Question

我想制作四个具有设置的宽度和高度的矩形。

var rects = {
  key1:  { x: base_x, y: base_y, fullname: "Seed", cnt: 3500, row: 1 },
  key2:  { x: 2*base_x, y: base_y, fullname: "Series A", cnt: 4700, row: 1 },
  key3:  { x: 3*base_x, y: base_y, fullname: "Series B", cnt: 2300, row: 1},
  key4:  { x: 4*base_x, y: base_y, fullname: "Series C", cnt: 1750, row: 1}}

我知道，如果我将{}替换为[]以将其变成一个数组，并且由于数组不能有键而放弃了键，那么我可以将它们变成矩形可以使用它。

var rectangles = svg.selectAll(".rect")
    .data(rects)
    .enter()
    .append("rect")

稍后，我想在需要键的函数中使用rects对象。

我想我可以像这样用rect创建一个数组

var rects_array = []
for (i = 0; i < d3.keys(rects).length; i++){
    rects_array.push(rects[d3.keys(rects)[i]])
}

我只是好奇我是否可以直接传递rects而不是通过它传递数组。

Answer 1

简短答案：否，不能。

D3 int main () { struct node { int data; struct node* next; }; struct node* top=NULL; int choice = 0; printMenu (); do { printf ("\nEnter a choice. "); scanf ("%d", &choice); switch (choice) { case 0: { exitProgram(); break; } case 1: { clrScreen(); break; } case 3: { pushpancake(); break; } default: printf("\nInvalid Choice!\n"); } } while (choice != 0); return 0; } void printMenu () { printf ("0) Exit program.\n\n"); printf ("1) Clear Screen.\n\n"); printf ("2) Display the pancake stack.\n\n"); printf ("3) Push a pancake.\n\n"); printf ("4) Pop a pancake.\n\n"); } void pushpancake() { char pancake[30]; int calories; printf("Enter pancake name and its calories:"); scanf("%s %d,", pancake, &calories); printf("The pancake has been added at the top of the stack."); struct node *temp=(struct node*)malloc(sizeof(struct node)); strcpy(temp->pancake,pancake); temp->data=calories; temp->next=top; top=temp; }方法仅接受三件事：

一个数组；
一个功能；
一无所有。

因此，您不能在此处使用对象。

关于最后一个代码片段，基于first()对象创建数组的最简单方法是使用Object.values：

library(tidyverse)

dat %>% 
  mutate(dist = abs(4-timepoint)) %>% 
  arrange(id, dist, timepoint) %>% 
  group_by(id) %>% 
  filter(timepoint %in% c(0, first(timepoint))) %>% 
  ungroup() %>% 
  arrange(id, timepoint)

或者，如果您想要D3方法，则可以使用d3.values，该方法在内部使用data()循环：

rects

var rects = {
  key1: {
    x: "foo",
    y: "bar",
    fullname: "Seed",
    cnt: 3500,
    row: 1
  },
  key2: {
    x: "foo",
    y: "bar",
    fullname: "Series A",
    cnt: 4700,
    row: 1
  },
  key3: {
    x: "foo",
    y: "bar",
    fullname: "Series B",
    cnt: 2300,
    row: 1
  },
  key4: {
    x: "foo",
    y: "bar",
    fullname: "Series C",
    cnt: 1750,
    row: 1
  }
};

console.log(Object.values(rects))

然后根据需要将数组传递到for...in，并在其他地方使用var rects = { key1: { x: "foo", y: "bar", fullname: "Seed", cnt: 3500, row: 1 }, key2: { x: "foo", y: "bar", fullname: "Series A", cnt: 4700, row: 1 }, key3: { x: "foo", y: "bar", fullname: "Series B", cnt: 2300, row: 1 }, key4: { x: "foo", y: "bar", fullname: "Series C", cnt: 1750, row: 1 } }; console.log(d3.values(rects))对象。

我可以使用字典作为矩形的“数据”对象吗？

1 个答案: