在python中创建漂亮的列输出
xeo*_*eor 95 python string list
我试图在python中创建一个很好的列列表,用于我创建的命令行管理工具.
基本上,我想要一个列表,如:
[['a', 'b', 'c'], ['aaaaaaaaaa', 'b', 'c'], ['a', 'bbbbbbbbbb', 'c']]
变成:
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
使用普通标签不会在这里做,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\naaaaaaaaaa b c\na bbbbbbbbbb c"
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
$ echo -e "a b c\naaaaaaaaaa b c\na bbbbbbbbbb c" | column -t
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
我已经四处寻找各种python库来做到这一点,但找不到任何有用的东西.
Kur*_*tal 120
从Python 2.6+开始,您可以按以下方式使用格式字符串将列设置为最少20个字符,并将文本对齐.
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
- 到目前为止是迄今为止最好的解决方案 (4认同)
- 从KurzedMetal添加解决方案,使用上面显示的格式说明符; {:> 20},>表示右对齐.通过使用{:<20},您将获得左对齐列,并使用{:^ 20},您将获得居中对齐的列. (4认同)
- f 字符串的适配： `print(f'{i :>50} {j :>25} {k :>5}')` （每行有 i、j、k 三个值） (4认同)
- 如果您想将其与其他格式说明符（例如“.2f”）结合使用，语法为“{:>20.2f}”。 (2认同)
- 我不认为这回答了问题——听起来OP想让每一行的宽度不超过容纳其内容所需的宽度。这只是设置了固定宽度 20。 (2认同)
Sha*_*hin 111
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
这样做是计算最长的数据条目以确定列宽,然后用于.ljust()在打印每列时添加必要的填充.
- 这将使所有列的宽度相同，这不是 `column -t` 所做的。 (3认同)
- 名称“longest”具有误导性，因为它不是最长的元素，而是 max_length。顺便说一句，最长的可能是这样的：`max((w for sub in data for w in sub), key=len)`。[PS 我不是投反对票的人] (2认同)
ant*_*tak 41
我带着相同的要求来到这里,但@lvc和@Preet的答案似乎更符合column -t那些列中产生的宽度不同的内容:
>>> 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
- 尼斯.这是实际遵循原始"规范"的最清晰的解决方案. (2认同)
- 这是对我有用的解决方案.其他解决方案产生了柱状输出,但是这个解决方案对填充具有最大的控制以及精确的柱宽. (2认同)
- 美丽的解决方案。对于任何不是字符串的列，只需添加一个额外的映射：`map(len, map(str, col))`。 (2认同)
Max*_*art 19
这个聚会有点晚了，我写的一个包的无耻插件，但你也可以查看Columnar包。
它需要一个输入列表列表和一个标题列表，并输出一个表格格式的字符串。此代码段创建了一个 docker-esque 表：
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)
或者你可以用颜色和边框来更漂亮一点。
要阅读有关列大小算法的更多信息并查看 API 的其余部分，您可以查看上面的链接或查看Columnar GitHub Repo
哇，只有17个答案。python的禅宗说“应该有一种——最好只有一种——明显的方法来做到这一点。”
所以这是第 18 种方法：tabulate包支持一堆可以显示为表格的数据类型，这是一个改编自他们文档的简单示例：
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)"]))
哪个输出
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
转换像这样的列是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')]
要查找每列所需的长度,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'
为了得到更像这样的表
---------------------------------------------------
| 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('\n') 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) + \
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, \
prefix \
+ delim.join([justify(str(item),width) for (item,width) in zip(row,maxWidths)]) \
+ 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 (\n).
"""
return reduce(lambda line, word, width=width: '%s%s%s' %
(line,
' \n'[(len(line[line.rfind('\n')+1:])
+ len(word.split('\n',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 '\n'.join([ text[width*i:width*(i+1)] \
for i in xrange(int(math.ceil(1.*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\n'
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)\n' % (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 |
#---------------------------------------------
在Python的配方页面包含了它一些改进.
pandas 基于创建数据框架的解决方案:
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输出,您可以使用to_string方法:
result = df.to_string(index=False, header=False)
print(result)
a b c
aaaaaaaaaa b c
a bbbbbbbbbb c
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