如何在Jupyter Notebook中显示numpy数组的全部结果值？

我编写了如下基于NumPy库的Python代码，用于生成晶格坐标数组coords1和coords2：

import numpy as np
import matplotlib.pyplot as plt

d0 = 0.3330630630630631
a0 = 0.15469469469469468

theta = 2
nmax=15

# lattice vectors sublattice 1
a1= np.array([3/2*a0,3**0.5/2*a0,0])
a2= np.array([3/2*a0,-3**0.5/2*a0,0])

# lattice vectors sublattice 2
b1 = np.array([ np.cos(theta) * a1[0] - np.sin(theta) * a1[1], np.sin(theta) * a1[0] + np.cos(theta) * a1[1], 0 ])
b2 = np.array([ np.cos(theta) * a2[0] - np.sin(theta) * a2[1], np.sin(theta) * a2[0] + np.cos(theta) * a2[1], 0 ])

## coordinates for the unrotated layer sublattice a&b
coords1a = np.array([i * a1 + j * a2 for i in range(-nmax-1, nmax+1) for j in range(-nmax-1, nmax+1)])
coords1b = np.array([i * a1 + j * a2 + [a0,0.,0.] for i in range(-nmax-1, nmax+1) for j in range(-nmax-1, nmax+1)])

## coordinates for the rotated layer sublattice a&b
coords2a = np.array([i * b1 + j * b2 + [0.,0.,d0]  for i in range(-nmax-1, nmax+1) for j in range(-nmax-1, nmax+1)])
coords2b = np.array([i * b1 + j * b2 + [np.cos(theta)*a0,np.sin(theta)*a0,d0] for i in range(-nmax-1, nmax+1) for j in range(-nmax-1, nmax+1)])

coords1 = np.concatenate((coords1a, coords1b))
coords2 = np.concatenate((coords2a, coords2b))

但在Jupyter Notebook中查看coords1和coords2数组时，仅显示6个元素及省略号，无法查看全部值，请问该如何设置才能显示所有结果值？

方法1：修改NumPy全局打印配置 #

通过设置NumPy的打印阈值为无穷大，让它显示数组的所有元素：

import numpy as np
# 设置显示所有元素
np.set_printoptions(threshold=np.inf)
# 直接打印数组
print(coords1)
print(coords2)

如果不想全局修改配置，可使用上下文管理器临时生效：

import numpy as np
with np.printoptions(threshold=np.inf):
    print(coords1)

方法2：转换为Python列表查看 #

将NumPy数组转换为普通Python列表，默认会显示所有元素：

print(coords1.tolist())

方法3：结合Jupyter的显示工具 #

使用IPython的display函数配合NumPy配置，在Notebook中完整显示数组：

from IPython.display import display
import numpy as np
np.set_printoptions(threshold=np.inf)
display(coords1)

内容的提问来源于stack exchange，提问作者didem