# 在python中对齐三个时间序列align three time series in python

``````from pandas import *
import pandas as pd
s1 = Series(vec1)
s2 = Series(vec2)
s3 = s1.align(s2,join='inner')

s1 = np.array(s1)
s2 = np.array(s2)
s3 = np.array(s3)

plt.plot(t,s1)
plt.plot(t,s2)
plt.plot(t,s3)
plt.show()
``````

## 2 个回复2

### #1楼 票数：6 已采纳

``````import matplotlib.pyplot as plt  #making toy data
x = np.arange(0, 2, .05)

s = []
s.append(np.sin(x*3))
s.append(np.cos(x*4+0.3))
s.append(np.sin(x)*np.exp(2-x)*-1)
plt.clf()
plt.plot(x, s[0], x, s[1], x, s[2])
plt.show()
plt.clf()
``````

``````mindexes = [i.argmin() for i in s]  #finding the minima
shifts = mindexes - min(mindexes)

def shifted(shift, x, s):
if shift == 0:   #x[:-0] was not what I wanted.
return x, s
else:
return x[:-1*shift], s[shift:]

for i in (0,1,2):
px, py = shifted(shifts[i], x, s[i])
plt.plot(px, py)
plt.show()
``````

### #2楼 票数：1

``````# assuming you are shifting s2 to the left
# cut off at 800 s
s1 = s1[s1.index>=800]
s2 = s2[s2.index>=800]

while s2.index[s2==s2.min()]>s1.index[s1==s1.min()]:
s2 = s2[801:]
s2.index = np.arange(800,800 + s2.shape[0])
``````

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