Rah*_*hah 6 r ggplot2 dataframe dplyr
我有一个数据框,看起来像:
DF_1>
T_id D1 D2 Num type type_2 fig
xt-1 2017-05-01 2017-03-25 12:11:45 10 A X 25.20
xt-2 2017-05-01 2017-03-25 21:05:25 20 A Y 20.15
xt-3 2017-05-01 2017-03-25 08:10:55 25 B X 15.11
xt-4 2017-05-03 2017-03-25 07:19:35 30 B Y 22.56
xt-5 2017-05-03 2017-03-25 13:12:56 45 C Z 35.45
xt-6 2017-05-03 2017-03-25 18:14:44 20 D Z 27.21
xt-7 2017-04-06 2017-03-25 19:21:35 15 A Z 23.20
xt-8 2017-04-06 2017-03-25 21:11:15 40 C X 21.40
xt-9 2017-04-08 2017-02-25 22:25:04 20 A A 27.50
xt-10 2017-04-06 2017-02-25 16:04:08 30 A Y 32.20
xt-11 2017-04-05 2017-02-25 18:15:25 20 C Z 30.20
xt-12 2017-04-01 2017-01-25 19:22:25 50 A Z 33.15
xt-13 2017-04-02 2017-01-25 23:19:05 15 A A 30.12
xt-14 2017-03-03 2017-01-25 14:25:09 15 D Y 31.25
xt-15 2017-03-10 2017-01-25 23:25:36 40 A X 25.45
Run Code Online (Sandbox Code Playgroud)
从上面的数据帧我想要下面提到的两个矩阵:
1. Date (Last Three Date from `sys.date()`)
D1 count sum mean_num total_sum count_A sum_A count_other sum_other mean_fig mean_TAT
2017-05-03 3 95 31.66 6 0 0 3 95 28.40
2017-05-02 0 0 0 3 0 0 0 0 0.00
2017-05-01 3 55 18.33 3 2 30 1 25 20.15
Run Code Online (Sandbox Code Playgroud)
mean_TAT:减去D2- D1并且根据count同一日期的值取当天的平均值. total_sum 将从该月的第一个日期累积.count_A并sum_A基于type作为A针对特定的一天.count_other而sum_other对于那些type以外的人A.2.基于月份(根据数据框架的最后三个月)
type_2基于特定月份的计数的前三个.increase_%将在上个月计算(即,如果count5月17日是50比4月17日100,那么将是-50%,并且基于其前一个月count和其他5行相同sum.A对于type_2"A" 的值,第四个将是每个月不变的.Other个type_2是上面提到的那个以外的第五个.Total将按照列,count并将sum有添加,因为mean将有意思.似乎我无法正确解释,希望数据框架可以理解矩阵.
期待一些帮助.
这是第一部分:
library(lubridate)
library(dplyr)
df2 <- df1 %>%
mutate(ym = year(D1)*100+month(D1)) %>%
arrange(D1) %>%
group_by(D1,ym) %>%
summarize(count = n(),
sum=sum(Num),
mean_num=mean(Num),
count_A=sum(type=='A'),
sum_A=sum(Num * (type=='A')),
count_other=sum(type!='A'),
sum_other=sum(Num * (type!='A')),
mean_fig = mean(fig),
mean_TAT = mean(D2-D1)) %>%
group_by(ym) %>%
mutate(total_sum=cumsum(count)) %>%
ungroup %>%
arrange(desc(D1)) %>%
select(D1,count,sum,mean_num,total_sum,count_A,sum_A,count_other,sum_other,mean_fig,mean_TAT)
# # A tibble: 9 x 11
# D1 count sum mean_num total_sum count_A sum_A count_other sum_other mean_fig mean_TAT
# <date> <int> <int> <dbl> <int> <int> <int> <int> <int> <dbl> <time>
# 1 2017-05-03 3 95 31.66667 6 0 0 3 95 28.40667 -39.00000 days
# 2 2017-05-01 3 55 18.33333 3 2 30 1 25 20.15333 -37.00000 days
# 3 2017-04-08 1 20 20.00000 7 1 20 0 0 27.50000 -42.00000 days
# 4 2017-04-06 3 85 28.33333 6 2 45 1 40 25.60000 -21.33333 days
# 5 2017-04-05 1 20 20.00000 3 0 0 1 20 30.20000 -39.00000 days
# 6 2017-04-02 1 15 15.00000 2 1 15 0 0 30.12000 -67.00000 days
# 7 2017-04-01 1 50 50.00000 1 1 50 0 0 33.15000 -66.00000 days
# 8 2017-03-10 1 40 40.00000 2 1 40 0 0 25.45000 -44.00000 days
# 9 2017-03-03 1 15 15.00000 1 0 0 1 15 31.25000 -37.00000 days
Run Code Online (Sandbox Code Playgroud)
数据
df1 <- read.table(text="T_id D1 D2 Num type type_2 fig
xt-1 2017-05-01 '2017-03-25 12:11:45' 10 A X 25.20
xt-2 2017-05-01 '2017-03-25 21:05:25' 20 A Y 20.15
xt-3 2017-05-01 '2017-03-25 08:10:55' 25 B X 15.11
xt-4 2017-05-03 '2017-03-25 07:19:35' 30 B Y 22.56
xt-5 2017-05-03 '2017-03-25 13:12:56' 45 C Z 35.45
xt-6 2017-05-03 '2017-03-25 18:14:44' 20 D Z 27.21
xt-7 2017-04-06 '2017-03-25 19:21:35' 15 A Z 23.20
xt-8 2017-04-06 '2017-03-25 21:11:15' 40 C W 21.40
xt-9 2017-04-08 '2017-02-25 22:25:04' 20 A Q 27.50
xt-10 2017-04-06 '2017-02-25 16:04:08' 30 A W 32.20
xt-11 2017-04-05 '2017-02-25 18:15:25' 20 C V 30.20
xt-12 2017-04-01 '2017-01-25 19:22:25' 50 A Z 33.15
xt-13 2017-04-02 '2017-01-25 23:19:05' 15 A Z 30.12
xt-14 2017-03-03 '2017-01-25 14:25:09' 15 D Y 31.25
xt-15 2017-03-10 '2017-01-25 23:25:36' 40 A X 25.45",h=T,strin=F)
df1$D1 <- as.Date(df1$D1,"%Y-%m-%d")
df1$D2 <- as.Date(df1$D2,"%Y-%m-%d")
expected_output <- read.table(text="D1 count sum mean_num total_sum count_A sum_A count_other sum_other mean_fig
2017-05-03 3 95 31.66 6 0 0 3 95 28.40
2017-05-02 0 0 0 3 0 0 0 0 0.00
2017-05-01 3 55 18.33 3 2 30 1 25 20.15")
Run Code Online (Sandbox Code Playgroud)
第2部分的一些提示:
如果不重新处理你的问题,我就无法创造奇迹(在这里提供准确的可重复输出是必要的).但是这里有一种接近的方式,希望如此:
df_month <- df1 %>%
mutate(ym = year(D1)*100+month(D1)) %>%
arrange(D1) %>%
group_by(ym) %>%
summarize(count = n(),
sum=sum(Num),
mean_num=mean(Num),
count_A=sum(type=='A'),
sum_A=sum(Num * (type=='A')),
count_other=sum(type!='A'),
sum_other=sum(Num * (type!='A')),
mean_fig = mean(fig),
mean_TAT = mean(D2-D1)) %>%
mutate(type_2=paste0(month.abb[ym%% 100],"-",ym %/% 100 -2000)) %>%
select(ym,type_2,count,sum,mean_num,count_A,sum_A,count_other,sum_other,mean_fig,mean_TAT)
df_top3 <- df1 %>%
filter(type_2 !="A") %>%
mutate(ym = year(D1)*100+month(D1)) %>%
arrange(desc(ym)) %>%
group_by(ym,type_2) %>%
summarize(count = n(),
sum=sum(Num),
mean_num=mean(Num),
count_A=sum(type=='A'),
sum_A=sum(Num * (type=='A')),
count_other=sum(type!='A'),
sum_other=sum(Num * (type!='A')),
mean_fig = mean(fig),
mean_TAT = mean(D2-D1)) %>%
group_by(ym) %>%
arrange(desc(count)) %>%
slice(1:3) %>%
ungroup %>%
select(ym,type_2,count,sum,mean_num,count_A,sum_A,count_other,sum_other,mean_fig,mean_TAT)
df_A <- df1 %>%
filter(type_2 == "A") %>%
mutate(ym = year(D1)*100+month(D1)) %>%
arrange(desc(ym)) %>%
group_by(ym,type_2) %>%
summarize(count = n(),
sum=sum(Num),
mean_num=mean(Num),
count_A=sum(type=='A'),
sum_A=sum(Num * (type=='A')),
count_other=sum(type!='A'),
sum_other=sum(Num * (type!='A')),
mean_fig = mean(fig),
mean_TAT = mean(D2-D1)) %>%
select(ym,type_2,count,sum,mean_num,count_A,sum_A,count_other,sum_other,mean_fig,mean_TAT)
df_other <- df1 %>%
mutate(ym = year(D1)*100+month(D1)) %>%
anti_join(bind_rows(df_top3,df_A),by = c("ym","type_2")) %>%
mutate(type_2="Other") %>%
arrange(desc(ym)) %>%
group_by(ym,type_2) %>%
summarize(count = n(),
sum=sum(Num),
mean_num=mean(Num),
count_A=sum(type=='A'),
sum_A=sum(Num * (type=='A')),
count_other=sum(type!='A'),
sum_other=sum(Num * (type!='A')),
mean_fig = mean(fig),
mean_TAT = mean(D2-D1)) %>%
select(ym,type_2,count,sum,mean_num,count_A,sum_A,count_other,sum_other,mean_fig,mean_TAT)
# it's empty with your example data
bind_rows(df_month,df_top3,df_A,df_other) %>%
arrange(ym) %>%
select(-ym) %>%
rename(Month = type_2)
Run Code Online (Sandbox Code Playgroud)