Python 如何向散点图添加一条最佳拟合线
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/37234163/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to add a line of best fit to scatter plot
提问by JavascriptLoser
I'm currently working with Pandas and matplotlib to perform some data visualization and I want to add a line of best fit to my scatter plot.
我目前正在使用 Pandas 和 matplotlib 来执行一些数据可视化，我想为散点图添加一条最适合的线。
Here is my code:
这是我的代码：
import matplotlib
import matplotlib.pyplot as plt
import pandas as panda
import numpy as np
def PCA_scatter(filename):
matplotlib.style.use('ggplot')
data = panda.read_csv(filename)
data_reduced = data[['2005', '2015']]
data_reduced.plot(kind='scatter', x='2005', y='2015')
plt.show()
PCA_scatter('file.csv')
How do I go about this?
我该怎么做？
回答by Robert Calhoun
回答by Stefan
You can use np.polyfit()and np.poly1d(). Estimate a first degree polynomial using the same xvalues, and add to the axobject created by the .scatter()plot. Using an example:
您可以使用np.polyfit()和np.poly1d()。使用相同的x值估计一阶多项式，并将其添加到绘图ax创建的对象中.scatter()。使用示例：
import numpy as np
2005 2015
0 18882 21979
1 1161 1044
2 482 558
3 2105 2471
4 427 1467
5 2688 2964
6 1806 1865
7 711 738
8 928 1096
9 1084 1309
10 854 901
11 827 1210
12 5034 6253
Estimate first-degree polynomial:
估计一次多项式：
z = np.polyfit(x=df.loc[:, 2005], y=df.loc[:, 2015], deg=1)
p = np.poly1d(z)
df['trendline'] = p(df.loc[:, 2005])
2005 2015 trendline
0 18882 21979 21989.829486
1 1161 1044 1418.214712
2 482 558 629.990208
3 2105 2471 2514.067336
4 427 1467 566.142863
5 2688 2964 3190.849200
6 1806 1865 2166.969948
7 711 738 895.827339
8 928 1096 1147.734139
9 1084 1309 1328.828428
10 854 901 1061.830437
11 827 1210 1030.487195
12 5034 6253 5914.228708
and plot:
和情节：
ax = df.plot.scatter(x=2005, y=2015)
df.set_index(2005, inplace=True)
df.trendline.sort_index(ascending=False).plot(ax=ax)
plt.gca().invert_xaxis()
To get:
要得到：
Also provides the the line equation:
还提供了线方程：
'y={0:.2f} x + {1:.2f}'.format(z[0],z[1])
y=1.16 x + 70.46
回答by Alex Williams
Another option (using np.linalg.lstsq):
另一种选择（使用np.linalg.lstsq）：
# generate some fake data
N = 50
x = np.random.randn(N, 1)
y = x*2.2 + np.random.randn(N, 1)*0.4 - 1.8
plt.axhline(0, color='r', zorder=-1)
plt.axvline(0, color='r', zorder=-1)
plt.scatter(x, y)
# fit least-squares with an intercept
w = np.linalg.lstsq(np.hstack((x, np.ones((N,1)))), y)[0]
xx = np.linspace(*plt.gca().get_xlim()).T
# plot best-fit line
plt.plot(xx, w[0]*xx + w[1], '-k')
回答by user702846
This is covering the plotlyapproach
这涵盖了plotly方法
#load the libraries
import pandas as pd
import numpy as np
import plotly.express as px
import plotly.graph_objects as go
# create the data
N = 50
x = pd.Series(np.random.randn(N))
y = x*2.2 - 1.8
# plot the data as a scatter plot
fig = px.scatter(x=x, y=y)
# fit a linear model
m, c = fit_line(x = x,
y = y)
# add the linear fit on top
fig.add_trace(
go.Scatter(
x=x,
y=m*x + c,
mode="lines",
line=go.scatter.Line(color="red"),
showlegend=False)
)
# optionally you can show the slop and the intercept
mid_point = x.mean()
fig.update_layout(
showlegend=False,
annotations=[
go.layout.Annotation(
x=mid_point,
y=m*mid_point + c,
xref="x",
yref="y",
text=str(round(m, 2))+'x+'+str(round(c, 2)) ,
)
]
)
fig.show()
where fit_lineis
这里fit_line是
def fit_line(x, y):
# given one dimensional x and y vectors - return x and y for fitting a line on top of the regression
# inspired by the numpy manual - https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.lstsq.html
x = x.to_numpy() # convert into numpy arrays
y = y.to_numpy() # convert into numpy arrays
A = np.vstack([x, np.ones(len(x))]).T # sent the design matrix using the intercepts
m, c = np.linalg.lstsq(A, y, rcond=None)[0]
return m, c
