Matplotlib中绘制的点之间的连接线

Question

我有4个X和Y列表，分别用matplotlib.pyplot绘制，即point1 [x1，y1]，point2 [x2，y2]，point3 [x3，y3]和point4 [x4，y4]。 我要在图中执行的操作是将point1连接到point2，将point2连接到point3，依此类推，直到所有4个点都连接在一起，在我的情况下这将代表一个正方形。 该数据是我正在使用的矩形泵的动态x和y排量的数据，该数据显示了是否超过了船舶月池内部的排量限制。

这是我到目前为止的代码，它给出了以下图和生成的图：

## SSLP displacement time histories to be plotted
point1 = (3.61, 4, -3)
point2 = (3.61, -4, -3)
point3 = (-3.61, -4, -3)
point4 = (-3.61, 4, -3)
SSLPXPoint1 = SSLP.TimeHistory('X', 1, objectExtra=OrcFxAPI.oeBuoy(point1))
SSLPYPoint1 = SSLP.TimeHistory('Y', 1, objectExtra=OrcFxAPI.oeBuoy(point1))
SSLPXPoint2 = SSLP.TimeHistory('X', 1, objectExtra=OrcFxAPI.oeBuoy(point2))
SSLPYPoint2 = SSLP.TimeHistory('Y', 1, objectExtra=OrcFxAPI.oeBuoy(point2))
SSLPXPoint3 = SSLP.TimeHistory('X', 1, objectExtra=OrcFxAPI.oeBuoy(point3))
SSLPYPoint3 = SSLP.TimeHistory('Y', 1, objectExtra=OrcFxAPI.oeBuoy(point3))
SSLPXPoint4 = SSLP.TimeHistory('X', 1, objectExtra=OrcFxAPI.oeBuoy(point4))
SSLPYPoint4 = SSLP.TimeHistory('Y', 1, objectExtra=OrcFxAPI.oeBuoy(point4))

# setup plot
caseName = os.path.splitext(info.modelFileName)[0]
point1Plot = [3.61, 4]
point2Plot = [3.61, -4]
point3Plot = [-3.61, -4]
point4Plot = [-3.61, 4]
vesselPointsX = [90.89, 100.89, 100.89, 90.89, 90.89]
vesselPointsY = [5, 5, -5, -5, 5]
moonpoolCLX = [89, 103]
moonpoolCLY = [0, 0]
fig = plt.figure(figsize=(20, 15))
ax = fig.add_subplot(1, 1, 1)
plt.plot(vesselPointsX, vesselPointsY, 'r', lw=2, label='OCV Moonpool Limits')
plt.plot(moonpoolCLX, moonpoolCLY, 'k--', label='Moonpool CL')
plt.plot(SSLPXPoint1, SSLPYPoint1, 'k')
plt.plot(SSLPXPoint2, SSLPYPoint2, 'k')
plt.plot(SSLPXPoint3, SSLPYPoint3, 'k')
plt.plot(SSLPXPoint4, SSLPYPoint4, 'k')
ax.set_title("SSLP Maximum Offsets Inside Moonpool for {}".format(caseName), fontsize=20)
ax.set_xlabel('Distance Along OCV from Stern [m]', fontsize=15)
ax.set_ylabel('Distance from Moonpool Centerline, (+) Towards Portside [m]', fontsize=15)
ax.set_xlim(89, 103)
ax.set_ylim(-7, 7)
plt.gca().set_aspect('equal', adjustable='box')
plt.draw()
plt.legend()
plt.tight_layout()
plt.show()

任何帮助将不胜感激。

谢谢，

布赖恩

Answer 1

这应该对您有帮助。 进行更改。

from matplotlib.pyplot import plot, show
plot([SSLPXPoint1, SSLPXPoint2], [SSLPYPoint1, SSLPYPoint2])
plot([SSLPXPoint3, SSLPXPoint2], [SSLPYPoint3, SSLPYPoint2])
plot([SSLPXPoint3, SSLPXPoint4], [SSLPYPoint3, SSLPYPoint4])
show()

已进行编辑，因为上一个连接了所有点。