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<p>我试图用值格式化一个表,但感觉我做得不好。例如,在对字符串进行除法和传递时,必须将int字典转换为int。而且%f似乎打印了加载的0。我只需要一个小数点的除法,而不是后面所有的0</p>
<pre><code>band1= {'channel1': 10564, 'channel2': 10589, 'channel3': 10612, 'channel4': 10637,'channel5': 10662, 'channel6': 10687,
'channel7': 10712, 'channel8': 10737, 'channel9': 10762, 'channel10': 10787,'channel11': 10812, 'channel12': 10837, }
print '%10s %10s %10s %10s %10s %10s %10s %10s %10s %10s %10s %10s' % ("Channel 1", "Channel 2", "Channel 3",
"Channel 4", "Channel 5", "Channel 6", "Channel 7","Channel 8", "Channel 9", "Channel 10", "Channel 11", "Channel 12")
print '%10s %10s %10s %10s %10s %10s %10s %10s %10s %10s %10s %10s' % (band1['channel1'], band1['channel2'], band1['channel3'],
band1['channel4'], band1['channel5'], band1['channel6'], band1['channel7'], band1['channel8'], band1['channel9'],
band1['channel10'], band1['channel11'], band1['channel12'])
print '%5f %5f %5f %5f %5f %5f %5f %5f %5f %5f %5f %5f' % (int(band1['channel1'])/5.0, int(band1['channel2'])/5.0, int(band1['channel3'])/5.0,
int(band1['channel4'])/5.0, int(band1['channel5'])/5.0, int(band1['channel6'])/5.0, int(band1['channel7'])/5.0, int(band1['channel8'])/5.0, int(band1['channel9'])/5.0,
int(band1['channel10'])/5.0, int(band1['channel11'])/5.0, int(band1['channel12'])/5.0)
</code></pre>
<p>这样可以得到:</p>
^{pr2}$