在 python 中创建漂亮的列输出
- 2025-01-07 08:45:00
- admin 原创
- 116
问题描述：
我正在尝试在 python 中创建一个漂亮的列表，以便与我创建的命令行管理工具一起使用。
基本上，我想要一个像这样的列表：
[['a', 'b', 'c'], ['aaaaaaaaaa', 'b', 'c'], ['a', 'bbbbbbbbbb', 'c']]
变成：
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
使用普通的标签在这里不起作用，因为我不知道每行中最长的数据。
这与 Linux 中的“column -t”行为相同。
$ echo -e "a b c
aaaaaaaaaa b c
a bbbbbbbbbb c"
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
$ echo -e "a b c
aaaaaaaaaa b c
a bbbbbbbbbb c" | column -t
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
我寻找过各种 Python 库来执行此操作，但找不到任何有用的东西。
解决方案 1：
从 Python 2.6+ 开始，您可以按照以下方式使用格式字符串将列设置为至少 20 个字符并将文本右对齐。
table_data = [
['a', 'b', 'c'],
['aaaaaaaaaa', 'b', 'c'],
['a', 'bbbbbbbbbb', 'c']
]
for row in table_data:
print("{: >20} {: >20} {: >20}".format(*row))
输出：
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
解决方案 2：
data = [['a', 'b', 'c'], ['aaaaaaaaaa', 'b', 'c'], ['a', 'bbbbbbbbbb', 'c']]
col_width = max(len(word) for row in data for word in row) + 2 # padding
for row in data:
print "".join(word.ljust(col_width) for word in row)
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
这样做的目的是计算最长的数据条目以确定列宽，然后.ljust()在打印每列时添加必要的填充。
解决方案 3：
我带着同样的要求来到这里，但@lvc 和@Preet 的答案似乎更符合column -t产生的结果，因为列有不同的宽度：
>>> rows = [ ['a', 'b', 'c', 'd']
... , ['aaaaaaaaaa', 'b', 'c', 'd']
... , ['a', 'bbbbbbbbbb', 'c', 'd']
... ]
...
>>> widths = [max(map(len, col)) for col in zip(*rows)]
>>> for row in rows:
... print " ".join((val.ljust(width) for val, width in zip(row, widths)))
...
a b c d
aaaaaaaaaa b c d
a bbbbbbbbbb c d
解决方案 4：
这有点晚了，而且这是对我编写的包的无耻推销，但您也可以查看Columnar包。
它接受输入列表和标题列表，并输出表格格式的字符串。此代码片段创建了一个类似 docker 的表格：
from columnar import columnar
headers = ['name', 'id', 'host', 'notes']
data = [
['busybox', 'c3c37d5d-38d2-409f-8d02-600fd9d51239', 'linuxnode-1-292735', 'Test server.'],
['alpine-python', '6bb77855-0fda-45a9-b553-e19e1a795f1e', 'linuxnode-2-249253', 'The one that runs python.'],
['redis', 'afb648ba-ac97-4fb2-8953-9a5b5f39663e', 'linuxnode-3-3416918', 'For queues and stuff.'],
['app-server', 'b866cd0f-bf80-40c7-84e3-c40891ec68f9', 'linuxnode-4-295918', 'A popular destination.'],
['nginx', '76fea0f0-aa53-4911-b7e4-fae28c2e469b', 'linuxnode-5-292735', 'Traffic Cop'],
]
table = columnar(data, headers, no_borders=True)
print(table)
或者您可以使用颜色和边框使其更加美观。 
要了解有关列大小算法的更多信息并查看其余 API，您可以查看上面的链接或查看Columnar GitHub Repo
解决方案 5：
哇，只有 17 个答案。Python 的禅宗说“应该有一个——最好只有一个——明显的方法来做这件事。”
因此，这里有第 18 种方法：tabulate包支持一组可以显示为表格的数据类型，这里有一个改编自其文档的简单示例：
from tabulate import tabulate
table = [["Sun",696000,1989100000],
["Earth",6371,5973.6],
["Moon",1737,73.5],
["Mars",3390,641.85]]
print(tabulate(table, headers=["Planet","R (km)", "mass (x 10^29 kg)"]))
输出
Planet R (km) mass (x 10^29 kg)
-------- -------- -------------------
Sun 696000 1.9891e+09
Earth 6371 5973.6
Moon 1737 73.5
Mars 3390 641.85
解决方案 6：
像这样转置列是 zip 的工作：
>>> a = [['a', 'b', 'c'], ['aaaaaaaaaa', 'b', 'c'], ['a', 'bbbbbbbbbb', 'c']]
>>> list(zip(*a))
[('a', 'aaaaaaaaaa', 'a'), ('b', 'b', 'bbbbbbbbbb'), ('c', 'c', 'c')]
要找到每列所需的长度，您可以使用max：
>>> trans_a = zip(*a)
>>> [max(len(c) for c in b) for b in trans_a]
[10, 10, 1]
您可以使用合适的填充来构造要传递给的字符串print：
>>> col_lenghts = [max(len(c) for c in b) for b in trans_a]
>>> padding = ' ' # You might want more
>>> padding.join(s.ljust(l) for s,l in zip(a[0], col_lenghts))
'a b c'
解决方案 7：
你必须通过 2 次传递来完成此操作：
获取每列的最大宽度。
str.ljust()使用第一遍中的最大宽度知识来格式化列str.rjust()
解决方案 8：
为了得到更漂亮的桌子，比如
---------------------------------------------------
| First Name | Last Name | Age | Position |
---------------------------------------------------
| John | Smith | 24 | Software |
| | | | Engineer |
---------------------------------------------------
| Mary | Brohowski | 23 | Sales |
| | | | Manager |
---------------------------------------------------
| Aristidis | Papageorgopoulos | 28 | Senior |
| | | | Reseacher |
---------------------------------------------------
你可以使用这个Python 配方：
'''
From http://code.activestate.com/recipes/267662-table-indentation/
PSF License
'''
import cStringIO,operator
def indent(rows, hasHeader=False, headerChar='-', delim=' | ', justify='left',
separateRows=False, prefix='', postfix='', wrapfunc=lambda x:x):
"""Indents a table by column.
- rows: A sequence of sequences of items, one sequence per row.
- hasHeader: True if the first row consists of the columns' names.
- headerChar: Character to be used for the row separator line
(if hasHeader==True or separateRows==True).
- delim: The column delimiter.
- justify: Determines how are data justified in their column.
Valid values are 'left','right' and 'center'.
- separateRows: True if rows are to be separated by a line
of 'headerChar's.
- prefix: A string prepended to each printed row.
- postfix: A string appended to each printed row.
- wrapfunc: A function f(text) for wrapping text; each element in
the table is first wrapped by this function."""
# closure for breaking logical rows to physical, using wrapfunc
def rowWrapper(row):
newRows = [wrapfunc(item).split('
') for item in row]
return [[substr or '' for substr in item] for item in map(None,*newRows)]
# break each logical row into one or more physical ones
logicalRows = [rowWrapper(row) for row in rows]
# columns of physical rows
columns = map(None,*reduce(operator.add,logicalRows))
# get the maximum of each column by the string length of its items
maxWidths = [max([len(str(item)) for item in column]) for column in columns]
rowSeparator = headerChar * (len(prefix) + len(postfix) + sum(maxWidths) + \n len(delim)*(len(maxWidths)-1))
# select the appropriate justify method
justify = {'center':str.center, 'right':str.rjust, 'left':str.ljust}[justify.lower()]
output=cStringIO.StringIO()
if separateRows: print >> output, rowSeparator
for physicalRows in logicalRows:
for row in physicalRows:
print >> output, \n prefix \n + delim.join([justify(str(item),width) for (item,width) in zip(row,maxWidths)]) \n + postfix
if separateRows or hasHeader: print >> output, rowSeparator; hasHeader=False
return output.getvalue()
# written by Mike Brown
# http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/148061
def wrap_onspace(text, width):
"""
A word-wrap function that preserves existing line breaks
and most spaces in the text. Expects that existing line
breaks are posix newlines (
).
"""
return reduce(lambda line, word, width=width: '%s%s%s' %
(line,
'
'[(len(line[line.rfind('
')+1:])
+ len(word.split('
',1)[0]
) >= width)],
word),
text.split(' ')
)
import re
def wrap_onspace_strict(text, width):
"""Similar to wrap_onspace, but enforces the width constraint:
words longer than width are split."""
wordRegex = re.compile(r'S{'+str(width)+r',}')
return wrap_onspace(wordRegex.sub(lambda m: wrap_always(m.group(),width),text),width)
import math
def wrap_always(text, width):
"""A simple word-wrap function that wraps text on exactly width characters.
It doesn't split the text in words."""
return '
'.join([ text[width*i:width*(i+1)] \n for i in xrange(int(math.ceil(1.*len(text)/width))) ])
if __name__ == '__main__':
labels = ('First Name', 'Last Name', 'Age', 'Position')
data = \n '''John,Smith,24,Software Engineer
Mary,Brohowski,23,Sales Manager
Aristidis,Papageorgopoulos,28,Senior Reseacher'''
rows = [row.strip().split(',') for row in data.splitlines()]
print 'Without wrapping function
'
print indent([labels]+rows, hasHeader=True)
# test indent with different wrapping functions
width = 10
for wrapper in (wrap_always,wrap_onspace,wrap_onspace_strict):
print 'Wrapping function: %s(x,width=%d)
' % (wrapper.__name__,width)
print indent([labels]+rows, hasHeader=True, separateRows=True,
prefix='| ', postfix=' |',
wrapfunc=lambda x: wrapper(x,width))
# output:
#
#Without wrapping function
#
#First Name | Last Name | Age | Position
#-------------------------------------------------------
#John | Smith | 24 | Software Engineer
#Mary | Brohowski | 23 | Sales Manager
#Aristidis | Papageorgopoulos | 28 | Senior Reseacher
#
#Wrapping function: wrap_always(x,width=10)
#
#----------------------------------------------
#| First Name | Last Name | Age | Position |
#----------------------------------------------
#| John | Smith | 24 | Software E |
#| | | | ngineer |
#----------------------------------------------
#| Mary | Brohowski | 23 | Sales Mana |
#| | | | ger |
#----------------------------------------------
#| Aristidis | Papageorgo | 28 | Senior Res |
#| | poulos | | eacher |
#----------------------------------------------
#
#Wrapping function: wrap_onspace(x,width=10)
#
#---------------------------------------------------
#| First Name | Last Name | Age | Position |
#---------------------------------------------------
#| John | Smith | 24 | Software |
#| | | | Engineer |
#---------------------------------------------------
#| Mary | Brohowski | 23 | Sales |
#| | | | Manager |
#---------------------------------------------------
#| Aristidis | Papageorgopoulos | 28 | Senior |
#| | | | Reseacher |
#---------------------------------------------------
#
#Wrapping function: wrap_onspace_strict(x,width=10)
#
#---------------------------------------------
#| First Name | Last Name | Age | Position |
#---------------------------------------------
#| John | Smith | 24 | Software |
#| | | | Engineer |
#---------------------------------------------
#| Mary | Brohowski | 23 | Sales |
#| | | | Manager |
#---------------------------------------------
#| Aristidis | Papageorgo | 28 | Senior |
#| | poulos | | Reseacher |
#---------------------------------------------
Python 食谱页面包含一些改进。
解决方案 9：
pandas基于创建数据框的解决方案：
import pandas as pd
l = [['a', 'b', 'c'], ['aaaaaaaaaa', 'b', 'c'], ['a', 'bbbbbbbbbb', 'c']]
df = pd.DataFrame(l)
print(df)
0 1 2
0 a b c
1 aaaaaaaaaa b c
2 a bbbbbbbbbb c
要删除索引和标题值以创建所需的输出，可以使用to_string方法：
result = df.to_string(index=False, header=False)
print(result)
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
解决方案 10：
Scolp是一个新的库，它可以让您轻松地打印流式列数据，同时自动调整列宽。
（免责声明：我是作者）
解决方案 11：
对于懒人来说
使用Python 3.和Pandas/Geopandas；通用简单的类内方法（对于“普通”脚本只需删除self*）：
函数着色：
def colorize(self,s,color):
s = color+str(s)+"[0m"
return s
标头：
print('{0:<23} {1:>24} {2:>26} {3:>26} {4:>11} {5:>11}'.format('Road name','Classification','Function','Form of road','Length','Distance') )
然后是来自 Pandas/Geopandas 数据框的数据：
for index, row in clipped.iterrows():
rdName = self.colorize(row['name1'],"[32m")
rdClass = self.colorize(row['roadClassification'],"[93m")
rdFunction = self.colorize(row['roadFunction'],"[33m")
rdForm = self.colorize(row['formOfWay'],"[94m")
rdLength = self.colorize(row['length'],"[97m")
rdDistance = self.colorize(row['distance'],"[96m")
print('{0:<30} {1:>35} {2:>35} {3:>35} {4:>20} {5:>20}'.format(rdName,rdClass,rdFunction,rdForm,rdLength,rdDistance) )
含义{0:<30} {1:>35} {2:>35} {3:>35} {4:>20} {5:>20}：
0, 1, 2, 3, 4, 5-> 列，本例中一共有 6 列
30, 35, 20-> 列宽（注意，你必须添加长度[96m- 对于 Python 来说这也是一个字符串），只需实验一下 :)
>, <-> 对齐：右对齐、左对齐（=也可以用零填充）
如果您想要区分最大值，则必须切换到特殊的 Pandas 样式函数，但假设这足以在终端窗口上显示数据。
结果：
解决方案 12：
与上一个答案略有不同（我没有足够的代表来评论它）。格式库允许您指定元素的宽度和对齐方式，但不能指定元素的起始位置，即，您可以说“宽度为 20 列”，但不能说“从第 20 列开始”。这导致了这个问题：
table_data = [
['a', 'b', 'c'],
['aaaaaaaaaa', 'b', 'c'],
['a', 'bbbbbbbbbb', 'c']
]
print("first row: {: >20} {: >20} {: >20}".format(*table_data[0]))
print("second row: {: >20} {: >20} {: >20}".format(*table_data[1]))
print("third row: {: >20} {: >20} {: >20}".format(*table_data[2]))
输出
first row: a b c
second row: aaaaaaaaaa b c
third row: a bbbbbbbbbb c
答案当然是也格式化文字字符串，这与格式结合起来有点奇怪：
table_data = [
['a', 'b', 'c'],
['aaaaaaaaaa', 'b', 'c'],
['a', 'bbbbbbbbbb', 'c']
]
print(f"{'first row:': <20} {table_data[0][0]: >20} {table_data[0][1]: >20} {table_data[0][2]: >20}")
print("{: <20} {: >20} {: >20} {: >20}".format(*['second row:', *table_data[1]]))
print("{: <20} {: >20} {: >20} {: >20}".format(*['third row:', *table_data[1]]))
输出
first row: a b c
second row: aaaaaaaaaa b c
third row: aaaaaaaaaa b c
解决方案 13：
这将根据其他答案中使用的最大度量设置独立的、最佳拟合列宽。
data = [['a', 'b', 'c'], ['aaaaaaaaaa', 'b', 'c'], ['a', 'bbbbbbbbbb', 'c']]
padding = 2
col_widths = [max(len(w) for w in [r[cn] for r in data]) + padding for cn in range(len(data[0]))]
format_string = "{{:{}}}{{:{}}}{{:{}}}".format(*col_widths)
for row in data:
print(format_string.format(*row))
解决方案 14：
格式化为表格需要正确的填充。一个通用的解决方案是使用名为的 Python 包prettytable。虽然不依赖库会更好，但这个包处理了所有边缘情况，并且很简单，没有任何进一步的依赖关系。
x = PrettyTable()
x.field_names =["field1", "field2", "field3"]
x.add_row(["col1_content", "col2_content", "col3_content"])
print(x)
解决方案 15：
从其他一些答案的基础上，我认为我有一个相当可读且强大的解决方案：
data = [['a', 'b', 'c'], ['aaaaaaaa', 'b', 'c'], ['a', 'bbbbbbbbbb', 'ccc']]
padding = 4
# build a format string of max widths
# ex: '{:43}{:77}{:104}'
num_cols = len(data[0])
widths = [0] * num_cols
for row in data:
for i, value in enumerate(row):
widths[i] = max(widths[i], len(str(value)))
format_string = "".join([f'{{:{w+padding}}}' for w in widths])
# print the data
for row in data:
print(format_string.format(*[str(x) for x in row]))
这也支持NoneType记录，通过将一些东西包装在str()
解决方案 16：
这是 Shawn Chin 答案的变体。宽度是每列固定的，而不是所有列的宽度。第一行下方和列之间也有边框。（icontract库用于执行合同。）
@icontract.pre(
lambda table: not table or all(len(row) == len(table[0]) for row in table))
@icontract.post(lambda table, result: result == "" if not table else True)
@icontract.post(lambda result: not result.endswith("
"))
def format_table(table: List[List[str]]) -> str:
"""
Format the table as equal-spaced columns.
:param table: rows of cells
:return: table as string
"""
cols = len(table[0])
col_widths = [max(len(row[i]) for row in table) for i in range(cols)]
lines = [] # type: List[str]
for i, row in enumerate(table):
parts = [] # type: List[str]
for cell, width in zip(row, col_widths):
parts.append(cell.ljust(width))
line = " | ".join(parts)
lines.append(line)
if i == 0:
border = [] # type: List[str]
for width in col_widths:
border.append("-" * width)
lines.append("-+-".join(border))
result = "
".join(lines)
return result
以下是一个例子：
>>> table = [['column 0', 'another column 1'], ['00', '01'], ['10', '11']]
>>> result = packagery._format_table(table=table)
>>> print(result)
column 0 | another column 1
---------+-----------------
00 | 01
10 | 11
解决方案 17：
更新了@Franck Dernoncourt 的花式配方，使其符合 python 3 和 PEP8 标准
import io
import math
import operator
import re
import functools
from itertools import zip_longest
def indent(
rows,
has_header=False,
header_char="-",
delim=" | ",
justify="left",
separate_rows=False,
prefix="",
postfix="",
wrapfunc=lambda x: x,
):
"""Indents a table by column.
- rows: A sequence of sequences of items, one sequence per row.
- hasHeader: True if the first row consists of the columns' names.
- headerChar: Character to be used for the row separator line
(if hasHeader==True or separateRows==True).
- delim: The column delimiter.
- justify: Determines how are data justified in their column.
Valid values are 'left','right' and 'center'.
- separateRows: True if rows are to be separated by a line
of 'headerChar's.
- prefix: A string prepended to each printed row.
- postfix: A string appended to each printed row.
- wrapfunc: A function f(text) for wrapping text; each element in
the table is first wrapped by this function."""
# closure for breaking logical rows to physical, using wrapfunc
def row_wrapper(row):
new_rows = [wrapfunc(item).split("
") for item in row]
return [[substr or "" for substr in item] for item in zip_longest(*new_rows)]
# break each logical row into one or more physical ones
logical_rows = [row_wrapper(row) for row in rows]
# columns of physical rows
columns = zip_longest(*functools.reduce(operator.add, logical_rows))
# get the maximum of each column by the string length of its items
max_widths = [max([len(str(item)) for item in column]) for column in columns]
row_separator = header_char * (
len(prefix) + len(postfix) + sum(max_widths) + len(delim) * (len(max_widths) - 1)
)
# select the appropriate justify method
justify = {"center": str.center, "right": str.rjust, "left": str.ljust}[
justify.lower()
]
output = io.StringIO()
if separate_rows:
print(output, row_separator)
for physicalRows in logical_rows:
for row in physicalRows:
print( output, prefix + delim.join(
[justify(str(item), width) for (item, width) in zip(row, max_widths)]
) + postfix)
if separate_rows or has_header:
print(output, row_separator)
has_header = False
return output.getvalue()
# written by Mike Brown
# http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/148061
def wrap_onspace(text, width):
"""
A word-wrap function that preserves existing line breaks
and most spaces in the text. Expects that existing line
breaks are posix newlines (
).
"""
return functools.reduce(
lambda line, word, i_width=width: "%s%s%s"
% (
line,
"
"[
(
len(line[line.rfind("
") + 1 :]) + len(word.split("
", 1)[0])
>= i_width
)
],
word,
),
text.split(" "),
)
def wrap_onspace_strict(text, i_width):
"""Similar to wrap_onspace, but enforces the width constraint:
words longer than width are split."""
word_regex = re.compile(r"S{" + str(i_width) + r",}")
return wrap_onspace(
word_regex.sub(lambda m: wrap_always(m.group(), i_width), text), i_width
)
def wrap_always(text, width):
"""A simple word-wrap function that wraps text on exactly width characters.
It doesn't split the text in words."""
return "
".join(
[
text[width * i : width * (i + 1)]
for i in range(int(math.ceil(1.0 * len(text) / width)))
]
)
if __name__ == "__main__":
labels = ("First Name", "Last Name", "Age", "Position")
data = """John,Smith,24,Software Engineer
Mary,Brohowski,23,Sales Manager
Aristidis,Papageorgopoulos,28,Senior Reseacher"""
rows = [row.strip().split(",") for row in data.splitlines()]
print("Without wrapping function
")
print(indent([labels] + rows, has_header=True))
# test indent with different wrapping functions
width = 10
for wrapper in (wrap_always, wrap_onspace, wrap_onspace_strict):
print("Wrapping function: %s(x,width=%d)
" % (wrapper.__name__, width))
print(
indent(
[labels] + rows,
has_header=True,
separate_rows=True,
prefix="| ",
postfix=" |",
wrapfunc=lambda x: wrapper(x, width),
)
)
# output:
#
# Without wrapping function
#
# First Name | Last Name | Age | Position
# -------------------------------------------------------
# John | Smith | 24 | Software Engineer
# Mary | Brohowski | 23 | Sales Manager
# Aristidis | Papageorgopoulos | 28 | Senior Reseacher
#
# Wrapping function: wrap_always(x,width=10)
#
# ----------------------------------------------
# | First Name | Last Name | Age | Position |
# ----------------------------------------------
# | John | Smith | 24 | Software E |
# | | | | ngineer |
# ----------------------------------------------
# | Mary | Brohowski | 23 | Sales Mana |
# | | | | ger |
# ----------------------------------------------
# | Aristidis | Papageorgo | 28 | Senior Res |
# | | poulos | | eacher |
# ----------------------------------------------
#
# Wrapping function: wrap_onspace(x,width=10)
#
# ---------------------------------------------------
# | First Name | Last Name | Age | Position |
# ---------------------------------------------------
# | John | Smith | 24 | Software |
# | | | | Engineer |
# ---------------------------------------------------
# | Mary | Brohowski | 23 | Sales |
# | | | | Manager |
# ---------------------------------------------------
# | Aristidis | Papageorgopoulos | 28 | Senior |
# | | | | Reseacher |
# ---------------------------------------------------
#
# Wrapping function: wrap_onspace_strict(x,width=10)
#
# ---------------------------------------------
# | First Name | Last Name | Age | Position |
# ---------------------------------------------
# | John | Smith | 24 | Software |
# | | | | Engineer |
# ---------------------------------------------
# | Mary | Brohowski | 23 | Sales |
# | | | | Manager |
# ---------------------------------------------
# | Aristidis | Papageorgo | 28 | Senior |
# | | poulos | | Reseacher |
# ---------------------------------------------
解决方案 18：
table = [['a', 'b', 'c'],
['aaaaaaaaaa', 'b', 'c'],
['a', 'bbbbbbbbbb', 'c']]
def print_table(table):
def get_fmt(table):
fmt = ''
for column, row in enumerate(table[0]):
fmt += '{{!s:<{}}} '.format(
max(len(str(row[column])) for row in table) + 2)
return fmt
fmt = get_fmt(table)
for row in table:
print(fmt.format(*row))
print_table(table)
解决方案 19：
这是一个有趣的小项目...
列.py
from __future__ import annotations
from typing import TYPE_CHECKING
if TYPE_CHECKING:
from typing import Iterable, Iterator, Sequence, Sized
Matrix = Sequence[Sequence]
def all_elem_same_length(list: Sequence) -> bool:
length = len(list[0])
for elem in list:
if not len(elem) == length:
return False
return True
def get_col(matrix: Matrix, col_i: int) -> Iterator[Sized]:
return (row[col_i] for row in matrix)
def get_cols(matrix: Matrix) -> Iterator[Iterable[Sized]]:
return (get_col(matrix, col_i) for col_i in range(len(matrix[0])))
def get_longest_elem(list: Iterable[Sized]) -> Sized:
return max(list, key=len)
def get_longest_elem_per_column(matrix: Matrix) -> Iterator[Sized]:
return (get_longest_elem(col) for col in get_cols(matrix))
def get_word_pad_fstr(element: Sized, padding: int) -> str:
return f"{{:{len(element)+padding}}}"
def get_row_elem_pad_strings(matrix: Matrix, padding: int) -> Iterator[str]:
return (
get_word_pad_fstr(word, padding) for word in get_longest_elem_per_column(matrix)
)
def print_table(matrix: Matrix, padding=4) -> None:
if not all_elem_same_length(matrix):
raise ValueError("Table rows must all have the same length.")
format_string = "".join(get_row_elem_pad_strings(matrix, padding))
for row in matrix:
print(format_string.format(*(str(e) for e in row)))
if __name__ == "__main__":
data = [["a", "b", "c"], ["aaaaaaaa", "b", "c"], ["a", "bbbbbbbbbb", "ccc"]]
print_table(data)
解决方案 20：
我意识到这个问题已经很老了，但是我不明白 Antak 的答案并且不想使用库，所以我提出了自己的解决方案。
解决方案假设记录是一个二维数组，记录的长度都相同，并且字段都是字符串。
def stringifyRecords(records):
column_widths = [0] * len(records[0])
for record in records:
for i, field in enumerate(record):
width = len(field)
if width > column_widths[i]: column_widths[i] = width
s = ""
for record in records:
for column_width, field in zip(column_widths, record):
s += field.ljust(column_width+1)
s += "
"
return s
解决方案 21：
我发现这个答案超级有用且优雅，最初来自这里：
matrix = [["A", "B"], ["C", "D"]]
print('
'.join([' '.join([str(cell) for cell in row]) for row in matrix]))
输出
A B
C D
解决方案 22：
您可以准备数据并将其传递给真正的column实用程序。
假设您已将数据打印到文件 /tmp/filename.txt 中，并使用制表符作为分隔符。然后您可以像这样将其列化：
import subprocess
result = subprocess.run("cat /tmp/filename.txt | column -N \"col_1,col_2,col_3\" -t -s' ' -R 2,3", shell=True, stdout=subprocess.PIPE)
print(result.stdout.decode("utf-8"))
如您所见，您可以使用列实用程序的功能，例如右对齐。
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