我想用我的大四学生在Java中做的一些改变 . 由于我不懂那种语言,我想在Matlab中做 . 我在这里附上了数据(文本格式)和情节 . 它是项目进度表的输出,其中值表示与任务(链接)相关的成本及其持续时间(比如任务2从第6个月到第9个月完成,成本为0.665m $) .
值0表示特定任务(行)在该时间段内没有工作 . Java中的图表显示了每项任务的时间和成本变化 .
我一直试图在Matlab中制作类似的情节(圆圈情况下为条形),但不能 .
Link# 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
1 0 0 0 0 0 0 0 0 0 0 0 0 0.34 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0.665 0.665 0.665 0.665 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3 0 0 0.34 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
4 0.36 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
5 0.14 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
6 0 0 0 0 0 0 0 0 0 0 0.58 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
7 0 0 0 0.57 0.57 0.57 0.57 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
8 0.64 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
9 0.58 0.58 0.58 0.58 0.58 0.58 0.58 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
10 0.24 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
11 0.56 0.56 0.56 0.56 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
12 0.2 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
13 0 0 0 0 0 0 0 0 0.56 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.62 0 0 0 0 0 0 0 0 0 0 0
15 0 0 0 0.582857143 0.582857143 0.582857143 0.582857143 0.582857143 0.582857143 0.582857143 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
16 0.536 0.536 0.536 0.536 0.536 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
17 0.2 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
18 0 0.6 0.6 0.6 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
19 0 0 0 0 0.0000 0 0 0 0 0 0 0 0.606666667 0.606666667 0.606666667 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
20 0 0 0 0.633333333 0.633333333 0.633333333 0.633333333 0.633333333 0.633333333 0.633333333 0.633333333 0.633333333 0.633333333 0.633333333 0.633333333 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
21 0 0 0.606666667 0.606666667 0.606666667 0.606666667 0.606666667 0.606666667 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
22 0 0 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0.643076923 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
23 0 0 0 0 0 0 0 0.58 0.58 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
24 0 0 0 0 0 0 0 0 0 0.651428571 0.651428571 0.651428571 0.651428571 0.651428571 0.651428571 0.651428571 0 0 0 0 0 0 0 0 0 0 0 0 0 0
25 0 0 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
26 0 0 0 0 0 0 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
27 0.546666667 0.546666667 0.546666667 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
28 0 0.655 0.655 0.655 0.655 0.655 0.655 0.655 0.655 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
29 0.5 0.5 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
30 0 0 0 0 0 0 0 0 0 0.44 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
31 0.54 0.54 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
32 0 0.64 0.64 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
33 0 0 0 0 0 0 0 0 0 0.611428571 0.611428571 0.611428571 0.611428571 0.611428571 0.611428571 0.611428571 0 0 0 0 0 0 0 0 0 0 0 0 0 0
34 0 0 0 0 0 0 0 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0.62 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
35 0.36 0.36 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
36 0 0 0 0 0 0 0 0 0 0 0.65 0.65 0.65 0.65 0.65 0.65 0.65 0.65 0 0 0 0 0 0 0 0 0 0 0 0
37 0.38 0.38 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
38 0.613333333 0.613333333 0.613333333 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
39 0.58 0.58 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
40 0 0 0 0 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0.647272727 0 0 0 0
1 回答
这是一个解决方案的示例,假设
data
是图表的所有 within (即没有轴刻度标签):结果: