ran*_*ane 7 r date list unnest mutate
我想在变量包含不同的YEAR时拆分行,也拆分col:"Price"用均匀的除以日期的数字 - > count(";")+1
有一个表尚未拆分变量.
# Dataset call df
Price Date
500 2016-01-01
400 2016-01-03;2016-01-09
1000 2016-01-04;2017-09-01;2017-08-10;2018-01-01
25 2016-01-04;2017-09-01
304 2015-01-02
238 2018-01-02;2018-02-02
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欲望展望
# Targeted df
Price Date
500 2016-01-01
400 2016-01-03;2016-01-09
250 2016-01-04
250 2017-09-01
250 2017-08-10
250 2018-01-01
12.5 2016-01-04
12.5 2017-09-01
304 2015-01-02
238 2018-01-02;2018-02-02
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一旦变量包含不同的年份定义,下面是操作必须做的.(这只是一个例子.)
mutate(Price = ifelse(DIFFERENT_DATE_ROW,
as.numeric(Price) / (str_count(Date,";")+1),
as.numeric(Price)),
Date = ifelse(DIFFERENT_DATE_ROW,
strsplit(as.character(Date),";"),
Date)) %>%
unnest()
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我遇到了一些不能使用dplyr函数的约束,"if_else"因为
否则无法识别NO操作.只有ifelse正常工作.
如何找出一个变量中的年份差异来PROVOKE分割线和拆分价格计算?
到目前为止分裂元素的操作就像
unlist(lapply(unlist(strsplit(df1$noFDate[8],";")),FUN = year))
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无法解决问题.
我是编码的初学者,请考虑真实数据超过200万行和50列,请随意更改上述所有操作.
这可能不是最有效的方法,但可以用来获得所需的答案。
#Get the row indices which we need to separate
inds <- sapply(strsplit(df$Date, ";"), function(x)
#Format the date into year and count number of unique values
#Return TRUE if number of unique values is greater than 1
length(unique(format(as.Date(x), "%Y"))) > 1
)
library(tidyverse)
library(stringr)
#Select those indices
df[inds, ] %>%
# divide the price by number of dates in that row
mutate(Price = Price / (str_count(Date,";") + 1)) %>%
# separate `;` delimited values in separate rows
separate_rows(Date, sep = ";") %>%
# bind the remaining rows as it is
bind_rows(df[!inds,])
# Price Date
#1 250.0 2016-01-04
#2 250.0 2017-09-01
#3 250.0 2017-08-10
#4 250.0 2018-01-01
#5 12.5 2016-01-04
#6 12.5 2017-09-01
#7 500.0 2016-01-01
#8 400.0 2016-01-03;2016-01-09
#9 304.0 2015-01-02
#10 238.0 2018-01-02;2018-02-02
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