假设我有以下数据
df = data.frame(name=c("A", "B", "C", "D"), score = c(10, 10, 9, 8))
我想在排名中添加一个新列.这就是我正在做的事情:
df %>% mutate(ranking = rank(score, ties.method = 'first'))
# name score ranking
# 1 A 10 3
# 2 B 10 4
# 3 C 9 2
# 4 D 8 1
但是,我想要的结果是:
# name score ranking
# 1 A 10 1
# 2 B 10 1
# 3 C 9 2
# 4 D 8 3
显然rank没有做我的想法.我应该使用什么功能？
注意:此问题的标题已经过编辑,使其成为plyr功能掩盖其dplyr对应项时的问题的规范问题.问题的其余部分保持不变.
假设我有以下数据:
dfx <- data.frame(
group = c(rep('A', 8), rep('B', 15), rep('C', 6)),
sex = sample(c("M", "F"), size = 29, replace = TRUE),
age = runif(n = 29, min = 18, max = 54)
)
有了旧的,plyr我可以使用以下代码创建一个总结我的数据的小表:
require(plyr)
ddply(dfx, .(group, sex), summarize,
mean = round(mean(age), 2),
sd = round(sd(age), 2))
输出看起来像这样:
group sex mean sd
1 A F 49.68 5.68
2 A M 32.21 6.27
3 B F 31.87 9.80
4 B M 37.54 …我的传单地图看起来像这样:
library(sp)
library(leaflet)
circleFun <- function(center = c(0,0),diameter = 1, npoints = 100){
r = diameter / 2
tt <- seq(0,2*pi,length.out = npoints)
xx <- center[1] + r * cos(tt)
yy <- center[2] + r * sin(tt)
Sr1 = Polygon(cbind(xx, yy))
Srs1 = Polygons(list(Sr1), "s1")
SpP = SpatialPolygons(list(Srs1), 1:1)
return(SpP)
}
Circle.Town <- circleFun(c(1,-1),2.3,npoints = 100)
df1 <- data.frame(long=c(0.6,1,1.4), lat=c(-2, -.8, -0.2), other=c('a', 'b', 'c'), VAM=c(10,8,6),
type=c('Public', 'Public', 'Private'), id=c(1:3)) %>%
mutate(X=paste0('<strong>id: </strong>',
id,
'<br><strong>type</strong>: ',
type,
'<br><strong>VAM</strong>: ',
VAM)) …我的数据看起来像这样
df = data.frame(name=c("A","A","B","B"),
group=c("g1","g2","g1","g2"),
V1=c(10,40,20,30),
V2=c(6,3,1,7))
我想重塑它看起来像这样:
df = data.frame(name=c("A", "B"),
V1.g1=c(10,20),
V1.g2=c(40,30),
V2.g1=c(6,1),
V2.g2=c(3,7))
用tidyR可以做到吗？
我可以用重塑来做到这一点
reshape(df, idvar='name', timevar='group', direction='wide')
但是学习新东西总是好的.
我正在尝试运行一些R代码,因为内存而崩溃.我得到的错误是:
Error in sendMaster(try(lapply(X = S, FUN = FUN, ...), silent = TRUE)) :
long vectors not supported yet: memory.c:3100
产生麻烦的功能如下:
StationUserX <- function(userNDX){
lat1 = deg2rad(geolocation$latitude[userNDX])
long1 = deg2rad(geolocation$longitude[userNDX])
session_user_id = as.character(geolocation$session_user_id[userNDX])
#Find closest station
Distance2Stations <- unlist(lapply(stationNDXs, Distance2StationX, lat1, long1))
# Return index for closest station and distance to closest station
stations_userX = data.frame(session_user_id = session_user_id,
station = ghcndstations$ID[stationNDXs],
Distance2Station = Distance2Stations)
stations_userX = stations_userX[with(stations_userX, order(Distance2Station)), ]
stations_userX = stations_userX[1:100,] #only the 100 closest stations...
row.names(stations_userX)<-NULL
return(stations_userX)
} …我想运行一些回归并用stargazer创建一个表.例如
linear.1 <- lm(rating ~ complaints + privileges + learning + raises + critical, data=attitude)
linear.2 <- lm(rating ~ complaints + privileges + learning, data=attitude)
stargazer(linear.1, linear.2, type="text", keep.stat=c("n"))
对于某些规格,我想在表中添加几行"X".那是:
rating
(1) (2)
-----------------------------------------------------------------
complaints 0.692*** 0.682***
(0.149) (0.129)
privileges -0.104 -0.103
(0.135) (0.129)
learning 0.249 0.238*
(0.160) (0.139)
raises -0.033
(0.202)
critical 0.015
(0.147)
Constant 11.011 11.258
(11.704) (7.318)
Note 1 X
Note 2 X
-----------------------------------------------------------------
Observations 30 30
=================================================================
Note: *p<0.1; **p<0.05; ***p<0.01
我怎样才能做到这一点？
谢谢!
我想用它lubridate来计算出生日期和今天的日期.现在我有这个:
library(lubridate)
today<-mdy(08312015)
dob<-mdy(09071982)
today-dob
这给了我他们几天的年龄.
我有以下图表:
library(ggplot2)
library(scales)
library(magrittr)
df1 <-
structure(
list(
x = structure(
1:5, .Label = c("5", "4", "3", "2",
"1"), class = "factor"
), y = c(
0.166666666666667, 0.361111111111111,
0.0833333333333333, 0.222222222222222, 0.291666666666667
)
), .Names = c("x",
"y"), row.names = c(NA,-5L), class = c("tbl_df", "tbl", "data.frame"), drop = TRUE
)
df1 %>% ggplot(aes(x , y )) + geom_bar(stat = "identity") +
scale_y_continuous(labels = percent)
我想在5和1之下添加带有粗体文本的两行注释.例如,'最高\nvalue'低于5,'最低\n值'低于1.
我试过geom_text但我不能把文字放在我想要的地方.
我想创建一个空列表,以便我可以用其他列表替换它的元素.
例如
simulations = 10
seeds = sample(10000:99999, simulations, replace=F)
test_function <- function(seedX){
lista = list(seedX=seedX,
dataframe=data.frame(x=runif(10, -5.0, 5.0),y=rnorm(10,mean = 0,sd = 1)))
return(lista)
}
results <- vector("list", simulations)
results[1] = test_function(seedX = seeds[1])
我收到以下错误:
Warning message:
In results[1] = test_function(seedX = seeds[1]) :
number of items to replace is not a multiple of replacement length
我究竟做错了什么？
谢谢!