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我有一本字典,我想将其格式化为表格:
band3 = \
{'channel1': [10564, 2112, 1922],
'channel10': [10787, 2157, 1967],
'channel11': [10812, 2162, 1972],
'channel12': [10837, 2167, 1977],
'channel2': [10589, 2117, 1927],
'channel3': [10612, 2122, 1932],
'channel4': [10637, 2127, 1937],
'channel5': [10662, 2132, 1942],
'channel6': [10687, 2137, 1947],
'channel7': [10712, 2142, 1952],
'channel8': [10737, 2147, 1957],
'channel9': [10762, 2152, 1962]}
我是这样做的:
table = [[], [], [], []]
# can't just sort the channel names because 'channel11' < 'channel2'
channel_numbers = []
for channel_name in band3.keys():
if channel_name.startswith('channel'):
channel_number = int(channel_name[7:])
channel_numbers.append(channel_number)
else:
raise ValueError("channel name doesn't follow pattern")
channel_numbers.sort()
for channel_number in channel_numbers:
channel_data = band2['channel%d' % channel_number]
column =[
'Channel %d' % channel_number,
str(channel_data[0]),
'%s/%s' % (channel_data[1], channel_data[2]),
str(channel_data[3])
]
cell_widths = map(len, column) #9 5 2 9
column_widths = max(cell_widths) # 9 or 10
for i in range(len(cell_widths)): #4
cell = column[i]
padded_cell = cell + ' '*(column_widths-len(cell))
table[i].append(padded_cell)
for line in table:
print(' '.join(line))
这给出:
Channel 1 Channel 2 Channel 3 Channel 4 Channel 5 Channel 6 Channel 7 Channel 8 Channel 9 Channel 10 Channel 11 Channel 12
10564 10589 10612 10637 10662 10687 10712 10737 10762 10787 10812 10837
2112/1922 2117/1927 2122/1932 2127/1937 2132/1942 2137/1947 2142/1952 2147/1957 2152/1962 2157/1967 2162/1972 2167/1977
20 0 0 26 32 0 26 0 0 0 0 15
但是现在我想给行命名:
Channel 1 Channel 2 Channel 3 Channel 4 Channel 5 Channel 6 Channel 7 Channel 8 Channel 9 Channel 10 Channel 11 Channel 12
UARFCN 10564 10589 10612 10637 10662 10687 10712 10737 10762 10787 10812 10837
DL/UL 2112/1922 2117/1927 2122/1932 2127/1937 2132/1942 2137/1947 2142/1952 2147/1957 2152/1962 2157/1967 2162/1972 2167/1977
RSSI 20 0 0 26 32 0 26 0 0 0 0 15
这很简单,我将打印循环更改为:
print " ",
print(' '.join(table[0]))
print "UARFCN",
print(' '.join(table[1]))
print "DL/UL ",
print(' '.join(table[2]))
print "RSSI ",
print(' '.join(table[3]))
但是我想知道一些更好的方法来打印这些列名。如上所述,我可能会因为填充计算而变得冗长,但我想知道什么是一种干净、简单的好方法。
编辑:格式尝试:
print('{0:6s} {1}'.format("", ' '.join(table[0])))
print('{0:2s} {1}'.format("UARFCN", ' '.join(table[1])))
print('{0:6s} {1}'.format("DL/UL", ' '.join(table[2])))
print('{0:6s} {1}'.format("RSSI", ' '.join(table[3])))
编辑:另一种方式
print('{0} {1}'.format("".ljust(6), ' '.join(table[0])))
print('{0} {1}'.format("UARFCN".ljust(6), ' '.join(table[1])))
print('{0} {1}'.format("DL/UL".ljust(6), ' '.join(table[2])))
print('{0} {1}'.format("RSSI".ljust(6), ' '.join(table[3])))
改进建议?
最佳答案
与这么多python任务一样,通过导入神奇的模块可以实现更优雅的代码。在这种情况下,该模块称为 prettytable .
这是一些工作代码:
from prettytable import PrettyTable
band3 = \
{'channel1': [10564, 2112, 1922],
'channel10': [10787, 2157, 1967],
'channel11': [10812, 2162, 1972],
'channel12': [10837, 2167, 1977],
'channel2': [10589, 2117, 1927],
'channel3': [10612, 2122, 1932],
'channel4': [10637, 2127, 1937],
'channel5': [10662, 2132, 1942],
'channel6': [10687, 2137, 1947],
'channel7': [10712, 2142, 1952],
'channel8': [10737, 2147, 1957],
'channel9': [10762, 2152, 1962]}
band3_new = {}
for key in band3.keys():
band3_new[ int( key.split('channel')[1] ) ] = band3[key]
x = PrettyTable()
x.add_column("", ["UARFCN", "DL/UL"])
for channel in band3_new.keys():
x.add_column("channel " + str(channel), [band3_new[channel][0], str(band3_new[channel][1]) + "/" + str(band3_new[channel][2]) ] )
print x
及其输出:
+--------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+------------+------------+------------+
| | channel 1 | channel 2 | channel 3 | channel 4 | channel 5 | channel 6 | channel 7 | channel 8 | channel 9 | channel 10 | channel 11 | channel 12 |
+--------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+------------+------------+------------+
| UARFCN | 10564 | 10589 | 10612 | 10637 | 10662 | 10687 | 10712 | 10737 | 10762 | 10787 | 10812 | 10837 |
| DL/UL | 2112/1922 | 2117/1927 | 2122/1932 | 2127/1937 | 2132/1942 | 2137/1947 | 2142/1952 | 2147/1957 | 2152/1962 | 2157/1967 | 2162/1972 | 2167/1977 |
+--------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+------------+------------+------------+
不过,我不知道你从哪里得到 RSSI 行值。
关于python - 表格格式化,打印出行名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16215764/
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