功能的一个for循环以产生具有值1的列:N由另一列匹配间隔
问题描述:
我有如下所示功能的一个for循环以产生具有值1的列:N由另一列匹配间隔
my_df=data.frame(x=runif(100, min = 0,max = 60),
y=runif(100, min = 0,max = 60)) #x and y in cm
有了这个,我需要从1值的新列的数据帧调节至36,匹配x
和y
每10厘米。例如,如果0<=x<=10 & 0<=y<=10
,放1,那么如果10<=x<=20 & 0<=y<=10
,把2等等放到6,然后0<=x<=10 & 10<=y<=20
开始7到12,等我试图做一个if
函数重复的x
6倍的时间间隔,并在每次迭代中增加y
的时间间隔。下面是函数
#my miscarried function 'zones'
>zones= function(x,y) {
i=vector(length = 6)
n=vector(length = 6)
z=vector(length = 36)
i[1]=0
z[1]=0
n[1]=1
for (t in 1:6) {
if (0<=x & x<10 & i[t]<=y & y<i[t]+10) { z[t] = n[t]} else
if (10<=x & x<20 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+1} else
if (20<=x & x<30 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+2} else
if (30<=x & x<40 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+3} else
if (40<=x & x<50 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+4}else
if (50<=x & x<=60 & i[t]<=y & y<i[t]+10) {z[t]=n[t]+5}
else {i[t+1]=i[t]+10
n[t+1]=n[t]+6}
}
return(z)
}
>xy$z=zones(x=xy$x,y=xy$y)
和我
There were 31 warnings (use warnings() to see them)
>xy$z
[1] 0 0 0 0 25 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
请帮助我之前,我孤独终老!
答
我认为这是一个窍门。
a <- cut(my_df$x, (0:6) * 10)
b <- cut(my_df$y, (0:6) * 10)
z <- interaction(a, b)
levels(z)
[1] "(0,10].(0,10]" "(10,20].(0,10]" "(20,30].(0,10]" "(30,40].(0,10]"
[5] "(40,50].(0,10]" "(50,60].(0,10]" "(0,10].(10,20]" "(10,20].(10,20]"
[9] "(20,30].(10,20]" "(30,40].(10,20]" "(40,50].(10,20]" "(50,60].(10,20]"
[13] "(0,10].(20,30]" "(10,20].(20,30]" "(20,30].(20,30]" "(30,40].(20,30]"
[17] "(40,50].(20,30]" "(50,60].(20,30]" "(0,10].(30,40]" "(10,20].(30,40]"
[21] "(20,30].(30,40]" "(30,40].(30,40]" "(40,50].(30,40]" "(50,60].(30,40]"
[25] "(0,10].(40,50]" "(10,20].(40,50]" "(20,30].(40,50]" "(30,40].(40,50]"
[29] "(40,50].(40,50]" "(50,60].(40,50]" "(0,10].(50,60]" "(10,20].(50,60]"
[33] "(20,30].(50,60]" "(30,40].(50,60]" "(40,50].(50,60]" "(50,60].(50,60]"
如果这类型的级别不适合您的口味,然后如下更改:
levels(z) <- 1:36
答
这是你所追求的?所得到的数字是res
列:
# Get bin index for x values and y values
my_df$bin1 <- as.numeric(cut(my_df$x, breaks = seq(0, max(my_df$x) + 10, by = 10)));
my_df$bin2 <- as.numeric(cut(my_df$y, breaks = seq(0, max(my_df$x) + 10, by = 10)));
# Multiply bin indices
my_df$res <- my_df$bin1 * my_df$bin2;
> head(my_df)
x y bin1 bin2 res
1 49.887499 47.302849 5 5 25
2 43.169773 50.931357 5 6 30
3 10.626466 43.673533 2 5 10
4 43.401454 3.397009 5 1 5
5 7.080386 22.870539 1 3 3
6 39.094724 24.672907 4 3 12
我打破用于说明目的的步骤;您可能不想保留中间栏bin1
和bin2
。
答
我们可能需要显示X,Y和Z之间的关系的表。之后,我们可以定义一个函数来进行连接。
该解决方案与此帖相关并受到启发(R dplyr join by range or virtual column)。您也可能会发现其他解决方案很有用。
# Set seed for reproducibility
set.seed(1)
# Create example data frame
my_df <- data.frame(x=runif(100, min = 0,max = 60),
y=runif(100, min = 0,max = 60))
# Load the dplyr package
library(dplyr)
# Create a table to show the relationship between x, y, and z
r <- expand.grid(x_from = seq(0, 50, 10), y_from = seq(0, 50, 10)) %>%
mutate(x_to = x_from + 10, y_to = y_from + 10, z = 1:n())
# Define a function for dynamic join
dynamic_join <- function(d, r){
if (!("z" %in% colnames(d))){
d[["z"]] <- NA_integer_
}
d <- d %>%
mutate(z = ifelse(x >= r$x_from & x < r$x_to & y >= r$y_from & y < r$y_to,
r$z, z))
return(d)
}
re_dynamic_join <- function(d, r){
r_list <- split(r, r$z)
for (i in 1:length(r_list)){
d <- dynamic_join(d, r_list[[i]])
}
return(d)
}
# Apply the function
re_dynamic_join(my_df, r)
x y z
1 15.930520 39.2834357 20
2 22.327434 21.1918363 15
3 34.371202 16.2156088 10
4 54.492467 59.5610437 36
5 12.100916 38.0095959 20
6 53.903381 12.7924881 12
7 56.680516 7.7623409 6
8 39.647868 28.6870821 16
9 37.746843 55.4444682 34
10 3.707176 35.9256580 19
11 12.358474 58.5702417 32
12 10.593405 43.9075507 26
13 41.221371 21.4036147 17
14 23.046223 25.8884214 15
15 46.190485 8.8926936 5
16 29.861955 0.7846545 3
17 43.057110 42.9339640 29
18 59.514366 6.1910541 6
19 22.802111 26.7770609 15
20 46.646713 38.4060627 23
21 56.082314 59.5103172 36
22 12.728551 29.7356147 14
23 39.100426 29.0609715 16
24 7.533306 10.4065401 7
25 16.033240 45.2892567 26
26 23.166846 27.2337294 15
27 0.803420 30.6701870 19
28 22.943277 12.4527068 9
29 52.181451 13.7194886 12
30 20.420940 35.7427198 21
31 28.924807 34.4923319 21
32 35.973950 4.6238628 4
33 29.612478 2.1324348 3
34 11.173056 38.5677295 20
35 49.642399 55.7169120 35
36 40.108004 35.8855453 23
37 47.654392 33.6540449 23
38 6.476618 31.5616634 19
39 43.422657 59.1057134 35
40 24.676466 30.4585093 21
41 49.256778 40.9672847 29
42 38.823612 36.0924731 22
43 46.975966 14.3321207 11
44 33.182179 15.4899556 10
45 31.783175 43.7585774 28
46 47.361374 27.1542499 17
47 1.399872 10.5076061 7
48 28.633804 44.8018962 27
49 43.938824 6.2992584 5
50 41.563893 51.8726969 35
51 28.657177 36.8786983 21
52 51.672569 33.4295723 24
53 26.285826 19.7266391 9
54 14.687837 27.1878867 14
55 4.240743 30.0264584 19
56 5.967970 10.8519817 7
57 18.976302 31.7778362 20
58 31.118056 4.5165447 4
59 39.720305 16.6653560 10
60 24.409811 12.7619712 9
61 54.772555 17.0874289 12
62 17.616202 53.7056462 32
63 27.543944 26.7741194 15
64 19.943680 46.7990934 26
65 39.052228 52.8371421 34
66 15.481007 24.7874526 14
67 28.712715 3.8285088 3
68 45.978640 20.1292495 17
69 5.054815 43.4235568 25
70 52.519280 20.2569200 18
71 20.344376 37.8248473 21
72 50.366421 50.4368732 36
73 20.801009 51.3678999 33
74 20.026496 23.4815569 15
75 28.581075 22.8296331 15
76 53.531900 53.7267256 36
77 51.860368 38.6589458 24
78 23.399373 44.4647189 27
79 46.639242 36.3182068 23
80 57.637080 54.1848967 36
81 26.079569 17.6238093 9
82 42.750881 11.4756066 11
83 23.999662 53.1870566 33
84 19.521129 30.2003691 20
85 45.425229 52.6234526 35
86 12.161535 11.3516173 8
87 42.667273 45.4861831 29
88 7.301515 43.4699336 25
89 14.729311 56.6234891 32
90 8.598263 32.8587952 19
91 14.377765 42.7046321 26
92 3.536063 23.3343060 13
93 38.537296 6.0523876 4
94 52.576153 55.6381253 36
95 46.734881 16.9939500 11
96 47.838530 35.4343895 23
97 27.316467 6.6216363 3
98 24.605045 50.4304219 33
99 48.652215 19.0778211 11
100 36.295997 46.9710802 28
既然你不会独自死亡,你应该接受其中一个答案。 :) – Suren
@ Headpoint对不起!我正在一家酒吧里死去,但我看到了灯光。谢谢你,你提供的解决方案工作得很好。 – jealcalat