为什么Dash上传文件时会出现解析错误?

时间:2019-08-27 15:10:56

标签: python plotly plotly-dash

上传Excel或CSV会导致错误。我遵循了Dash演示,但是一旦我尝试对其进行扩展以进行诸如绘图之类的操作时,它将无法正常工作。我不想只显示一张桌子。 Dash_Table函数已更新,因此以前使用Dash_Table_Experiments的示例不再起作用

我整夜都在交换堆栈,修改代码和阅读其他解决方案。下面提供了完整的工作代码。我还想添加一个下拉回调函数,以通过分类变量“过滤”数据。

 import base64
 import datetime
 import io
 import plotly.graph_objs as go
 import dash
 from dash.dependencies import Input, Output, State
 import dash_core_components as dcc
 import dash_html_components as html
 import dash_table

 import pandas as pd


 external_stylesheets = ['https://codepen.io/chriddyp/pen/bWLwgP.css']

 app = dash.Dash(__name__, external_stylesheets=external_stylesheets)

 app.layout = html.Div([
  dcc.Upload(
      id='upload-data',
      children=html.Div([
        'Drag and Drop or ',
        html.A('Select Files')
    ]),
    style={
        'width': '100%',
        'height': '60px',
        'lineHeight': '60px',
        'borderWidth': '1px',
        'borderStyle': 'dashed',
        'borderRadius': '5px',
        'textAlign': 'center',
        'margin': '10px'
    },
    # Allow multiple files to be uploaded
    multiple=False
  ),
  html.Div(id='output-data-upload'),
  dcc.Graph(id='graph1')
 ])


 def parse_contents(contents, filename):
   content_type, content_string = contents.split(',')

   decoded = base64.b64decode(content_string)

   try:
       if 'csv' in filename:
         # Assume that the user uploaded a CSV file
          df = pd.read_csv(
            io.StringIO(decoded.decode('utf-8')))
       elif 'xls' in filename:
          # Assume that the user uploaded an excel file
          df = pd.read_excel(io.BytesIO(decoded))
   except Exception as e:
       print(e)
       return html.Div([
           'There was an error processing this file.'
       ])

return html.Div([
    html.H5(filename),
    # html.H6(datetime.datetime.fromtimestamp(date)),

    dash_table.DataTable(
        data=df.to_dict('records'),
        columns=[{'name': i, 'id': i} for i in df.columns]
    ),

    html.Hr(),  # horizontal line
    # For debugging, display the raw contents provided by the web browser
    html.Div('Raw Content'),
    html.Pre(contents[0:200] + '...', style={
        'whiteSpace': 'pre-wrap',
        'wordBreak': 'break-all'
    })
 ])


 @app.callback(Output('output-data-upload', 'children'),
          [Input('upload-data', 'contents')],
          [State('upload-data', 'filename')])
 def update_output(list_of_contents, list_of_names):
   if list_of_contents is not None:
       children = [
           parse_contents(c, n) for c, n in
           zip(list_of_contents, list_of_names)]
      return children

@app.callback(
   Output('graph1', 'figure'),
   [Input('upload-data', 'contents'),
   Input('upload-data', 'filename')])

def plot_graph(contents, filename):
   df = parse_contents(contents, filename)
   trace1 = go.Bar(
        x=df['Quarter'],
        y=df['Score'],
    )

layout = go.Layout(
    title='graph1'

)
fig = go.Figure(data = [trace1], layout=layout)
return fig

if __name__ == '__main__':
   app.run_server(debug=True)

我得到的错误是:回调错误,更新了output-data-upload.children:ValueError:没有足够的值要解压(预期2,得到1)

AttributeError:'NoneType'对象没有属性'split'

问题似乎出在python处理解析器的方式上:

  def parse_contents(contents, filename):
     content_type, content_string = contents.split(',')

   decoded = base64.b64decode(content_string)

但是,所有解决方案似乎都无法解决问题。

请帮助。既然有这么多人为此而苦苦挣扎(看来),那么如果我们能够解决它并发布一个能完成Shiny可以轻松完成的功能的代码(Github?),那就太好了。

2 个答案:

答案 0 :(得分:0)

您的代码:

   children = [
       parse_contents(c, n) for c, n in
       zip(list_of_contents, list_of_names)]

然后

def parse_contents(contents, filename):
    content_type, content_string = contents.split(',')
    decoded = base64.b64decode(content_string)
    ...

如果c处于None(无),则contents的{​​{1}}参数为None(无),并且发生“ None has no .split”错误,就会发生错误。

如果parse_contents不是None而是只有一个单词,则c仅返回一个元素,并且发生“没有足够的值要解压”错误。

我将对其进行过滤:

contents.split()

您还可以考虑在pairs = zip(list_of_contents, list_of_names) children = [parse_contents(c, n) for (c, n) in pairs if c and (len(c.split(',')) == 2)] 外部进行拆分,并更改其周围的代码。

我还将尝试记录内容错误的文件名,例如parse_contents

答案 1 :(得分:0)

解决了。在这里发帖供其他人使用:

def parse_contents(contents, filename):
if contents is not None:
    content_type, content_string = contents.split(',')

    decoded = base64.b64decode(content_string)

    try:
        if 'csv' in filename:
            # Assume that the user uploaded a CSV file
            df = pd.read_csv(
                io.StringIO(decoded.decode('utf-8')))
        elif 'xlsx' in filename:
            # Assume that the user uploaded an excel file
            df = pd.read_excel(io.BytesIO(decoded))
    except Exception as e:
        print(e)
        return html.Div([
            'There was an error processing this file.'
        ])

    return df
else:
    return [{}]


@app.callback(Output('table', 'data'),
              [Input('upload-data', 'contents'),
              Input('upload-data', 'filename')])
def update_output(contents, filename):
    if contents is not None:
        df = parse_contents(contents, filename)
        if df is not None:
            return df.to_dict('records')
        else:
            return [{}]
    else:
        return [{}]
@app.callback(
    Output('graph1', 'figure'),
    [Input('upload-data', 'contents'),
     Input('upload-data', 'filename')])

def plot_graph(contents, filename):
    df = parse_contents(contents, filename)
    trace1 = go.Bar(
            x=df['Quarter'],
            y=df['Score'],
        )

    layout = go.Layout(
        title='graph1'

)
fig = go.Figure(data = [trace1], layout=layout)
return fig