我有一个用例,我有多个线图(带图例),我需要根据列条件更新线图。下面是两个数据集的示例,基于国家/地区,列数据源更改。但我面临的问题是,数据源的列数并不固定,甚至类型也可能不同。因此,当我在选择新国家时基于回调更新数据源时,我收到此错误:
Error: attempted to retrieve property array for nonexistent field 'pay_conv_7d.content'.
我猜是因为在新的数据源中,pay_conv_7d.content
列不存在,但在我的情节中,这些行已经存在。我一直试图通过各种方法解决这个问题(为所有国家选择制作公共列 - 在回调中添加数据源中缺少的列,但仍然会遇到问题。
有没有什么干净的方法可以使用回调更新多个线图,而不是做很多hackish方式?任何见解或帮助将非常感激。非常感谢! :)
def setup_multiline_plots(x_axis, y_axis, title_text, data_source, plot):
num_categories = len(data_source.data['categories'])
legends_list = list(data_source.data['categories'])
colors_list = Spectral11[0:num_categories]
# xs = [data_source.data['%s.'%x_axis].values] * num_categories
# ys = [data_source.data[('%s.%s')%(y_axis,column)] for column in data_source.data['categories']]
# data_source.data['x_series'] = xs
# data_source.data['y_series'] = ys
# plot.multi_line('x_series', 'y_series', line_color=colors_list,legend='categories', line_width=3, source=data_source)
plot_list = []
for (colr, leg, column) in zip(colors_list, legends_list, data_source.data['categories']):
xs, ys = '%s.'%x_axis, ('%s.%s')%(y_axis,column)
plot.line(xs,ys, source=data_source, color=colr, legend=leg, line_width=3, name=ys)
plot_list.append(ys)
data_source.data['plot_names'] = data_source.data.get('plot_names',[]) + plot_list
plot.title.text = title_text
def update_plot(country, timeseries_df, timeseries_source,
aggregate_df, aggregate_source, category,
plot_pay_7d, plot_r_pay_90d):
aggregate_metrics = aggregate_df.loc[aggregate_df.country == country]
aggregate_metrics = aggregate_metrics.nlargest(10, 'cost')
category_types = list(aggregate_metrics[category].unique())
timeseries_df = timeseries_df[timeseries_df[category].isin(category_types)]
timeseries_multi_line_metrics = get_multiline_column_datasource(timeseries_df, category, country)
# len_series = len(timeseries_multi_line_metrics.data['time.'])
# previous_legends = timeseries_source.data['plot_names']
# current_legends = timeseries_multi_line_metrics.data.keys()
# common_legends = list(set(previous_legends) & set(current_legends))
# additional_legends_list = list(set(previous_legends) - set(current_legends))
# for legend in additional_legends_list:
# zeros = pd.Series(np.array([0] * len_series), name=legend)
# timeseries_multi_line_metrics.add(zeros, legend)
# timeseries_multi_line_metrics.data['plot_names'] = previous_legends
timeseries_source.data = timeseries_multi_line_metrics.data
aggregate_source.data = aggregate_source.from_df(aggregate_metrics)
def get_multiline_column_datasource(df, category, country):
df_country = df[df.country == country]
df_pivoted = pd.DataFrame(df_country.pivot_table(index='time', columns=category, aggfunc=np.sum).reset_index())
df_pivoted.columns = df_pivoted.columns.to_series().str.join('.')
categories = list(set([column.split('.')[1] for column in list(df_pivoted.columns)]))[1:]
data_source = ColumnDataSource(df_pivoted)
data_source.data['categories'] = categories
答案 0 :(得分:0)
最近我不得不更新Multiline
字形的数据。如果您想查看我的算法,请检查my question。
我认为您至少可以通过三种方式更新ColumnDataSource
:
您可以创建数据帧来实例化新CDS
cds = ColumnDataSource(df_pivoted)
data_source.data = cds.data
您可以创建字典并直接将其分配给数据属性
d = {
'xs0': [[7.0, 986.0], [17.0, 6.0], [7.0, 67.0]],
'ys0': [[79.0, 69.0], [179.0, 169.0], [729.0, 69.0]],
'xs1': [[17.0, 166.0], [17.0, 116.0], [17.0, 126.0]],
'ys1': [[179.0, 169.0], [179.0, 1169.0], [1729.0, 169.0]],
'xs2': [[27.0, 276.0], [27.0, 216.0], [27.0, 226.0]],
'ys2': [[279.0, 269.0], [279.0, 2619.0], [2579.0, 2569.0]]
}
data_source.data = d
如果您需要不同大小的列或空列,则可以使用NaN
值填充空白,以保持列大小。我认为这是你问题的解决方案:
import numpy as np
d = {
'xs0': [[7.0, 986.0], [17.0, 6.0], [7.0, 67.0]],
'ys0': [[79.0, 69.0], [179.0, 169.0], [729.0, 69.0]],
'xs1': [[17.0, 166.0], [np.nan], [np.nan]],
'ys1': [[179.0, 169.0], [np.nan], [np.nan]],
'xs2': [[np.nan], [np.nan], [np.nan]],
'ys2': [[np.nan], [np.nan], [np.nan]]
}
data_source.data = d
或者,如果您只需要修改一些值,则可以使用方法patch
。查看文档here。
以下示例说明如何修补整个列元素。在这种情况下,
source = ColumnDataSource(data=dict(foo=[10, 20, 30], bar=[100, 200, 300]))
patches = {
'foo' : [ (slice(2), [11, 12]) ],
'bar' : [ (0, 101), (2, 301) ],
}
source.patch(patches)
执行此操作后,source.data的值将为:
dict(foo=[11, 22, 30], bar=[101, 200, 301])
注意:一次性更新以避免性能问题非常重要