我需要在Kaplan Meier图上绘制at_risk
数字。
最终结果应与此类似:
我在渲染时遇到的困难是图形底部的No. of patients at risk
。此处显示的值与x轴上的值相对应。因此,从本质上讲,它就像与X平行绘制的Y轴。
我一直在尝试复制在这里找到的多轴(https://plot.ly/python/multiple-axes/)没有成功,并且还尝试了一个子图并隐藏除X轴以外的所有内容,但是其值与上图不一致。
最好的方法是什么?
最佳答案
您可以使用子图绘制Kaplan-Meier生存图,其中Plotly有风险的患者。第一个图具有生存率,第二个图是散点图,其中仅显示文本,即未显示标记。
这两个图的y轴都相同,并且将处于危险状态的患者分别绘制在x值上。
更多示例在这里:
https://github.com/Ashafix/Kaplan-Meier_Plotly
实施例1-男性和女性患者中的肺癌
import pandas as pd
import lifelines
import plotly
import numpy as np
plotly.offline.init_notebook_mode()
df = pd.read_csv('http://www-eio.upc.edu/~pau/cms/rdata/csv/survival/lung.csv')
fig = plotly.tools.make_subplots(rows=2, cols=1, print_grid=False)
kmfs = []
dict_sex = {1: 'Male', 2: 'Female'}
steps = 5 # the number of time points where number of patients at risk which should be shown
x_min = 0 # min value in x-axis, used to make sure that both plots have the same range
x_max = 0 # max value in x-axis
for sex in df.sex.unique():
T = df[df.sex == sex]["time"]
E = df[df.sex == sex]["status"]
kmf = lifelines.KaplanMeierFitter()
kmf.fit(T, event_observed=E)
kmfs.append(kmf)
x_max = max(x_max, max(kmf.event_table.index))
x_min = min(x_min, min(kmf.event_table.index))
fig.append_trace(plotly.graph_objs.Scatter(x=kmf.survival_function_.index,
y=kmf.survival_function_.values.flatten(),
name=dict_sex[sex]),
1, 1)
for s, sex in enumerate(df.sex.unique()):
x = []
kmf = kmfs[s].event_table
for i in range(0, int(x_max), int(x_max / (steps - 1))):
x.append(kmf.iloc[np.abs(kmf.index - i).argsort()[0]].name)
fig.append_trace(plotly.graph_objs.Scatter(x=x,
y=[dict_sex[sex]] * len(x),
text=[kmfs[s].event_table[kmfs[s].event_table.index == t].at_risk.values[0] for t in x],
mode='text',
showlegend=False),
2, 1)
# just a dummy line used as a spacer/header
t = [''] * len(x)
t[1] = 'Patients at risk'
fig.append_trace(plotly.graph_objs.Scatter(x=x,
y=[''] * len(x),
text=t,
mode='text',
showlegend=False),
2, 1)
# prettier layout
x_axis_range = [x_min - x_max * 0.05, x_max * 1.05]
fig['layout']['xaxis2']['visible'] = False
fig['layout']['xaxis2']['range'] = x_axis_range
fig['layout']['xaxis']['range'] = x_axis_range
fig['layout']['yaxis']['domain'] = [0.4, 1]
fig['layout']['yaxis2']['domain'] = [0.0, 0.3]
fig['layout']['yaxis2']['showgrid'] = False
fig['layout']['yaxis']['showgrid'] = False
plotly.offline.iplot(fig)
实施例2-不同治疗方法的结肠癌
df = pd.read_csv('http://www-eio.upc.edu/~pau/cms/rdata/csv/survival/colon.csv')
fig = plotly.tools.make_subplots(rows=2, cols=1, print_grid=False)
kmfs = []
steps = 5 # the number of time points where number of patients at risk which should be shown
x_min = 0 # min value in x-axis, used to make sure that both plots have the same range
x_max = 0 # max value in x-axis
for rx in df.rx.unique():
T = df[df.rx == rx]["time"]
E = df[df.rx == rx]["status"]
kmf = lifelines.KaplanMeierFitter()
kmf.fit(T, event_observed=E)
kmfs.append(kmf)
x_max = max(x_max, max(kmf.event_table.index))
x_min = min(x_min, min(kmf.event_table.index))
fig.append_trace(plotly.graph_objs.Scatter(x=kmf.survival_function_.index,
y=kmf.survival_function_.values.flatten(),
name=rx),
1, 1)
fig_patients = []
for s, rx in enumerate(df.rx.unique()):
kmf = kmfs[s].event_table
x = []
for i in range(0, int(x_max), int(x_max / (steps - 1))):
x.append(kmf.iloc[np.abs(kmf.index - i).argsort()[0]].name)
fig.append_trace(plotly.graph_objs.Scatter(x=x,
y=[rx] * len(x),
text=[kmfs[s].event_table[kmfs[s].event_table.index == t].at_risk.values[0] for t in x],
mode='text',
showlegend=False),
2, 1)
# just a dummy line used as a spacer/header
t = [''] * len(x)
t[1] = 'Patients at risk'
fig.append_trace(plotly.graph_objs.Scatter(x=x,
y=[''] * len(x),
text=t,
mode='text',
showlegend=False),
2, 1)
# prettier layout
x_axis_range = [x_min - x_max * 0.05, x_max * 1.05]
fig['layout']['xaxis2']['visible'] = False
fig['layout']['xaxis2']['range'] = x_axis_range
fig['layout']['xaxis']['range'] = x_axis_range
fig['layout']['yaxis']['domain'] = [0.4, 1]
fig['layout']['yaxis2']['domain'] = [0.0, 0.3]
fig['layout']['yaxis2']['showgrid'] = False
fig['layout']['yaxis']['showgrid'] = False
plotly.offline.iplot(fig)