将R中的组的中位数分别乘以条件 [英] multiply median for groups separately in R by condition
本文介绍了将R中的组的中位数分别乘以条件的处理方法，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
我有这个数据集
df=structure(list(Dt = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L,
9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L,
22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L,
35L, 36L, 37L, 38L, 39L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L,
10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L,
23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L,
36L, 37L, 38L, 39L), .Label = c("2018-02-20 00:00:00.000", "2018-02-21 00:00:00.000",
"2018-02-22 00:00:00.000", "2018-02-23 00:00:00.000", "2018-02-24 00:00:00.000",
"2018-02-25 00:00:00.000", "2018-02-26 00:00:00.000", "2018-02-27 00:00:00.000",
"2018-02-28 00:00:00.000", "2018-03-01 00:00:00.000", "2018-03-02 00:00:00.000",
"2018-03-03 00:00:00.000", "2018-03-04 00:00:00.000", "2018-03-05 00:00:00.000",
"2018-03-06 00:00:00.000", "2018-03-07 00:00:00.000", "2018-03-08 00:00:00.000",
"2018-03-09 00:00:00.000", "2018-03-10 00:00:00.000", "2018-03-11 00:00:00.000",
"2018-03-12 00:00:00.000", "2018-03-13 00:00:00.000", "2018-03-14 00:00:00.000",
"2018-03-15 00:00:00.000", "2018-03-16 00:00:00.000", "2018-03-17 00:00:00.000",
"2018-03-18 00:00:00.000", "2018-03-19 00:00:00.000", "2018-03-20 00:00:00.000",
"2018-03-21 00:00:00.000", "2018-03-22 00:00:00.000", "2018-03-23 00:00:00.000",
"2018-03-24 00:00:00.000", "2018-03-25 00:00:00.000", "2018-03-26 00:00:00.000",
"2018-03-27 00:00:00.000", "2018-03-28 00:00:00.000", "2018-03-29 00:00:00.000",
"2018-03-30 00:00:00.000"), class = "factor"), ItemRelation = c(158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L
), stuff = c(200L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 3600L,
0L, 0L, 0L, 0L, 700L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 1000L, 2600L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 400L, 700L,
200L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 3600L, 0L, 0L, 0L,
0L, 700L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1000L,
2600L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 400L, 700L), num = c(1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L), year = c(2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L), action = c(0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 1L, 1L, 1L, 1L)), .Names = c("Dt", "ItemRelation",
"stuff", "num", "year", "action"), class = "data.frame", row.names = c(NA,
-78L))
对该数据执行了下一个操作. 1.按事物计算动作的第一类和动作零类的中位数的操作(最后五个非零观测值). 2.然后从第一类别的中位数减去零类别的中位数. MKR的解决方案非常准确.
The next operation was performed on this data. 1. the operation of calculating the median for the first category of the action and the zero category of action by stuff ( last five non-zero observations). 2.then the median of the zero category was subtracted from the median in the first category. Solution of MKR is very accurate.
library(dplyr)
df %>% filter(stuff > 0) %>% #First filter out for stuff > 0 which of our interest
group_by(ItemRelation, num, year) %>%
mutate(m = median(stuff[action==1]),
m0 = median(tail(stuff[action==0], 5))) %>% # Calculate m and m0 for all rows
filter(action == 1) %>% # Now keep only rows with action == 1
mutate(m = m-m0) %>%
select(-Dt,-m0,-action
该如何将每个组的计算结果乘以一个动作的数量，但是仅对那些大于零的对象乘以一个动作. 例如，对于阶层
How to do that the calculated result for each group was multiplied by the number of ones by action, but only for those that are more than zero by stuff. For example, for stratum
ItemRelation num year
158043 1459 2018
我们有4个在操作中，只有2个在其他方面大于零 因此我们将计算结果(m)乘以2.
we have 4 ones in action, and only two ones by stuff more then zero so the calculated result (m) we multiply on two.
推荐答案
数据已经过滤了dplyr - chain中的stuff>0. n()代表每个组的计数，其中stuff>0和action ==1.因此，可以将m的最终值乘以n().最后，distinct将确保已删除重复的行.
Data is already filter for stuff>0 in dplyr - chain. Then() represent the count per group where stuff>0 and action ==1. Hence, one can multiply the final value of m with n(). At the end, distinct will ensure that duplicate rows has been removed.
library(dplyr)
df %>% filter(stuff > 0) %>% #First filter out for stuff > 0 which of our interest
group_by(ItemRelation, num, year) %>%
mutate(m = median(stuff[action==1]),
m0 = median(tail(stuff[action==0], 5))) %>% # Calculate m and m0 for all rows
filter(action == 1) %>% # Now keep only rows with action == 1
mutate(m = (m-m0)*n()) %>%
select(-Dt,-m0,-action, - stuff) %>% distinct()
# # A tibble: 2 x 4
# # Groups: ItemRelation, num, year [2]
# ItemRelation num year m
# <int> <int> <int> <dbl>
# 1 158043 1459 2018 -900
# 2 234 1459 2018 -900
这篇关于将R中的组的中位数分别乘以条件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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