [, 1] [, 2]
#[1,] "Rice" "Banana"
#[2,] "Peanuts" "Rice"
#[3,] "Rice" "Rice"
#[4,] "Rice" "Rice"
#[5,] "Banana" "Rice"
#[6,] "Rice" "Rice"
#[7,] "Rice" "Banana"
#[8,] "Rice" "Peanuts"
Flow2 <- table("From" = Flow2[, 1], "To" = Flow2[, 2])
# To
#From Banana Peanuts Rice
# Banana 0 0 1
# Peanuts 0 0 1
# Rice 2 1 3
as.data.frame(Flow2)
# From To Freq
#1 Banana Banana 0
#2 Peanuts Banana 0
#3 Rice Banana 2
#4 Banana Peanuts 0
#5 Peanuts Peanuts 0
#6 Rice Peanuts 1
#7 Banana Rice 1
#8 Peanuts Rice 1
#9 Rice Rice 3
答案 1 :(得分：0)
这是完整的tidyverse解决方案。
library(tidyverse)
data <-
tribble(~Name, ~Date, ~Food,
"Fred", "01/01/2018", "Peanuts",
"Jim", "03/02/2018", "Banana",
"Barney", "02/02/2018", "Rice",
"Fred", "06/03/2018", "Rice",
"Barry", "12/02/2018", "Peanuts",
"John", "04/04/2018", "Rice",
"Jim", "03/03/2018", "Rice",
"Fred", "20/04/2018", "Rice",
"Den", "12/02/2018", "Banana",
"Barney", "04/05/2018", "Banana",
"Jim", "05/06/2018", "Rice",
"John", "06/07/2018", "Peanuts",
"Jim", "30/06/2018", "Banana",
"Fred", "05/05/2018", "Rice")
首先，我们将日期转换为正确的格式。
data_clean <-
data %>%
mutate(Date = as.Date(Date, "%d/%m/%Y"))
data_clean
然后我们获得每个人与arrange，summarise和str_c(..., collapse = ", ")吃的食物清单。
list_of_food_by_person <-
data_clean %>%
group_by(Name) %>%
distinct(Name, Food) %>%
arrange(Food) %>%
summarise(List = str_c(Food, collapse = ", "))
list_of_food_by_person
# A tibble: 6 x 2
Name List
<chr> <chr>
1 Barney Banana, Rice
2 Barry Peanuts
3 Den Banana
4 Fred Peanuts, Rice
5 Jim Banana, Rice
6 John Peanuts, Rice
同样，我们使用str_c()来获得人均食品流量。
flow_of_food_per_person <-
data_clean %>%
arrange(Date) %>%
group_by(Name) %>%
summarise(Flow = str_c(Food, collapse = " to "))
flow_of_food_per_person
# A tibble: 6 x 2
Name Flow
<chr> <chr>
1 Barney Rice to Banana
2 Barry Peanuts
3 Den Banana
4 Fred Peanuts to Rice to Rice to Rice
5 Jim Banana to Rice to Rice to Banana
6 John Rice to Peanuts
最后，我们用group_by()和sequence(n())得到每人每个项目的顺序。我实际上并没有利用此顺序，但是您已经要求创建该顺序的方法。我所做的只是使用lag()来获取上一个食物，然后使用str_glue()将其放入字符串值。
flow_count <-
data_clean %>%
arrange(Date) %>%
group_by(Name) %>%
mutate(Order = sequence(n())) %>%
mutate(Previous = lag(Food),
Flow = str_glue("{Previous} to {Food}")) %>%
ungroup() %>%
filter(!is.na(Previous)) %>%
count(Flow)
flow_count
# A tibble: 5 x 2
Flow n
<chr> <int>
1 Banana to Rice 1
2 Peanuts to Rice 1
3 Rice to Banana 2
4 Rice to Peanuts 1
5 Rice to Rice 3