Python 3.8,Spyder IDE
我使用,
pip install schemdraw[matplotlib]
我正在使用下面的代码,但是我得到的只是一个空白的画布,
import schemdraw
import schemdraw.elements as elm
from schemdraw import logic
from schemdraw.parsing import logicparse
with schemdraw.Drawing() as d:
logicparse('not ((w and x) or (y and z))', outlabel='$\overline{Q}$')也试过了,
with schemdraw.Drawing() as d:
d = logicparse('not ((w and x) or (y and z))', outlabel='$\overline{Q}$')
d.draw()给了一个空白的画布。
也试过了,
with schemdraw.Drawing() as d:
d.add(logicparse('not ((w and x) or (y and z))', outlabel='$\overline{Q}$'))也试过了,
d = logicparse('not ((w and x) or (y and z))', outlabel='$\overline{Q}$')
d.draw()什么都不返回,甚至不返回空白画布。
这会引发以下错误,
Traceback (most recent call last):
File "D:\Python codes\temp.py", line 10, in <module>
d.add(logicparse('not ((w and x) or (y and z))', outlabel='$\overline{Q}$'))
File "D:\ProgramData\Anaconda3\lib\site-packages\schemdraw\schemdraw.py", line 233, in add
element = element(**kwargs)
TypeError: 'Drawing' object is not callable我也试过(在进口之后)刚刚,
logicparse('not ((w and x) or (y and z))', outlabel='$\overline{Q}$')
根本没有输出。我安装了最新版本的pyparse。
以下简单的电路工作得很好,
with schemdraw.Drawing() as d:
d.add(elm.Resistor())
d.add(elm.Capacitor())
d.add(elm.Diode())这里有什么明显的遗漏吗?
发布于 2022-05-11 18:46:38
试试这个:
from schemdraw.parsing import logicparse
drawing = logicparse('not ((w and x) or (y and z))', outlabel='$\overline{Q}$')
drawing.draw()发布于 2022-05-18 03:32:28
你可以试试:
with logicparse('not ((w and x) or (y and z))', outlabel='$\overline{Q}$'):
passlogicparse返回一个绘图,因此它不能将它放入一个with schemdraw.Drawing块中。但是,Spyder的控制台似乎并没有从Drawing.draw()获得这个数字。
https://stackoverflow.com/questions/72197709
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