我有一个看起来像这样的数据集：
Ord_ID Supplier Trans_Type Date
1 A PO 2/3/18
1 A Receipt 2/15/18
2 B PO 2/4/18
2 B Receipt 3/13/18
3 C PO 2/7/18
3 C Receipt 3/1/18
3 C Receipt 3/5/18
3 C Receipt 3/29/18
4 B PO 2/9/18
4 B Receipt 2/20/18
4 B Receipt 2/27/18
5 D PO 2/18/18
5 D Receipt 4/2/18
基本上，我需要将“日期”列分为3个不同的列。我需要一个PO_Date列，该列列出每个订单的最早收货日期，以及每个订单的最后收货日期。由于某些订单只有一个收货日期，因此第二列和第三列应该相同。我尝试使用spread()，但我想是因为每个无效的订单都有不同的收货日期。我该如何做到这一点？
所需结果：
Ord_ID Supplier PO_Date First_Receipt_Date Last_Receipt_Date
1 A 2/3/18 2/15/18 2/15/18
2 B 2/4/18 3/13/18 3/13/18
3 C 2/7/18 3/1/18 3/29/18
4 B 2/9/18 2/20/18 2/27/18
5 D 2/18/18 4/2/18 4/2/18
答案 0 :(得分：3)
使用dplyr。首先，确保列Date为日期格式。假设数据框被命名为mydata：
library(dplyr)
mydata <- mydata %>%
mutate(Date = as.Date(Date, "%m/%d/%y")
现在，您可以过滤Receipt，计算最大/最小日期，然后过滤PO的原始数据并将它们结合在一起：
mydata %>%
filter(Trans_Type == "Receipt") %>%
group_by(Ord_ID, Supplier) %>%
summarise(First_Receipt_Date = min(Date),
Last_Receipt_Date = max(Date)) %>%
ungroup() %>%
left_join(filter(mydata, Trans_Type == "PO")) %>%
select(Ord_ID, Supplier, PO_Date = Date, First_Receipt_Date, Last_Receipt_Date)
结果：
Ord_ID Supplier PO_Date First_Receipt_Date Last_Receipt_Date
<int> <chr> <date> <date> <date>
1 1 A 2018-02-03 2018-02-15 2018-02-15
2 2 B 2018-02-04 2018-03-13 2018-03-13
3 3 C 2018-02-07 2018-03-01 2018-03-29
4 4 B 2018-02-09 2018-02-20 2018-02-27
5 5 D 2018-02-18 2018-04-02 2018-04-02
答案 1 :(得分：2)
使用 <html>
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width,initial-scale=1.0">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.0/css/bootstrap.min.css">
<link rel="stylesheet" type="text/css" href="css/style.css">
<script src="https://code.jquery.com/jquery-2.1.3.min.js"></script>
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.1/js/bootstrap.min.js"></script>
<script src="js/jquery.min.js"></script>
<script src="js/bootstrap.min.js"></script>
<title></title>
</head>
<body>
<div class="container ">
<nav class="navbar navbar-expand-sm navbar-dark bg-dark">
<a href="#" class="navbar-brand"><img src="logo.png">The building co.</a>
<button type="button" class="navbar-toggler" data-toggle="collapse" data-target="#myMenu">
<span class="navbar-toggler-icon"></span>
</button>
<div class="collapse navbar-collapse" id="myMenu">
<ul class="navbar-nav">
<li class="nav-item"> <a href="#" class="navbar-link">Home </a> </li>
<li class="nav-item"> <a href="#" class="navbar-link">contact us </a> </li>
<li class="nav-item"> <a href="#" class="navbar-link">policy </a> </li>
<li class="nav-item"> <a href="#" class="navbar-link">terms </a> </li>
<li class="nav-item"> <a href="#" class="navbar-link">About us </a> </li>
</ul>
</div>
</nav>
</div>
</body>
</html> ，借用@divibisan的示例数据：
tidyverse如果数据未与示例数据中的排序相同，请添加{{ 1}}作为第一步。
答案 2 :(得分：1)
我将从以下内容开始：
data %>%
group_by(Supplier, Trans_Type) %>%
summarise(min_date = min(Date),
max_date = max(Date)
) %>%
ungroup()
然后，您可以使用gather和spread来检索所需的列。
答案 3 :(得分：0)
这是另一个基于tidyverse的解决方案，它避免了left_join。我不知道哪种方法在大型数据集上会更快，但是拥有更多选项总是很好：
df <- structure(list(Ord_ID = c(1L, 1L, 2L, 2L, 3L, 3L, 3L, 3L, 4L,
4L, 4L, 5L, 5L), Supplier = structure(c(1L, 1L, 2L, 2L, 3L, 3L,
3L, 3L, 2L, 2L, 2L, 4L, 4L), .Label = c("A", "B", "C", "D"), class = "factor"),
Trans_Type = c("PO", "Receipt", "PO", "Receipt", "PO", "Receipt",
"Receipt", "Receipt", "PO", "Receipt", "Receipt", "PO", "Receipt"
), Date = structure(c(17565, 17577, 17566, 17603, 17569,
17591, 17595, 17619, 17571, 17582, 17589, 17580, 17623), class = "Date")), row.names = c(NA,
-13L), class = "data.frame")
df %>%
group_by(Ord_ID, Supplier, Trans_Type) %>%
# Keep only min and max date values
filter(Date == min(Date) | Date == max(Date) | Trans_Type != 'Receipt') %>%
# Rename 2nd Receipt value Receipt_2 so there are no duplicated values
mutate(Trans_Type2 = if_else(Trans_Type == 'Receipt' & row_number() == 2,
'Receipt_2', Trans_Type)) %>%
# Drop Trans_Type variable (we can't replace in mutate since it's a grouping var)
ungroup(Trans_Type) %>%
select(-Trans_Type) %>%
# Spread the now unduplicated Trans_Type values
spread(Trans_Type2, Date) %>%
# Fill in Receipt_2 values where they're missing
mutate(Receipt_2 = if_else(is.na(Receipt_2), Receipt, Receipt_2))
# A tibble: 5 x 5
Ord_ID Supplier PO Receipt Receipt_2
<int> <fct> <date> <date> <date>
1 1 A 2018-02-03 2018-02-15 2018-02-15
2 2 B 2018-02-04 2018-03-13 2018-03-13
3 3 C 2018-02-07 2018-03-01 2018-03-29
4 4 B 2018-02-09 2018-02-20 2018-02-27
5 5 D 2018-02-18 2018-04-02 2018-04-02
答案 4 :(得分：0)
您可以仅使用dplyr更改采购订单日期以及第一个和最后一个收货日期的新列：
test1<-test %>%
mutate(Date = mdy(Date)) %>%
group_by(Ord_ID) %>%
mutate(PO_Date = ifelse(Trans_Type == "PO", Date, NA),
Receipt_Date_First = min(Date[Trans_Type=="Receipt"]),
Receipt_Date_Last = max(Date[Trans_Type=="Receipt"])) %>%
filter(!is.na(PO_Date)) %>%
mutate(PO_Date = as.Date(as.numeric(PO_Date)))
明细：
test1<-test %>%
#convert format of "Date" column to as.Date to identify min and max dates
mutate(Date = mdy(Date)) %>%
#group by the Order ID
group_by(Ord_ID) %>%
#PO_Date will be where the "Trans_Type" is "PO" --> since the column is in date format,
#dplyr will convert this to numeric, but can be fixed later
mutate(PO_Date = ifelse(Trans_Type == "PO", Date, NA),
#first receipt date is the minimum date of a receipt transaction
Receipt_Date_First = min(Date[Trans_Type=="Receipt"]),
#last receipt date is the maximum date of a receipt transaction
Receipt_Date_Last = max(Date[Trans_Type=="Receipt"])) %>%
#to remove duplicates
filter(!is.na(PO_Date)) %>%
#convert "PO_Date" column back to as.Date from numeric
mutate(PO_Date = as.Date(as.numeric(PO_Date)))