Question

假设我有3个数据框，其中包含对象的纬度，经度和时间标签信息，这样每列代表一个不同的对象，每一行代表一个时间点。

latitudes =
object          1            2            3                
0          -8.064267    -8.047483    -8.056339       
1          -8.064267    -8.047483    -8.056339     
2          -8.064267    -8.047483    -8.056339   
3          -8.064267    -8.047483    -8.056339   
4          -8.064267    -8.047483    -8.056339   
5          -8.064267    -8.047483    -8.056339   


longitudes =
object          1            2            3         
0         -34.878386   -34.904086   -34.889661     
1         -34.878386   -34.904086   -34.889661     
2         -34.878386   -34.904086   -34.889661    
3         -34.878386   -34.904086   -34.889661   
4         -34.878386   -34.904086   -34.889661    
5         -34.878386   -34.904086   -34.889661   

times =
object  1                      2                     3                   
0       2016-03-05 07:52:00   2016-03-05 16:26:00   2016-03-05 16:58:00   
1       2016-03-05 08:19:00   2016-03-05 16:42:00   2016-03-05 17:45:00   
2       2016-03-05 08:52:00   2016-03-05 17:06:00   2016-03-05 17:58:00   
3       2016-03-05 09:36:00   2016-03-05 18:21:00   2016-03-05 18:23:00   
4           NaT               2016-03-05 23:06:00   2016-03-05 22:38:00   
5           NaT               2016-03-05 23:09:00       NaT

我想绘制一个3D轨迹，z轴上有时间，x轴上有经度，y轴上有纬度，所有对象都在这里。轨迹在同一个地块上。我怎么能这样做？

这是我的尝试，但不起作用：

    from mpl_toolkits.mplot3d import Axes3D
    import matplotlib.pyplot as plt
    fig = plt.figure()
    ax = fig.add_subplot(111,projection='3d')
ax.plot(longitudes.values,latitudes.values,times.values)
    ax.set_xlabel('x')
    ax.set_ylabel('y')
    ax.set_zlabel('Time')
    plt.show()

错误：

ValueError: third arg must be a format string

谢谢

Answer 1

fig = plt.figure()
ax = fig.gca(projection='3d')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('Time')

for t in times.columns:
    ax.plot(longitudes[t].values, latitudes[t].values, times[t].values, label='t')

ax.legend()

Lm.png

Answer 2

您需要过滤掉NaT值，因为它们无法绘制。

latitudes = u"""                
-8.064267    -8.047483    -8.056339       
-8.064267    -8.047483    -8.056339     
-8.064267    -8.047483    -8.056339   
-8.064267    -8.047483    -8.056339   
-8.064267    -8.047483    -8.056339   
-8.064267    -8.047483    -8.056339"""  


longitudes = u"""        
-34.878386   -34.904086   -34.889661     
-34.878386   -34.904086   -34.889661     
-34.878386   -34.904086   -34.889661    
-34.878386   -34.904086   -34.889661   
-34.878386   -34.904086   -34.889661    
-34.878386   -34.904086   -34.889661""" 

times = u"""                
2016-03-05 07:52:00, 2016-03-05 16:26:00, 2016-03-05 16:58:00   
2016-03-05 08:19:00, 2016-03-05 16:42:00, 2016-03-05 17:45:00   
2016-03-05 08:52:00, 2016-03-05 17:06:00, 2016-03-05 17:58:00   
2016-03-05 09:36:00, 2016-03-05 18:21:00, 2016-03-05 18:23:00   
2016-03-05 09:36:00, 2016-03-05 23:06:00, 2016-03-05 22:38:00   
2016-03-05 09:36:00, 2016-03-05 23:09:00, 2016-03-05 09:36:00
"""

import io
import pandas as pd
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

kw = dict(delim_whitespace=True, header=None)
latitudes = pd.read_csv(io.StringIO(latitudes),**kw)
longitudes = pd.read_csv(io.StringIO(longitudes),**kw)
times = pd.read_csv(io.StringIO(times), delimiter=",",
                    infer_datetime_format=True,header=None)
times.iloc[4:,0] = None
times.iloc[5,2] = None

for i in times.columns:
    times[i] = pd.to_datetime(times[i], format="%Y-%m-%d %H:%M:%S")

fig = plt.figure()
ax = fig.add_subplot(111,projection='3d')

for t in times.columns:
    inx = times[t].values != None
    x = longitudes[t].values[inx]
    y = latitudes[t].values[inx]
    z = times[t].values[inx]
    ax.plot(x,y,z)

ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('Time')
plt.show()

如何制作3D时空轨迹图？

2 个答案: