首先，首先。以下是我的数据：
lat <- c(12, 12, 58, 58, 58, 58, 58, 45, 45, 45, 45, 45, 45, 64, 64, 64, 64, 64, 64, 64)
long <- c(-14, -14, 139, 139, 139, 139, 139, -68, -68, -68, -68, -68, 1, 1, 1, 1, 1, 1, 1, 1)
sex <- c("M", "M", "M", "M", "F", "M", "M", "F", "M", "M", "M", "F", "M", "F", "M", "F", "F", "F", "F", "M")
score <- c(2, 6, 3, 6, 5, 4, 3, 2, 3, 9, 9, 8, 6, 5, 6, 7, 5, 7, 5, 1)
data <- data.frame(lat, long, sex, score)
数据应如下所示：
lat long sex score
1 12 -14 M 2
2 12 -14 M 6
3 58 139 M 3
4 58 139 M 6
5 58 139 F 5
6 58 139 M 4
7 58 139 M 3
8 45 -68 F 2
9 45 -68 M 3
10 45 -68 M 9
11 45 -68 M 9
12 45 -68 F 8
13 45 1 M 6
14 64 1 F 5
15 64 1 M 6
16 64 1 F 7
17 64 1 F 5
18 64 1 F 7
19 64 1 F 5
20 64 1 M 1
我最终试图弄明白这一点。变量是纬度，经度，性别和分数。我希望每个地点都有相同数量的男性和女性（即经度和纬度相同）。例如，第二个位置（第3行到第7行）只有一个女性。应保留这名女性，并保留其余个体的一名男性（也许是随机抽样）。有些地方只有一种性别的信息，例如第一个位置（第1行和第2行）只有男性数据。应删除此位置的行（因为没有女性）。所有按计划进行的最终数据集应如下所示：
lat2 long2 sex2 score2
1 58 139 F 5
2 58 139 M 4
3 45 -68 F 2
4 45 -68 M 3
5 45 -68 M 9
6 45 -68 F 8
7 64 1 M 6
8 64 1 F 5
9 64 1 F 7
10 64 1 M 1
任何帮助都将不胜感激。
答案 0 :(得分：5)
以下是lapply的解决方案：
data[unlist(lapply(with(data, split(seq.int(nrow(data)), paste(lat, long))),
# 'split' splits the sequence of row numbers (indices) along the unique
# combinations of 'lat' and 'long'
# 'lapply' applies the following function to all sub-sequences
function(x) {
# which of the indices are for males:
male <- which(data[x, "sex"] == "M")
# which of the indices are for females:
female <- which(data[x, "sex"] == "F")
# sample from the indices of males:
s_male <- sample(male, min(length(male), length(female)))
# sample from the indices of females:
s_female <- sample(female, min(length(male), length(female)))
# combine both sampled indices:
x[c(s_male, s_female)]
})), ]
# The function 'lappy' returns a list of indices which is transformed to a vector
# using 'unlist'. These indices are used to subset the original data frame.
结果：
lat long sex score
9 45 -68 M 3
11 45 -68 M 9
12 45 -68 F 8
8 45 -68 F 2
7 58 139 M 3
5 58 139 F 5
20 64 1 M 1
15 64 1 M 6
19 64 1 F 5
16 64 1 F 7
答案 1 :(得分：2)
下面是一个快速的方法，包括创建一个lat-long组合的临时列。我们根据此列拆分DF，计算每个拆分中的M / F，适当地采样，然后重新组合。
# First, We call the dataframe something other than "data" ;)
mydf <- data.frame(lat, long, sex, score)
# create a new data frame with a temporary column, which concatenates the lat & long.
mydf.new <- data.frame(mydf, latlong=paste(mydf$lat, mydf$long, sep=","))
# Split the data frame according to the lat-long location
mydf.splat <- split(mydf.new, mydf.new$latlong)
# eg, taking a look at one of our tables:
mydf.splat[[4]]
sampled <-
lapply(mydf.splat, function(tabl) {
Ms <- sum(tabl$sex=="M")
Fs <- sum(tabl$sex=="F")
if(Fs == 0 || Ms ==0) # If either is zero, we drop that location
return(NULL)
if(Fs == Ms) # If they are both equal, no need to sample.
return(tabl)
# If number of Females less than Males, return all Females
# and sample from males in ammount equal to Females
if (Fs < Ms)
return(tabl[c(which(tabl$sex=="F"), sample(which(tabl$sex=="M"), Fs)), ])
if (Ms < Fs) # same as previous, but for Males < Femals
return(tabl[c(which(tabl$sex=="M"), sample(which(tabl$sex=="F"), Ms)), ])
stop("hmmm... something went wrong.") ## We should never hit this line, but just in case.
})
# Flatten into a single table
mydf.new <- do.call(rbind, sampled)
# Clean up
row.names(mydf.new) <- NULL # remove the row names that were added
mydf.new$latlong <- NULL # remove the temporary column that we added
mydf.new
# lat long sex score
# 1 45 -68 F 2
# 2 45 -68 F 8
# 3 45 -68 M 9
# 4 45 -68 M 3
# 5 58 139 F 5
# 6 58 139 M 3
# 7 64 1 M 6
# 8 64 1 M 1
# 9 64 1 F 7
# 10 64 1 F 5
答案 2 :(得分：2)
这将返回值作为列表元素：
spl <- split(data, interaction(data$lat, data$long) ,drop=TRUE)
# interaction creates all the two way pairs from those two vectors
# drop is needed to eliminate the dataframes with no representation
res <- lapply(spl, function(x) { #First find the nuber of each gender to select
N=min(table(x$sex)) # then sample each sex separately
rbind( x[ x$sex=="M" & row.names(x) %in% sample(row.names(x[x$sex=="M",] ), N) , ],
# One (or both) of these will be "sampling" all of that sex.
x[ x$sex=="F" & row.names(x) %in% sample(row.names(x[x$sex=="F", ]), N) , ] )
} )
res
#------------
$`45.-68`
lat long sex score
9 45 -68 M 3
11 45 -68 M 9
8 45 -68 F 2
12 45 -68 F 8
$`12.-14` # So there were no women in this group and zero could be matched
[1] lat long sex score
<0 rows> (or 0-length row.names)
$`45.1`
[1] lat long sex score
<0 rows> (or 0-length row.names)
$`64.1`
lat long sex score
15 64 1 M 6
20 64 1 M 1
16 64 1 F 7
17 64 1 F 5
$`58.139`
lat long sex score
7 58 139 M 3
5 58 139 F 5
,,,,但如果您想将其作为数据框，则可以使用do.call(rbind, res)：
> do.call(rbind, res)
lat long sex score
45.-68.10 45 -68 M 9
45.-68.11 45 -68 M 9
45.-68.8 45 -68 F 2
45.-68.12 45 -68 F 8
64.1.15 64 1 M 6
64.1.20 64 1 M 1
64.1.17 64 1 F 5
64.1.18 64 1 F 7
58.139.6 58 139 M 4
58.139.5 58 139 F 5