数据帧的下标通过向量的条件值 [英] r subscript of dataframe with condition values by a vector
问题描述
这似乎很简单，但是它一直保持我的忙碌。
this seems to be rather easy, but it keeps my busy since a while.
我有一个数据帧（df），n列和一个相同的向量数值（n）。
I have a dataframe (df) with n columns and a vector with the same number (n) of values.
向量中的值是数据框中列中的观察值的阈值。所以线索是，如何告诉R对每一列使用不同的阈值？
The values in the vector are thresholds for the observations in the columns in the dataframe. So the clue is, how to tell R to use different thresholds for each column?
我想保留数据帧中的所有观察结果，以满足每列的各种阈值（在上面或下面，在例子中没有关系）。不符合阈值标准的观察值应设置为0。
I want to keep all the observations in the dataframe which fulfill the various thresholds for each column (above or below, doesnt matter in the example). The observations which do not fulfill the threshold criterion should be set to 0.
我不想要数据帧的一部分。
I dont want a subset of the dataframe.
任何人都可以帮忙吗？非常感谢。
Can anyone help? Thanks a lot in advance.
推荐答案
给出一些示例数据和阈值
Given some example data and thresholds
set.seed(42)
dat <- data.frame(matrix(runif(100), ncol = 10))
## thresholds
thresh <- seq(0.5, 0.95, length.out = 10)
thresh
我们可以使用 mapply（）函数来确定每列（在此）中的哪些观察值大于或等于阈值。使用这些索引，我们可以通过以下方式替换与 0 相对应的值：
we can use the mapply() function to work out which observations in each column (in this) are greater than or equal to the threshold. Using those indices, we can replace the values corresponding to the indices with 0 via:
dat[mapply(">=", dat, thresh)] <- 0
以下是电话号码：
> dat
X1 X2 X3 X4 X5
1 0.9148060 0.4577418 0.90403139 0.737595618 0.37955924
2 0.9370754 0.7191123 0.13871017 0.811055141 0.43577158
3 0.2861395 0.9346722 0.98889173 0.388108283 0.03743103
4 0.8304476 0.2554288 0.94666823 0.685169729 0.97353991
5 0.6417455 0.4622928 0.08243756 0.003948339 0.43175125
6 0.5190959 0.9400145 0.51421178 0.832916080 0.95757660
7 0.7365883 0.9782264 0.39020347 0.007334147 0.88775491
8 0.1346666 0.1174874 0.90573813 0.207658973 0.63997877
9 0.6569923 0.4749971 0.44696963 0.906601408 0.97096661
10 0.7050648 0.5603327 0.83600426 0.611778643 0.61883821
X6 X7 X8 X9 X10
1 0.33342721 0.6756073 0.042988796 0.58160400 0.6674265147
2 0.34674825 0.9828172 0.140479094 0.15790521 0.0002388966
3 0.39848541 0.7595443 0.216385415 0.35902831 0.2085699569
4 0.78469278 0.5664884 0.479398564 0.64563188 0.9330341273
5 0.03893649 0.8496897 0.197410342 0.77582336 0.9256447486
6 0.74879539 0.1894739 0.719355838 0.56364684 0.7340943010
7 0.67727683 0.2712866 0.007884739 0.23370340 0.3330719834
8 0.17126433 0.8281585 0.375489965 0.08998052 0.5150633298
9 0.26108796 0.6932048 0.514407708 0.08561206 0.7439746463
10 0.51441293 0.2405447 0.001570554 0.30521837 0.6191592400
> dat[mapply(">=", dat, thresh)] <- 0
> dat
X1 X2 X3 X4 X5
1 0.0000000 0.4577418 0.00000000 0.000000000 0.37955924
2 0.0000000 0.0000000 0.13871017 0.000000000 0.43577158
3 0.2861395 0.0000000 0.00000000 0.388108283 0.03743103
4 0.0000000 0.2554288 0.00000000 0.000000000 0.00000000
5 0.0000000 0.4622928 0.08243756 0.003948339 0.43175125
6 0.0000000 0.0000000 0.51421178 0.000000000 0.00000000
7 0.0000000 0.0000000 0.39020347 0.007334147 0.00000000
8 0.1346666 0.1174874 0.00000000 0.207658973 0.63997877
9 0.0000000 0.4749971 0.44696963 0.000000000 0.00000000
10 0.0000000 0.0000000 0.00000000 0.611778643 0.61883821
X6 X7 X8 X9 X10
1 0.33342721 0.6756073 0.042988796 0.58160400 0.6674265147
2 0.34674825 0.0000000 0.140479094 0.15790521 0.0002388966
3 0.39848541 0.7595443 0.216385415 0.35902831 0.2085699569
4 0.00000000 0.5664884 0.479398564 0.64563188 0.9330341273
5 0.03893649 0.0000000 0.197410342 0.77582336 0.9256447486
6 0.74879539 0.1894739 0.719355838 0.56364684 0.7340943010
7 0.67727683 0.2712866 0.007884739 0.23370340 0.3330719834
8 0.17126433 0.0000000 0.375489965 0.08998052 0.5150633298
9 0.26108796 0.6932048 0.514407708 0.08561206 0.7439746463
10 0.51441293 0.2405447 0.001570554 0.30521837 0.6191592400
有意义的是注意什么 mapply（）在这种情况下返回：
It is instructive to notice what mapply() returns in this case:
> mapply(">=", dat, thresh)
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
[1,] TRUE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE
[2,] TRUE TRUE FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE
[3,] FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[4,] TRUE FALSE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE
[5,] TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE
[6,] TRUE TRUE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE
[7,] TRUE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
[8,] FALSE FALSE TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE
[9,] TRUE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE
[10,] TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
这些逻辑值用于选择满足的观察结果门槛。你可以使用一个不同的二进制运算符;有关各种选项，请参阅？>在编写 mapply（）调用时，请考虑二进制运算符的左侧和右侧，以便 mapply（） call会给出：
and it is those logical values that are used to select the observations that meet the threshold. You can a different binary operator to the one I used; see ?">" for the various options. When writing the mapply() call, think of it in terms of left-hand-side and right-hand-side of the binary operator, such that an mapply() call would give:
mapply(">", lhs, rhs)
我们可以在哪里写
lhs > rhs
更新：正如@DWin已经回答了关于两个阈值的评论将更新我的答案以匹配。
Update: As @DWin has answered the comment about two thresholds I will update my Answer to match.
thresh1 <- seq(0.05, 0.5, length.out = 10)
thresh2 <- seq(0.55, 0.95, length.out = 10)
set.seed(42)
dat <- data.frame(matrix(runif(100), ncol = 10))
l1 <- mapply(">", dat, thresh1)
l2 <- mapply("<", dat, thresh2)
我们可以看到哪些元素符合两个限制：
We can see which elements match both constraints:
> l1 & l2
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
[1,] FALSE TRUE FALSE FALSE TRUE TRUE TRUE FALSE TRUE TRUE
[2,] FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE
[3,] TRUE FALSE FALSE TRUE FALSE TRUE TRUE FALSE FALSE FALSE
[4,] FALSE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE
[5,] FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE TRUE TRUE
[6,] TRUE FALSE TRUE FALSE FALSE TRUE FALSE TRUE TRUE TRUE
[7,] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[8,] TRUE TRUE FALSE TRUE TRUE FALSE FALSE FALSE FALSE TRUE
[9,] FALSE TRUE TRUE FALSE FALSE FALSE TRUE TRUE FALSE TRUE
[10,] FALSE TRUE FALSE TRUE TRUE TRUE FALSE FALSE FALSE TRUE
并且可以使用相同的结构来选择匹配的元素：
and the same construct can be used to select those elements that match:
dat[l1 & l2] <- 0
dat
> dat
X1 X2 X3 X4 X5 X6 X7 X8
1 0.9148060 0.0000000 0.90403139 0.737595618 0.00000000 0.00000000 0.0000000 0.042988796
2 0.9370754 0.7191123 0.13871017 0.811055141 0.00000000 0.00000000 0.9828172 0.140479094
3 0.0000000 0.9346722 0.98889173 0.000000000 0.03743103 0.00000000 0.0000000 0.216385415
4 0.8304476 0.0000000 0.94666823 0.685169729 0.97353991 0.78469278 0.0000000 0.000000000
5 0.6417455 0.0000000 0.08243756 0.003948339 0.00000000 0.03893649 0.8496897 0.197410342
6 0.0000000 0.9400145 0.00000000 0.832916080 0.95757660 0.00000000 0.1894739 0.000000000
7 0.7365883 0.9782264 0.00000000 0.007334147 0.88775491 0.00000000 0.2712866 0.007884739
8 0.0000000 0.0000000 0.90573813 0.000000000 0.00000000 0.17126433 0.8281585 0.375489965
9 0.6569923 0.0000000 0.00000000 0.906601408 0.97096661 0.26108796 0.0000000 0.000000000
10 0.7050648 0.0000000 0.83600426 0.000000000 0.00000000 0.00000000 0.2405447 0.001570554
X9 X10
1 0.00000000 0.0000000000
2 0.15790521 0.0002388966
3 0.35902831 0.2085699569
4 0.00000000 0.0000000000
5 0.00000000 0.0000000000
6 0.00000000 0.0000000000
7 0.23370340 0.3330719834
8 0.08998052 0.0000000000
9 0.08561206 0.0000000000
10 0.30521837 0.0000000000
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