调用函数的 Bokeh 列布局
Bokeh column layout of called function
我有一个带有一些滑块和 Select 选项的散景脚本:
from bokeh.plotting import show
from bokeh.models import Select, Slider
from bokeh.layouts import row, column
def add_sliders():
# defining sliders with start values:
slider1 = Slider(title = 'Option1',start = 0, end = 10, step = 1, value = 6)
slider2 = Slider(title = 'Option2',start = -1, end = 1, step = 1, value = 0)
select1 = Select(title='Option3', value="In", options=["In","Out"])
return slider1, slider2, select1
slider1, slider2, select1= add_sliders()
l = column(slider1, slider2, select1 )
show(l)
现在,我的问题是:我不想说明 slider1, slider2, select1= add_sliders()
,而是想说明 l = column(add_sliders())
,所以被调用函数的所有输出参数的列布局 add_sliders.有办法吗?
您应该可以使用 column(children=add_sliders())
或 column(*add_sliders())
。
我有一个带有一些滑块和 Select 选项的散景脚本:
from bokeh.plotting import show
from bokeh.models import Select, Slider
from bokeh.layouts import row, column
def add_sliders():
# defining sliders with start values:
slider1 = Slider(title = 'Option1',start = 0, end = 10, step = 1, value = 6)
slider2 = Slider(title = 'Option2',start = -1, end = 1, step = 1, value = 0)
select1 = Select(title='Option3', value="In", options=["In","Out"])
return slider1, slider2, select1
slider1, slider2, select1= add_sliders()
l = column(slider1, slider2, select1 )
show(l)
现在,我的问题是:我不想说明 slider1, slider2, select1= add_sliders()
,而是想说明 l = column(add_sliders())
,所以被调用函数的所有输出参数的列布局 add_sliders.有办法吗?
您应该可以使用 column(children=add_sliders())
或 column(*add_sliders())
。