问题:在python中创建漂亮的列输出
我试图在python中创建一个不错的列列表,以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
使用普通标签不会解决问题,因为我不知道每一行中最长的数据。
这与Linux中的’column -t’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
我到处寻找各种python库来执行此操作,但找不到任何有用的方法。
回答 0
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.ljust()
在打印出每一列时用于添加必要的填充。
回答 1
从Python 2.6+开始,您可以通过以下方式使用格式字符串,以将列设置为至少20个字符,并将文本向右对齐。
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
回答 2
我来到这里有相同的要求,但@lvc和@Preet的答案似乎与column -t
该列中产生的内容具有不同的宽度更加一致:
>>> 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
回答 3
这对聚会来说有点晚了,是我写的一个无耻的插件,但是您也可以查看Columnar软件包。
它接受一个输入列表和一个标题列表,并输出一个表格式的字符串。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,您可以得到一些带有颜色和边框的爱好者。
要了解有关列大小调整算法的更多信息并查看API的其余部分,可以查看上面的链接或查看Columnar GitHub Repo
回答 4
您必须使用2个通行证来执行此操作:
- 获取每列的最大宽度。
- 使用格式化使用我们最宽的知识从第一遍列
str.ljust()
和str.rjust()
回答 5
像这样转换列是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,可以使用max
:
>>> trans_a = zip(*a)
>>> [max(len(c) for c in b) for b in trans_a]
[10, 10, 1]
您可以在适当的填充下使用它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'
回答 6
要获得更高级的表
---------------------------------------------------
| First Name | Last Name | Age | Position |
---------------------------------------------------
| John | Smith | 24 | Software |
| | | | Engineer |
---------------------------------------------------
| Mary | Brohowski | 23 | Sales |
| | | | Manager |
---------------------------------------------------
| Aristidis | Papageorgopoulos | 28 | Senior |
| | | | Reseacher |
---------------------------------------------------
您可以使用以下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。
回答 7
pandas
创建数据框的基于解决方案:
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
要删除索引和标头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
回答 8
Scolp是一个新的库,可让您在自动调整列宽的同时轻松打印流式列数据。
(免责声明:我是作者)
回答 9
这将基于其他答案中使用的最大度量设置独立的,最适合的列宽。
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))
回答 10
对于懒惰的人
使用Python 3. *和Pandas / Geopandas的代码;通用的简单类内方法(对于“普通”脚本,只需删除self):
函数着色:
def colorize(self,s,color):
s = color+str(s)+"\033[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'],"\033[32m")
rdClass = self.colorize(row['roadClassification'],"\033[93m")
rdFunction = self.colorize(row['roadFunction'],"\033[33m")
rdForm = self.colorize(row['formOfWay'],"\033[94m")
rdLength = self.colorize(row['length'],"\033[97m")
rdDistance = self.colorize(row['distance'],"\033[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,在这种情况下总共有6个
30, 35, 20
->列的宽度(请注意,您必须添加长度\033[96m
-对于Python来说,这也是一个字符串),只需进行实验即可:)
>, <
->对齐:右,左(也=
用于填充零)
如果要区分例如最大值,则必须切换到特殊的Pandas样式函数,但是假设这足以在终端窗口上显示数据。
结果:
回答 11
与先前的答案略有不同(我没有足够的代表对此发表评论)。格式库允许您指定元素的宽度和对齐方式,但不能指定元素的开始位置,即可以说“宽20列”,而不能说“从20列开始”。导致此问题:
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]))
输出量
first row: a b c
second row: aaaaaaaaaa b c
third row: a bbbbbbbbbb c
当然,答案是格式化文字字符串,它与格式有点奇怪地结合在一起:
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]]))
输出量
first row: a b c
second row: aaaaaaaaaa b c
third row: aaaaaaaaaa b c
回答 12
我发现这个答案超级有用且优雅,最初是从这里开始的:
matrix = [["A", "B"], ["C", "D"]]
print('\n'.join(['\t'.join([str(cell) for cell in row]) for row in matrix]))
输出量
A B
C D
回答 13
这是肖恩·钦(Shawn Chin)答案的一种变体。宽度是固定的,不是每列都固定。第一行下方和各列之间还有一个边框。(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("\n"))
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 = "\n".join(lines)
return result
这是一个例子:
>>> 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
回答 14
更新了@Franck Dernoncourt花式食谱以符合python 3和PEP8
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("\n") 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 (\n).
"""
return functools.reduce(
lambda line, word, i_width=width: "%s%s%s"
% (
line,
" \n"[
(
len(line[line.rfind("\n") + 1 :]) + len(word.split("\n", 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 "\n".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\n")
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)\n" % (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 |
# ---------------------------------------------
回答 15
我知道这个问题很旧,但是我不了解Antak的答案,也不想使用库,所以我推出了自己的解决方案。
解决方案假定记录是2D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 += "\n"
return s