本文介绍了缺少值组合的完整数据框的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个包含两个因素(distance
)和年份(years
)的数据框.我想将每个因子的所有 years
值都补 0.
I have a dataframe with two factors (distance
) and years (years
). I would like to complete all years
values for every factor by 0.
即从此:
distance years area
1 NPR 3 10
2 NPR 4 20
3 NPR 7 30
4 100 1 40
5 100 5 50
6 100 6 60
得到这个:
distance years area
1 NPR 1 0
2 NPR 2 0
3 NPR 3 10
4 NPR 4 20
5 NPR 5 0
6 NPR 6 0
7 NPR 7 30
8 100 1 40
9 100 2 0
10 100 3 0
11 100 4 0
12 100 5 50
13 100 6 60
14 100 7 0
我尝试应用 expand()
函数:
library(tidyr)
library(dplyr, warn.conflicts = FALSE)
expand(df, years = 1:7)
但这只会生成一列数据框,并不会扩展原始数据框:
but this just produces one column dataframe and does not expand the original one:
# A tibble: 7 x 1
years
<int>
1 1
2 2
3 3
4 4
5 5
6 6
7 7
或 expand.grid()
也不起作用:
require(utils)
expand.grid(df, years = 1:7)
Error in match.names(clabs, names(xi)) :
names do not match previous names
In addition: Warning message:
In format.data.frame(x, digits = digits, na.encode = FALSE) :
corrupt data frame: columns will be truncated or padded with NAs
是否有一种简单的方法来扩展
我的数据框?以及如何基于distance
和uniqueLoc
这两个类来展开?
Is there a simple way to expand
my dataframe? And how to expand it based on two categories: distance
and uniqueLoc
?
distance <- rep(c("NPR", "100"), each = 3)
years <-c(3,4,7, 1,5,6)
area <-seq(10,60,10)
uniqueLoc<-rep(c("a", "b"), 3)
df<-data.frame(uniqueLoc, distance, years, area)
> df
uniqueLoc distance years area
1 a NPR 3 10
2 b NPR 4 20
3 a NPR 7 30
4 b 100 1 40
5 a 100 5 50
6 b 100 6 60
推荐答案
你可以使用tidyr::complete
函数:
complete(df, distance, years = full_seq(years, period = 1), fill = list(area = 0))
# A tibble: 14 x 3
distance years area
<fct> <dbl> <dbl>
1 100 1. 40.
2 100 2. 0.
3 100 3. 0.
4 100 4. 0.
5 100 5. 50.
6 100 6. 60.
7 100 7. 0.
8 NPR 1. 0.
9 NPR 2. 0.
10 NPR 3. 10.
11 NPR 4. 20.
12 NPR 5. 0.
13 NPR 6. 0.
14 NPR 7. 30.
或更短:
complete(df, distance, years = 1:7, fill = list(area = 0))
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