以表格格式打印字典？

Question

我有一本字典，看起来像：

band2 = {'channel11': [10812, 2162, 1972, 0], 'channel10': [10787, 2157, 1967, 0], 'channel3': [10612, 2122, 1932, 0], 'channel2': [10589, 2117, 1927, 0], 'channel1': [10564, 2112, 1922, 20], 'channel7': [10712, 2142, 1952, 26], 'channel6': [10687, 2137, 1947, 0], 'channel5': [10662, 2132, 1942, 32], 'channel4': [10637, 2127, 1937, 26], 'channel9': [10762, 2152, 1962, 0], 'channel8': [10737, 2147, 1957, 0], 'channel12': [10837, 2167, 1977, 15]} 

我那么像这样排序是：

zipped = zip(*((key, value) for key, value in sorted(band2.items(), 
              key= lambda x: int(x[0][7:])))) 

这给出了这样的结果：

[('channel1', 'channel2', 'channel3', 'channel4', 'channel5', 'channel6', 'channel7', 'channel8', 'channel9', 'channel10', 'channel11', 'channel12'), ([10564, 2112, 1922, 20], [10589, 2117, 1927, 0], [10612, 2122, 1932, 0], [10637, 2127, 1937, 26], [10662, 2132, 1942, 32], [10687, 2137, 1947, 0], [10712, 2142, 1952, 26], [10737, 2147, 1957, 0], [10762, 2152, 1962, 0], [10787, 2157, 1967, 0], [10812, 2162, 1972, 0], [10837, 2167, 1977, 15])] 

我想知道如何打印这在一定的格式：

Channel 1 Channel 2 ...... Channel 12 
value[0] value[0] ........value[0] 
value[1]/value[2] value[1]/value[2].....value[1]/value[2] 
value[3] value[3]...........value[3] 

所以一个例子输出为：

Channel 1 Channel 2 <other values> Channel 12 
10564  10589  <other values> 10837 
2112/1922 2117/1927 <other values> 2167/1977 
20  0   <other values> 15 

我不确定如何去在一个整洁表打印这一点，我会用格式？

Answer 1

我最终没有使用zipped。

table = [[], [], [], []] 
# We can't just sort the channel names because 'channel11' < 'channel2' 
channel_numbers = [] 
for channel_name in band2.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] 
    table[0].append('Channel %d' % channel_number) 
    table[1].append(str(channel_data[0])) 
    table[2].append('%s/%s' % (channel_data[1], channel_data[2])) 
    table[3].append(str(channel_data[3])) 

for line in table: 
    print('\t'.join(line)) 

如果标签没有很好地工作，确定每一列的宽度和垫空间：

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) 
    column_width = max(cell_widths) 
    for i in range(len(column)): 
     cell = column[i] 
     padded_cell = cell + ' '*(column_width-len(cell)) 
     table[i].append(padded_cell) 

for line in table: 
    # If tabs don't make the columns align properly, try iterating through 
    # column in line, and padding it with the appropriate number of spaces. 
    print(' '.join(line)) 

Answer 2

你可以尝试pandas它提供了一个很好的结构，处理表：

熊猫的数据帧可以从项目建成：

In [69]: import pandas as pd 
In [70]: df = pd.DataFrame.from_items(sorted(band2.items(), key=lambda x: int(x[0][7:]))) 
In [71]: df = df.astype('string') 

的毗连第1行和r流2

In [72]: df.ix[1] = df.ix[1] + '/' + df.ix[2] 
In [73]: df = df.ix[[0, 1, 3]] 

打印它没有指数

In [74]: print df.to_string(index=None) 

    channel1 channel2 channel3 channel4 channel5 channel6 channel7 channel8 channel9 channel10 channel11 channel12 
    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