Python 排列数字列（以表格格式打印输出）
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Line up columns of numbers (print output in table format)
提问by pb100
I have data (numbers) saved in the following format (example):
我有以下格式保存的数据（数字）（示例）：
234 127 34 23 45567
23 12 4 4 45
23456 2 1 444 567
...
Is there any python-way method to line up the numbers and get them as
是否有任何 python-way 方法来排列数字并将它们作为
234 127 34 23 45567
23 12 4 4 45
23456 2 1 444 567
(I cannot predict the column size).
（我无法预测列大小）。
采纳答案by Kevin Jacobs
Here is a simple, self-contained example that shows how to format variable column widths:
这是一个简单的、自包含的示例，展示了如何设置可变列宽的格式：
data = '''\
234 127 34 23 45567
23 12 4 4 45
23456 2 1 444 567'''
# Split input data by row and then on spaces
rows = [ line.strip().split(' ') for line in data.split('\n') ]
# Reorganize data by columns
cols = zip(*rows)
# Compute column widths by taking maximum length of values per column
col_widths = [ max(len(value) for value in col) for col in cols ]
# Create a suitable format string
format = ' '.join(['%%%ds' % width for width in col_widths ])
# Print each row using the computed format
for row in rows:
print format % tuple(row)
which outputs:
输出：
234 127 34 23 45567 23 12 4 4 45 23456 2 1 444 567
234 127 34 23 45567 23 12 4 4 45 23456 2 1 444 567
回答by Rahul
Try this nice well documented example http://code.activestate.com/recipes/267662-table-indentation/
试试这个很好的有据可查的例子http://code.activestate.com/recipes/267662-table-indentation/
回答by gimel
You need some way of finding the column size, maybe by reading all the data and finding the maximum width.
您需要某种方法来查找列大小，可能是通过读取所有数据并找到最大宽度。
>>> line='234 127 34 23 45567'
>>> line.split()
['234', '127', '34', '23', '45567']
>>> max(map(len, line.split()))
5
Repeat over all lines, to find column size (e.g., 5).
Constructing a formatted line with percent formattingis straightforward.
在所有行上重复，以找到列大小（例如，5）。构造带格式的行percent formatting很简单。
>>> colsize = 5
>>> ' '.join(('%*s' % (colsize, i) for i in line.split()))
' 234 127 34 23 45567'
>>>
回答by miku
#!/usr/bin/env python
class ALIGN:
LEFT, RIGHT = '-', ''
class Column(list):
def __init__(self, name, data, align=ALIGN.RIGHT):
list.__init__(self, data)
self.name = name
width = max(len(str(x)) for x in data + [name])
self.format = ' %%%s%ds ' % (align, width)
class Table:
def __init__(self, *columns):
self.columns = columns
self.length = max(len(x) for x in columns)
def get_row(self, i=None):
for x in self.columns:
if i is None:
yield x.format % x.name
else:
yield x.format % x[i]
def get_rows(self):
yield ' '.join(self.get_row(None))
for i in range(0, self.length):
yield ' '.join(self.get_row(i))
def __str__(self):
return '\n'.join(self.get_rows())
For your example:
对于您的示例：
if __name__ == '__main__':
print Table(
Column("", [234, 32, 23456]),
Column("", [127, 12, 2]),
Column("", [34, 4, 1]),
Column("", [23, 4, 444]),
Column("", [45567, 45, 567])
)
It will yield:
它将产生：
234 127 34 23 45567
32 12 4 4 45
23456 2 1 444 567
Adapted from http://code.activestate.com/recipes/577202-render-tables-for-text-interface/
改编自http://code.activestate.com/recipes/577202-render-tables-for-text-interface/
回答by John La Rooy
>>> rows = """234 127 34 23 45567
... 23 12 4 4 45
... 23456 2 1 444 567"""
first convert the rows into a 2d array (list of lists)
首先将行转换为二维数组（列表列表）
>>> arr=[x.split() for x in rows.split("\n")]
now compute the space each field will need to fit into
现在计算每个字段需要适合的空间
>>> widths = [max(map(len,(f[i] for f in tab))) for i in range(len(arr[0]))]
and pad each element to fit into that space
并填充每个元素以适应该空间
>>> [[k.rjust(widths[i]) for i,k in enumerate(j)] for j in arr]
[[' 234', '127', '34', ' 23', '45567'], [' 23', ' 12', ' 4', ' 4', ' 45'], ['23456', ' 2', ' 1', '444', ' 567']]
finally join the array back into a string
最后将数组连接回一个字符串
>>> print "\n".join(" ".join(k.rjust(widths[i]) for i,k in enumerate(j)) for j in arr)
234 127 34 23 45567
23 12 4 4 45
23456 2 1 444 567
回答by Kody
Integer in front of the d is what column the integers will start after the previous number so you can line them up however you see fit
d 前面的整数是整数将在前一个数字之后开始的列，因此您可以将它们排成一行，但您认为合适
print("{0:4d} {1:4d} {2:4d} {3:4d} {4:4d}".format(234, 127, 34, 23, 45567))
print("{0:4d} {1:4d} {2:4d} {3:4d} {4:4d}".format(234, 127, 34, 23, 45567))
repeat as necessary
根据需要重复
回答by Gnubie
Kevin Jacobs's answer modified to allow a variable number of integers on each row:
凯文·雅各布斯 (Kevin Jacobs) 的答案已修改为允许每行有可变数量的整数：
def align(data, delimiter = '\t', is_left_align = True):
rows = [row.strip().split(delimiter) for row in data.split('\n')]
cols = map(lambda *row: [str(field) or '' for field in row], *rows)
widths = [max(len(field) for field in col) for col in cols]
format = ['%%%s%ds' % ('-' if is_left_align else '', width) for width in widths]
return '\n'.join([delimiter.join(format[:len(row)]) % tuple(row) for row in rows])
data = '''\
234 127 34 23 45567
23 12 4 4 45
23456 2 1 444 567'''
print(align(data, ' ', False))
