我试图将带有列名的对象传递给spread函数,但是它不是读取对象内部的值而只是尝试使用对象名本身
这里只是一个玩具示例
library(tidyr)
d = (1:4)
n = c("a"," a", "b","b")
s = c(1, 2,5,7)
df = data.frame(d,n, s)
Value <- n
data_wide <- spread(df, Value , s)
错误:键列&#39;值&#39;在输入中不存在。
虽然下面工作正常:
data_wide <- spread(df, n, s)
d a a b
1 1 NA 1 NA
2 2 2 NA NA
3 3 NA NA 5
4 4 NA NA 7
答案 0 :(得分:13)
我们可以使用spread_()
将变量名称作为字符串传递:
library(tidyr)
# dummy data
df1 <- data.frame(d = (1:4),
n = c("a", "a", "b", "b") ,
s = c(1, 2, 5, 7))
myKey <- "n"
myValue <- "s"
spread_(data = df1, key_col = myKey , value_col = myValue)
答案 1 :(得分:3)
使用data.table
library(data.table)
dcast(setDT(df), eval(as.name(myValue))~ eval(as.name(myKey)), value.var=myValue)
关于在tidyr
函数中传递名称,这个answer也可以提供帮助(已经发布了几个小时)。
答案 2 :(得分:0)
看起来tidyr
现在可以自动识别-它可以同时进行两种操作:输入列名或将其放入字符串
stocks <- data.frame(
time = as.Date('2009-01-01') + 0:9,
X = rnorm(10, 0, 1),
Y = rnorm(10, 0, 2),
Z = rnorm(10, 0, 4)
)
stocksm <- stocks %>% gather(stock, price, -time) # make it to skinny table
# make it wide table
stocksm %>% spread(stock, price)
# time X Y Z
# 1 2009-01-01 0.7444343 -0.07030219 0.9140019
# 2 2009-01-02 1.1988507 2.98659296 5.3044361
# 3 2009-01-03 -0.4344259 -0.11526884 -3.8380602
# 4 2009-01-04 0.8154400 2.08313458 -0.1152524
# 5 2009-01-05 1.1965647 -0.59055846 3.5647410
# ...
# it's same if put column names in strings
stocksm %>% spread('stock', 'price')
# time X Y Z
# 1 2009-01-01 0.7444343 -0.07030219 0.9140019
# 2 2009-01-02 1.1988507 2.98659296 5.3044361
# 3 2009-01-03 -0.4344259 -0.11526884 -3.8380602
# 4 2009-01-04 0.8154400 2.08313458 -0.1152524
# 5 2009-01-05 1.1965647 -0.59055846 3.5647410
# or put in string variables
col1 = 'stock'
col2 = 'price'
stocksm %>% spread(col1, col2)
# time X Y Z
# 1 2009-01-01 0.7444343 -0.07030219 0.9140019
# 2 2009-01-02 1.1988507 2.98659296 5.3044361
# 3 2009-01-03 -0.4344259 -0.11526884 -3.8380602
# 4 2009-01-04 0.8154400 2.08313458 -0.1152524
# 5 2009-01-05 1.1965647 -0.59055846 3.5647410