我有import QtQuick 2.0
Item {
id: root
function startOne() {
animOne.start()
}
function startTwo() {
animTwo.start()
}
property var listEvent: {
1000: root.startOne,
2000: root.startTwo
}
property int elapsed: 0
property int last_elapsed: 0
Rectangle {
id: one
x: 0
y: 0
width: 200
height: 200
color: "black"
}
Rectangle {
id: two
x: 600
y: 600
width: 200
height: 200
color: "black"
}
PropertyAnimation {
id: animOne
target: one
properties: "x"
to: 600
running: false
}
PropertyAnimation {
id: animTwo
target: two
properties: "x"
to: 0
running: false
}
function pop(last_elapsed, elapsed) {
for (var i = last_elapsed; i <= elapsed; i++) {
if (i in root.listEvent) {
listEvent[i]()
}
}
}
function scheduleEvent(delta) {
root.last_elapsed = root.elapsed
root.elapsed += delta
root.pop(root.last_elapsed, root.elapsed)
}
Timer {
id: scheduler
interval: 16
running: true
repeat: true
onTriggered: {
root.scheduleEvent(scheduler.interval)
console.log("timer: " + root.elapsed)
}
}
}
这种格式：
dictionary需要将此数据放在d = {'Name 1': list_of_links,'Name 2': list_of_links,'Name 3': list_of_links}
中，并带有两个DataFrame：
columns我试过这个：
Names and Links
Name 1 -> Link
Name 1 -> Link
... ...
Name 2 -> Link
Name 2 -> Link
... ....
Name 3 -> Link
Name 3 -> Link
然后匹配长度
links = []
names = []
for key in d:
names.append(key)
links.append(d[key])
最后将所有值复制到两个新列表中，但它似乎不是一个好的方法
答案 0 :(得分：2)
从两个名称dict和每个links开始：
d = {'Name 1': ['link{}'.format(l) for l in list(range(10))], 'Name 2': ['link{}'.format(l) for l in list(range(10, 20))]}
{'Name 1': ['link0', 'link1', 'link2', 'link3', 'link4', 'link5', 'link6', 'link7', 'link8', 'link9'], 'Name 2': ['link10', 'link11', 'link12', 'link13', 'link14', 'link15', 'link16', 'link17', 'link18', 'link19']}
您可以创建DataFrame .from_dict()，.stack()，并清理index：
df = pd.DataFrame.from_dict(d, orient='index').stack().reset_index(1, drop=True).to_frame().reset_index()
df.columns = ['name', 'link']
得到：
name link
0 Name 1 link0
1 Name 1 link1
2 Name 1 link2
3 Name 1 link3
4 Name 1 link4
5 Name 1 link5
6 Name 1 link6
7 Name 1 link7
8 Name 1 link8
9 Name 1 link9
10 Name 2 link10
11 Name 2 link11
12 Name 2 link12
13 Name 2 link13
14 Name 2 link14
15 Name 2 link15
16 Name 2 link16
17 Name 2 link17
18 Name 2 link18
19 Name 2 link19
答案 1 :(得分：1)
在这里，您还可以从元组构建DataFrame：
pd.DataFrame([(i,k) for i in d.keys() for k in d[i]],columns=['names','links'])