Question

我有一个字典，我想格式化成一个表：

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]}

我这样做：

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))

这给出了：

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

但是现在我想命名行：

       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]))

但是我想知道打印这些列名的一些更好的方法。我可以通过上面的填充计算得到详细信息，但我想知道什么是一个干净，简单的方法。

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？

Answer 1

与许多python任务一样，通过导入魔法模块可以实现更优雅的代码。在这种情况下，该模块称为prettytable。

以下是一些有效的代码：

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  |
+--------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+------------+------------+------------+

但我无法弄清楚从哪里获得RSSI行值。

Answer 2

我想如果您不介意订单，这将有助于打印..：

band3 = \
{'channel1': [10564, 2112, 1922],
...
 'channel9': [10762, 2152, 1962]}

types = ['','UARFCN','DL/UL','RSSI']

# ASSUME THE DICTIONARY VALUES ARE EQUAL IN LENGTH
li1 = [types[0]]+band3.keys() # the stuff on the side
li2 = zip(types[1:],*band3.values()) # the main data
li3 = zip(li1,*li2)
for i,x in enumerate(li3):
    x = map(str, x)
    li3[i] = [a.ljust(b) for b in (max(map(len,x)),) for a in x]

final = zip(*li3)
for i in final:
    print ' '.join(i)

输出：

       channel3 channel2 channel1 channel7 channel6 channel5 channel4 channel9 channel8 channel12 channel11 channel10
UARFCN 10612    10589    10564    10712    10687    10662    10637    10762    10737    10837     10812     10787    
DL/UL  2122     2117     2112     2142     2137     2132     2127     2152     2147     2167      2162      2157     
RSSI   1932     1927     1922     1952     1947     1942     1937     1962     1957     1977      1972      1967

表格式化，打印出行名称

2 个答案: