是否有一种简单的内置方法可以将 2D Python 列表打印为 2D 矩阵？
所以这：
[["A", "B"], ["C", "D"]]
会变成类似的东西
A B
C D
我找到了该pprint模块，但它似乎没有做我想要的。
问问题
123632 次
10 回答
90
为了让事情变得有趣，让我们尝试一个更大的矩阵：
matrix = [
["Ah!", "We do have some Camembert", "sir"],
["It's a bit", "runny", "sir"],
["Well,", "as a matter of fact it's", "very runny, sir"],
["I think it's runnier", "than you", "like it, sir"]
]
s = [[str(e) for e in row] for row in matrix]
lens = [max(map(len, col)) for col in zip(*s)]
fmt = '\t'.join('{{:{}}}'.format(x) for x in lens)
table = [fmt.format(*row) for row in s]
print '\n'.join(table)
输出：
Ah! We do have some Camembert sir
It's a bit runny sir
Well, as a matter of fact it's very runny, sir
I think it's runnier than you like it, sir
UPD：对于多行单元格，这样的东西应该可以工作：
text = [
["Ah!", "We do have\nsome Camembert", "sir"],
["It's a bit", "runny", "sir"],
["Well,", "as a matter\nof fact it's", "very runny,\nsir"],
["I think it's\nrunnier", "than you", "like it,\nsir"]
]
from itertools import chain, izip_longest
matrix = chain.from_iterable(
izip_longest(
*(x.splitlines() for x in y),
fillvalue='')
for y in text)
然后应用上面的代码。
于 2012-11-04T00:36:01.240 回答
50
如果您可以使用 Pandas（Python 数据分析库），您可以通过将其转换为 DataFrame 对象来漂亮地打印 2D 矩阵：
from pandas import *
x = [["A", "B"], ["C", "D"]]
print DataFrame(x)
0 1
0 A B
1 C D
于 2015-08-22T18:36:11.693 回答
40
对于没有任何第三方库的 Python 3：
matrix = [["A", "B"], ["C", "D"]]
print('\n'.join(['\t'.join([str(cell) for cell in row]) for row in matrix]))
输出
A B
C D
于 2018-05-09T16:06:03.893 回答
36
你总是可以使用numpy：
import numpy as np
A = [['A', 'B'], ['C', 'D']]
print(np.matrix(A))
输出：
[['A' 'B']
['C' 'D']]
于 2016-04-11T01:48:14.423 回答
8
只是为了提供一个更简单的替代方案print('\n'.join(\['\t'.join(\[str(cell) for cell in row\]) for row in matrix\]))：
matrix = [["A", "B"], ["C", "D"]]
for row in matrix:
print(*row)
解释
*rowunpacks row，因此print("A", "B")称为 when rowis ["A", "B"]，例如。
注意
只有当每列具有相同的宽度时，两个答案才会被很好地格式化。要更改分隔符，请使用sep关键字。例如，
for row in matrix:
print(*row, sep=', ')
将打印
A, B
C, D
反而。
没有 for 循环的单线
print(*(' '.join(row) for row in matrix), sep='\n')
' '.join(row) for row in matrix)为每一行返回一个字符串，例如A Bwhen rowis ["A", "B"]。
*(' '.join(row) for row in matrix), sep='\n')解包返回序列的生成器'A B', 'C D'，因此在给出print('A B', 'C D', sep='\n')的示例中调用它matrix。
于 2019-12-01T08:48:07.977 回答
6
pandas比使用prettytable模块更轻量级的方法
from prettytable import PrettyTable
x = [["A", "B"], ["C", "D"]]
p = PrettyTable()
for row in x:
p.add_row(row)
print p.get_string(header=False, border=False)
产量：
A B
C D
prettytable有很多选项可以以不同的方式格式化您的输出。
有关更多信息，请参阅https://code.google.com/p/prettytable/
于 2015-08-22T20:47:23.277 回答
1
如果您使用的是 Notebook/IPython 环境，那么 sympy 可以使用 IPython.display 打印令人愉悦的矩阵：
import numpy as np
from sympy import Matrix, init_printing
init_printing()
print(np.random.random((3,3)))
display(np.random.random((3,3)))
display(Matrix(np.random.random((3,3))))
于 2020-09-10T17:02:54.397 回答
1
我还推荐tabulate，它也可以选择打印标题：
from tabulate import tabulate
lst = [['London', 20],['Paris', 30]]
print(tabulate(lst, headers=['City', 'Temperature']))
：
City Temperature
------ -------------
London 20
Paris 30
于 2020-12-10T22:04:58.040 回答
0
You can update print's end=' ' so that it prints space instead of '\n' in the inner loop and outer loop can have print().
a=[["a","b"],["c","d"]]
for i in a:
for j in i:
print(j, end=' ')
print()
I found this solution from here.
于 2018-12-23T06:57:50.913 回答
-2
请参阅以下代码。
# Define an empty list (intended to be used as a matrix)
matrix = []
matrix.append([1, 2, 3, 4])
matrix.append([4, 6, 7, 8])
print matrix
# Now just print out the two rows separately
print matrix[0]
print matrix[1]
于 2018-01-13T14:30:43.260 回答