我有一些代码以某种格式打印字典。但是由于频道数量可能会有所不同，我想更改此格式。但我想知道它是否可以像我所拥有的那样轻松调整？这是代码：
band3={'channel11': [10812, 2162, 1972], 'channel10': [10787, 2157, 1967], 'channel3': [10612, 2122, 1932], 'channel2': [10589, 2117, 1927], 'channel1': [10564, 2112, 1922], 'channel7': [10712, 2142, 1952], 'channel6': [10687, 2137, 1947], 'channel5': [10662, 2132, 1942], 'channel4': [10637, 2127, 1937], 'channel9': [10762, 2152, 1962], 'channel8': [10737, 2147, 1957], 'channel12': [10837, 2167, 1977]}
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)
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])))
目前是输出：
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
RSSI 20 0 0 26 32 0 26 0 0 0 0 15
我想改变它，而不是在一条长行上列出所有数据，以便在每4个元素之后，下一位数据打印在一个新行上。
Channel 1 Channel 2 Channel 3 Channel 4
UARFCN 10564 10589 10612 10637
DL/UL 2112/1922 2117/1927 2122/1932 2127/1937
RSSI 20 0 0 26
Channel 5 Channel 6 Channel 7 Channel 8
UARFCN 10662 10687 10712 10737
DL/UL 2112/1922 2117/1927 2122/1932 2127/1937
RSSI 32 0 26 0
Channel 9 Channel 10 Channel 11 Channel 12
UARFCN 10762 10787 10812 10837
DL/UL 2132/1942 2137/1947 2142/1952 2147/1957
RSSI 0 0 0 15
编辑Mar：
names = ["", "UARFCN", "DL/UL", "RSSI"]
lwidth = max(len(l) for l in names)
for i in range(0, len(table[0]), 4):
for j, head in enumerate(names):
print(' {0:<{lwidth}} {1}'.format(head, ' '.join(table[j][i:i+4])))
print
答案 0 :(得分：2)
您知道str.format()可以为您辩护吗？添加<6作为格式化字符串会让format左侧调整文本，使用空格以匹配6个字符：'{0:<6} {1}'。您可以对表'列'的其余部分执行相同的操作，包括使用可变宽度格式。
此外，您可以在格式中使用描述性命名，并使用一些序列索引来形成整行：
table_row = '{label:<{lwidth}} {row[0]:<{cwidth}} {row[1]:<{cwidth}} {row[2]:<{cwidth}} {row[3]:<{cwidth}}'
创建一个四列格式，其列宽为参数cwidth。我也使标签宽度变量，因此您可以稍后修改标签（添加更多，使用详细标签等）而不会破坏您的布局。
接下来，将每个标签的数据分组，以便于处理：
labels = ('', 'UARFCN', 'DL/UL', 'RSSI')
lwidth = max(len(l) for l in labels)
table = []
channel_numbers = [int(cname[7:]) if cname.startswith('channel') else None for cname in band3]
if None in channel_numbers:
raise ValueError("channel name doesn't follow pattern")
channel_numbers.sort()
cwidth = 0
for channel_number in channel_numbers:
channel_data = band2['channel{}'.format(channel_number)]
entry = dict(zip(labels, (
'Channel {}'.format(channel_number),
channel_data[0],
'{}/{}'.format(*channel_data[1:3]),
channel_data[3]
)))
cwidth = max(cwidth, max(len(str(v)) for v in entry.values()))
table.append(entry)
现在我们将列表中的表数据作为字典条目;以这种方式将它们分组更容易。接下来我们使用itertools grouper recipe：
from itertools import izip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
# use a empty dictionary as filler at the end
for group in grouper(table, 4, dict.fromkeys(labels, '')):
for label in labels:
print(table_row.format(label=label, cwidth=cwidth, lwidth=lwidth,
row=[entry[label] for entry in group]))
print()
然后输出（使用重建的band2和band3输入）：
Channel 1 Channel 2 Channel 3 Channel 4
UARFCN 10564 10589 10612 10637
DL/UL 2112/1922 2117/1927 2122/1932 2127/1937
RSSI 20 0 0 26
Channel 5 Channel 6 Channel 7 Channel 8
UARFCN 10662 10687 10712 10737
DL/UL 2132/1942 2137/1947 2142/1952 2147/1957
RSSI 32 0 26 0
Channel 9 Channel 10 Channel 11 Channel 12
UARFCN 10762 10787 10812 10837
DL/UL 2152/1962 2157/1967 2162/1972 2167/1977
RSSI 0 0 0 15
dict.fromkeys(labels, '')默认值确保如果表中有一行少于4个条目，则最后一列用空字符串填充：
Channel 1 Channel 2 Channel 3 Channel 4
UARFCN 10564 10589 10612 10637
DL/UL 2112/1922 2117/1927 2122/1932 2127/1937
RSSI 20 0 0 26
Channel 5 Channel 6 Channel 7 Channel 8
UARFCN 10662 10687 10712 10737
DL/UL 2132/1942 2137/1947 2142/1952 2147/1957
RSSI 32 0 26 0
Channel 9 Channel 10 Channel 11
UARFCN 10762 10787 10812
DL/UL 2152/1962 2157/1967 2162/1972
RSSI 0 0 0
答案 1 :(得分：1)
names = ["", "UARFCN", "DL/UL", "RSSI"]
for i in range(0, len(table[0]), 4):
for j, head in enumerate(names):
print('{0} {1}'.format(head.ljust(6), ' '.join(table[j][i:i+4])) )
print()