在JupyterLab和Jupyter Notebook中,您都可以使用ctrl + Enter以下命令执行单元:
码:
print('line 1')
print('line 2')
print('line 3')
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单元格和输出:
但是,你怎么能运行只 line 2?甚至在不运行整个单元的情况下选择单元中的线?当然,您可以只插入带有单行或选择的行的单元格,但这确实非常麻烦并且很快。那么,有没有更好的方法可以做到这一点?
我已经实现了简单的代码来gradient boosting regression (GBR)预测one output并且效果很好,但是当我尝试预测时two outputs出现错误,如下所示:
ValueError Traceback (most recent call last)
<ipython-input-5-bb1f191ee195> in <module>()
4 }
5 gradient_boosting_regressor = ensemble.GradientBoostingRegressor(**params)
----> 6 gradient_boosting_regressor.fit(train_data,train_targets)
7 # 'learning_rate': 0.01
D:\Anoconda\lib\site-packages\sklearn\ensemble\gradient_boosting.py in fit(self, X, y, sample_weight, monitor)
977
978 # Check input
--> 979 X, y = check_X_y(X, y, accept_sparse=['csr', 'csc', 'coo'], dtype=DTYPE)
980 n_samples, self.n_features_ = X.shape
981 if sample_weight is None:
D:\Anoconda\lib\site-packages\sklearn\utils\validation.py in check_X_y(X, y, accept_sparse, dtype, order, copy, force_all_finite, ensure_2d, allow_nd, multi_output, ensure_min_samples, …Run Code Online (Sandbox Code Playgroud) 我想用 R 中的绘图来表示以原点为根的 2D 向量。此外,我想根据分类变量为向量着色。问题是我可以创建颜色编码但没有箭头的线条:
library(plotly)
library(dplyr)
v <- c(1, 1)
b1 <- c(1, 0)
b2 <- c(0, 1)
df <- data.frame(
x = c(v[1], b1[1], b2[1]),
y = c(v[2], b1[2], b2[2]),
is_basis = c(FALSE, TRUE, TRUE)
)
df %>%
plot_ly(x = ~x, y = ~y, color = ~is_basis) %>%
add_segments(xend = ~x, yend = ~y, x = 0, y = 0, colors = c("red","black"))
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或者使用箭头但不使用颜色编码:
df %>%
plot_ly(x = ~x, y = ~y, color = ~is_basis) %>%
add_annotations(x = …Run Code Online (Sandbox Code Playgroud) 我正在尝试将趋势线添加到绘制的条形图中plotly
代码:
import plotly.express as px
fig = px.bar(count, x="date", y="count",trendline="ols")
fig.update_layout(
xaxis_title="Date",
yaxis_title = "Count"
)
fig.show()
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错误:
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-129-8b01de219d3c> in <module>
----> 1 fig = px.bar(count, x="date", y="count",trendline="ols")
2
3 fig.update_layout(
4 xaxis_title="Date",
5 yaxis_title = "Count"
TypeError: bar() got an unexpected keyword argument 'trendline'
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这是数据
如何成功地将趋势线添加到该图中?
我喜欢使用 Plotly 来可视化所有内容,我试图通过 Plotly 来可视化混淆矩阵,这是我的代码:
def plot_confusion_matrix(y_true, y_pred, class_names):
confusion_matrix = metrics.confusion_matrix(y_true, y_pred)
confusion_matrix = confusion_matrix.astype(int)
layout = {
"title": "Confusion Matrix",
"xaxis": {"title": "Predicted value"},
"yaxis": {"title": "Real value"}
}
fig = go.Figure(data=go.Heatmap(z=confusion_matrix,
x=class_names,
y=class_names,
hoverongaps=False),
layout=layout)
fig.show()
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结果是
python visualization machine-learning confusion-matrix plotly
抱歉,帖子太长了。我是 python 和 plotly 的新手,所以请耐心等待。
我正在尝试制作带有趋势线的散点图,以向我展示包括回归参数在内的图例图例,但出于某种原因,我不明白为什么px.scatter不向我展示我的轨迹图例。这是我的代码
fig1 = px.scatter(data_frame = dataframe,
x="xdata",
y="ydata",
trendline = 'ols')
fig1.layout.showlegend = True
fig1.show()
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这会显示散点图和趋势线,但即使我试图覆盖它也没有图例。
我曾经pio.write_json(fig1, "fig1.plotly")将它导出到 jupyterlab plotly 图表工作室并手动添加图例,但即使我启用了它,它也不会显示在图表工作室中。
我打印了变量print(fig1)以查看发生了什么,这是(部分)结果
(Scatter({
'hovertemplate': '%co=%{x}<br>RPM=%{y}<extra></extra>',
'legendgroup': '',
'marker': {'color': '#636efa', 'symbol': 'circle'},
'mode': 'markers',
'name': '',
'showlegend': False,
'x': array([*** some x data ***]),
'xaxis': 'x',
'y': array([*** some y data ***]),
'yaxis': 'y'
}), Scatter({
'hovertemplate': ('<b>OLS trendline</b><br>RPM = ' ... ' <b>(trend)</b><extra></extra>'),
'legendgroup': '',
'marker': {'color': '#636efa', …Run Code Online (Sandbox Code Playgroud) python data-visualization plotly plotly-python plotly-express
我的数据框看起来像这样:
user age gender
0 23 12 male
1 24 13 male
2 25 15 female
3 26 20 male
4 27 21 male
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并使用
px.sunburst(df, path=["gender", "age"])
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给我正确的森伯斯特图,其中性别位于饼图的中间部分,并且对于每个性别,它都有相关的年龄。
我想使用 graph_objects 而不是 plotly express 来做到这一点,因为我希望两个旭日图并排。
从 df 我上面有我如何在graph_objects中使用它。我不明白要向标签、父母、ID 等添加什么值...
fig = go.Figure()
fig.add_trace(
go.Sunburst(
lables = df.age,
parents = df.gender,
domain=dict(column=0)
)
)
fig.show()
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我已经阅读了文档,但是我无法理解它是如何工作的。如果有人知道,请告诉我如何使用我上面的 df 使用 graph_object 创建森伯斯特图。
不确定我是否在这里遗漏了一些明显的东西,但是当我在带有注释的文本中插入一个中断(<br>)时,它似乎忽略了它。我已经尝试过fig.add_annotations,但同样的事情发生了。
你知道为什么这不起作用吗?
import pandas as pd
import plotly.graph_objects as go
import numpy as np
df = pd.DataFrame({"Growth_Type": ["Growing Fast", "Growing", "Stable", "Dropping", "Dropping Fast"],
"Accounts": [407,1275,3785,1467,623],
"Gain_Share": [1.20,8.1,34.4,6.5,0.4],
"Keep_Share": [16.5, 101.2, 306.3, 107.2, 27.7]})
df2 = pd.concat([pd.DataFrame({"Growth_Type":df["Growth_Type"],
"Opportunity_Type": np.repeat("Gain Share", 5),
"Wallet_Share": df["Gain_Share"]}),
pd.DataFrame({"Growth_Type":df["Growth_Type"],
"Opportunity_Type": np.repeat("Keep Share", 5),
"Wallet_Share": df["Keep_Share"]})])
fig = go.Figure()
fig.add_trace(go.Bar(x = df2["Wallet_Share"],
y = df2["Growth_Type"],
orientation = "h"
))
fig.update_layout(font = dict(size = 12, color = "#A6ACAF"),
xaxis_tickprefix = "$",
plot_bgcolor …Run Code Online (Sandbox Code Playgroud) 我希望我的等高线图网格引用相同的颜色条,但我得到了四个堆叠的颜色条。我怎样才能只有一个颜色条,其数值引用所有图中的数据?或者,换句话说,我的绘图颜色如何引用相同的颜色条?
这是测试代码:
import plotly.graph_objects as go
from plotly.subplots import make_subplots
z1 = [[2, 4, 7, 12, 13, 14, 15, 16],
[3, 1, 6, 11, 12, 13, 16, 17],
[4, 2, 7, 7, 11, 14, 17, 18],
[5, 3, 8, 8, 13, 15, 18, 19],
[7, 4, 10, 9, 16, 18, 20, 19],
[9, 10, 5, 27, 23, 21, 21, 21],
[11, 14, 17, 26, 25, 24, 23, 22]]
z2 = [[20, 44, 7, 120, 1, 1, 5, 16],
[3, 10, …Run Code Online (Sandbox Code Playgroud) 我想更改plotly python 中3D 曲面图的配色方案。Plotly 默认分配配色方案,如下图所示。
这是我的代码
import import plotly.graph_objects as go
import pandas as pd
data = pd.read_csv('\Data.csv')
data.set_index("years", inplace = True)
figure = go.Figure(data=[go.Surface(z=data.values)])
figure.update_layout(
scene = dict(
xaxis = dict(
title = 'Months',
#nticks = 5,
autorange='reversed',
showgrid=True,
gridwidth=1,
gridcolor='Blue',
ticktext = data.columns,
tickvals= list(range(0,data.shape[1]))),
yaxis = dict(
title = 'years',
showgrid=True,
gridwidth=1,
gridcolor='Blue',
ticktext = data.index,
tickvals= list(range(0,data.shape[0]))),
zaxis = dict(
title = 'Discharge (Cumecs)',
#showgrid=True,
gridwidth=1,
gridcolor='Blue')),
tilte = 'Plot 1'
)
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