python - 使用 for 循环在子图中绘制箱形图

Question

我有一个数据框(df)。我想绘制每列的箱线图，并将它们显示为子图。但是，下面的代码会出错。

from plotly.subplots import make_subplots
import plotly.express as px
import pandas as pd

fig = make_subplots(rows = 5, cols = 2)
for i, column in enumerate(df.select_dtypes(include = np.number).columns.tolist()):
    fig.add_trace(px.box(df, y = [column]), row = i%5, col = i//5)
    fig.update_layout(width = 800, height = 800)
    fig.show()

错误:“----> 5 Fig.add_trace(px.box(df, y = [column]), row = i%5, col =i//5)”

Answer 1

在我这边，您的代码片段会触发:

The 'data' property is a tuple of trace instances that may be specified as:

• A list or tuple of trace instances (e.g. [Scatter(...), Bar(...)])
• A single trace instance

这是因为 fig.add_trace 被设计为将 trace 对象 作为输入，例如 go.Box ，而不是您在此处使用的 px.box 返回的 figure 对象。因此，您可以使用类似以下内容来代替:

# plotly setup
plot_rows=6
plot_cols=6
fig = make_subplots(rows=plot_rows, cols=plot_cols)

# add traces
x = 0
for i in range(1, plot_rows + 1):
    for j in range(1, plot_cols + 1):
        #print(str(i)+ ', ' + str(j))
        fig.add_trace(go.Box(y=df[df.columns[x]].values,
                                 name = df.columns[x],
                            ),
                     row=i,
                     col=j)

        x=x+1

并得到:

完整代码:

## imports
from plotly.subplots import make_subplots
import plotly.graph_objs as go
import pandas as pd
import numpy as np

# data
np.random.seed(123)
frame_rows = 10
n_plots = 36
frame_columns = ['B_'+str(e) for e in list(range(n_plots+1))]
df = pd.DataFrame(np.random.uniform(-10,10,size=(frame_rows, len(frame_columns))),
                  index=pd.date_range('1/1/2020', periods=frame_rows),
                    columns=frame_columns)
df=df.cumsum()+100
df.iloc[0]=100

# plotly setup
plot_rows=6
plot_cols=6
fig = make_subplots(rows=plot_rows, cols=plot_cols)

# add traces
x = 0
for i in range(1, plot_rows + 1):
    for j in range(1, plot_cols + 1):
        #print(str(i)+ ', ' + str(j))
        fig.add_trace(go.Box(y=df[df.columns[x]].values,
                                 name = df.columns[x],
                            ),
                     row=i,
                     col=j)

        x=x+1

# Format and show fig
fig.update_layout(height=1200, width=1200)
fig.show()