我试图根据其他两列的值更改一列的值。 到目前为止，这让我有些头疼，而且我不确定是否可能。
我的数据集如下所示。一栏是时间，其他两栏反映了后代的父母关系。在类似时间点1的奇怪情况下 我有后代“ D”在数据集中首次弹出，但尚未出现 在上一个时间点在那里同时充当后代和父亲。
数据
structure(list(time = c(0L, 0L, 0L, 1L, 1L, 1L, 2L, 2L, 2L),
offspring = c("A", "B", "C", "A", "D", "E", "A", "F", "G"
), parent = c(NA, NA, NA, "A", "B", "D", "A", "A", "F")), class = "data.frame", row.names = c(NA,
-9L))
我想提供什么帮助
查找一次性存在的所有后代，而不是前一个 （不考虑时间点0），并且像 D和F
找到它们后，我想将一个准确的时间点减少0.5
time offspring parent
0 A NA
0 B NA
0 C NA
1 A A
0.5 D B
1 E D
2 A A
1.5 F A
2 G F
在此问题上的任何帮助或指导将不胜感激。
答案 0 :(得分：1)
创建2个数据框，以查找每只动物作为父母和后代的初次出现。
找到两个合并列中出现的时间和动物，然后在原始数据框中更新时间。
df <-structure(list(time = c(0L, 0L, 0L, 1L, 1L, 1L, 2L, 2L, 2L),
offspring = c("A", "B", "C", "A", "D", "E", "A", "F", "G"),
parent = c(NA, NA, NA, "A", "B", "D", "A", "A", "F")), class = "data.frame",
row.names = c(NA, -9L))
library(dplyr)
#find the row where each Letter First appears as both a parent and offspring
parents <-df %>% filter(complete.cases(.)) %>% group_by(parent) %>% slice(1) %>% select(time, parent)
offsprings <- df %>% group_by(offspring) %>% slice(1) %>% select(time, offspring)
combined <- full_join(offsprings, parents)
#rows where the names match for both parent and offspring
matchingrows <-which(combined$parent == combined$offspring)
#update the times
for (i in matchingrows){
row = which(df$time == combined$time[i] & df$offspring == combined$offspring[i])
df$time[row] <- df$time[row] - 0.5
}
df
答案 1 :(得分：0)
如果需要，在data.table中：
library(data.table)
DT <- data.table(time = c(0,0,0,1,1,1,2,2,2),
offspring = c('A', 'B', 'C', 'A', 'D', 'E', 'A', 'F', 'G'),
parent = c(NA, NA, NA, 'A', 'B', 'D', 'A', 'A', 'F'))
for (i in seq_len(nrow(DT))) {
DT[i, time := fifelse(time != 0 & offspring %chin% DT[, parent] & !(offspring %chin% DT[seq_len(i-1), offspring]),
time - 0.5,
time)]
}
> DT
time offspring parent
1: 0.0 A <NA>
2: 0.0 B <NA>
3: 0.0 C <NA>
4: 1.0 A A
5: 0.5 D B
6: 1.0 E D
7: 2.0 A A
8: 1.5 F A
9: 2.0 G F
与dplyr：
library(dplyr)
library(tibble)
tbl <- tibble(time = c(0,0,0,1,1,1,2,2,2),
offspring = c('A', 'B', 'C', 'A', 'D', 'E', 'A', 'F', 'G'),
parent = c(NA, NA, NA, 'A', 'B', 'D', 'A', 'A', 'F'))
for (i in seq_len(nrow(tbl))) {
tbl[i,][['time']] <- tbl[i, ] %>% mutate(time = if_else(time != 0 &
offspring %in% tbl[['parent']] &
!(offspring %in% tbl[seq_len(i-1),][['offspring']]),
time - 0.5,
time)) %>% pull(time)
}
> tbl
# A tibble: 9 x 3
time offspring parent
<dbl> <chr> <chr>
1 0 A NA
2 0 B NA
3 0 C NA
4 1 A A
5 0.5 D B
6 1 E D
7 2 A A
8 1.5 F A
9 2 G F
答案 2 :(得分：0)
我的解决方案可能不是最简洁，但是我能够使它起作用，并且可以推广到更大的数据集。我敢肯定有很多方法可以对此进行改进，所以我很好奇其他人提出的建议。首先，我在0下标方面遇到了麻烦，因此我在时间列中加了2 2，最后又减去了。
这个想法是我遍历所有行，找到当年（第0年之后）而不是前一年的后代。然后，我检查当年哪些个人也是父母。我将这些时间段内这些人的后代编译为向量，因为稍后我们将删除它们。然后，我用时间-.5，该后代及其父代创建新行。我将它们编译到一个新的数据框中，它将替换要删除的行。
因为每个时间戳都有重复，所以我将要删除的行向量和要添加的行df设为唯一。然后，我对原始数据帧进行删除和添加，使数据类型一致。
parent_offspring <- data.frame(
"time" = c( rep(0,3), rep(1,3), rep(2,3)),
"offspring" = c("A","B","C","A","D","E","A","F","G"),
"parent" = c(NA, NA, NA, "A","B","D","A","A","F")
)
po<- parent_offspring
po$time <- po$time+2
delete_vec <- vector()
df_to_add <- data.frame()
for (i in seq_along(po$time)) {
q <- po$time[[i]] # Value of "Time" variable for the row
a <- which(po$time == q) # Rows sharing that value of "Time"
offspring_curr <- po$offspring[a] # Offspring at that time
b <- which(po$time==(q-1)) # Rows of offspring at Time-1
offspring_prev <- po$offspring[b] # Identities of offspring at Time-1
f<- offspring_curr[offspring_curr %in% offspring_prev == FALSE] # Which offspring at Time were not offspring at Time-1
if (length(f) == 0) {
next
} else { ##skip ahead if none of the offspring at Time were not offspring at Time - 1
parents_curr <- po$parent[which(po$time == q)] # Parents at current time
parent_and_offpsring_curr <- intersect(f,parents_curr) # Which individuals are both parents and offspring at the current time
if (length(parent_and_offpsring_curr) == 0) {
next
} else { ## skip ahead if no individuals are both parents and offspring
g<- which(po$time==q & po$offspring==parent_and_offpsring_curr) # which offspring row is occupied by an individual who is both a parent and offspring at the current time
delete_vec <- append(delete_vec,g) #we'll be deleting those rows in the end so we'll keep track of them and save them in a vector
h<- po$parent[g] # this is the parent for the offspring/parent individual in the current time.
add_row<-c((q-.5), parent_and_offpsring_curr, h) # make a new row with the fractional time, parent/offspring individual, and their parent for row when the parent/offspring individual is an offspring
df_to_add <- rbind(df_to_add,add_row) ## we'll add these rows at the end
}
}
}
delete_vec<-unique(delete_vec) ## iteration gave us duplicates
df_to_add <- unique(df_to_add) ## same as above
colnames(df_to_add) <- colnames(po) ## fix column names for new df
po<- po[-delete_vec,] ## remove the offspring rows for the parent/offspring individuals
po<-rbind(po,df_to_add) ## add the rows with fractional times
rownames(po) <- c(1:nrow(po)) ## fix the row numbers
po$time<- as.numeric(po$time) ## time was converted to character when put into a vector with letters
po$time <- po$time-2 ## back to the original time values
po
time offspring parent
1 0.0 A <NA>
2 0.0 B <NA>
3 0.0 C <NA>
4 1.0 A A
5 1.0 E D
6 2.0 A A
7 2.0 G F
8 0.5 D B
9 1.5 F A
然后您可以使用dplyr :: arrange将行按时间升序排列