当我运行下面的代码时,program_exe.update_data()函数将执行两次。避免这种情况的最佳方法是什么?该函数执行比较耗时,因此两次运行它并不理想。任何建议将不胜感激!
app = dash.Dash(__name__)
server = app.server
dict_main = program_exe.update_data() #this creates a nested dictionary
rpm = list(dict_main.keys())
channels = dict_main[rpm[0]]
app.layout = html.Div(
[
html.Div([
dcc.Dropdown(
id='rpm-dropdown',
options=[{'label': speed, 'value': speed} for speed in rpm],
value=list(dict_main.keys())[0],
# I removed the multi=True because it requires a distinction between the columns in the next dropdown...
searchable=False
),
], style={'width': '20%', 'display': 'inline-block'}),
html.Div([
dcc.Dropdown(
id='channel-dropdown',
multi=True
),
], style={'width': '20%', 'display': 'inline-block'}
),
html.Div([
dcc.Graph(
id='Main-Graph' # the initial graph is in the callback
),
], style={'width': '98%', 'display': 'inline-block'}
)
]
)
@app.callback(
Output('channel-dropdown', 'options'),
[Input('rpm-dropdown', 'value')])
def update_date_dropdown(speed):
return [{'label': i, 'value': i} for i in dict_main[speed]]
@app.callback(
Output('Main-Graph', 'figure'),
[Input('channel-dropdown', 'value')],
[State('rpm-dropdown', 'value')]) # This is the way to inform the callback which dataframe is to be charted
def updateGraph(channels, speed):
if channels:
# return the entire figure with the different traces
return go.Figure(data=[go.Scatter(x=dict_main[speed]['Manager'], y=dict_main[speed][i]) for i in channels])
else:
# at initialization the graph is returned empty
return go.Figure(data=[])
if __name__ == '__main__':
app.run_server(debug=True)
答案 0 :(得分:1)
您可以使用缓存仅一次点击该功能。有关更多详细信息,请参见this page。
简短示例:
from flask_caching import Cache
cache = Cache(app.server, config={
'CACHE_TYPE': 'filesystem',
'CACHE_DIR': 'cache-directory',
})
@cache.memoize(timeout=6000)
def my_cached_function():
return program_exe.update_data()
我没有您的func,但是当我在本地使用占位符对其进行测试时,它最初被调用了两次,但是在添加缓存后才被调用一次。