83

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。

4

10 回答 10

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。

另见http://pypi.python.org/pypi/texttable

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))))

在此处输入图像描述

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