题主让我有点不知道是不头晕感觉，if判断句应该在循环体内，险答案外，以下代码可能符合题主想法

d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59 }
tds=[]
for name,score in d.iteritems():
	if score>=60:
		tds.append('<tr><td>%s</td><td>%s</td></tr>' % (name,score))
	else:
		tds.append('<tr><td>%s</td><td style="color:red">%s</td></tr>' % (name,score))
print '<table border="1">'
print '<tr><th>Name</th><th>Score</th></tr>'
print '\n'.join(tds)
print '</table>'

你的代码中，

代码块1实现了把生成了一个列表对象tds, 具体内容如下：

['<tr><td>%Adam</td><td>%95</td></tr>' , '<tr><td>%Lisa</td><td>%85</td></tr>' , '<tr><td>%Bart</td><td>%59</td></tr>']

······················································································································································································

代码块2，我实在不知道有什么用，if语句中判断的score有具体指向吗？

score首先不是一个对象，不能直接对其判断，其次，score在这个题目中是一个用在for循环中的临时抽象的名称，不能不直接判断吧。

···················································································································································································

代码块3  join()函数的功能，是把一个list对象拼接成一个字符串对象

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正确的代码的逻辑应该是这样的： 在通过for循环生成dts列表对象的时候，使对应分数是59的那个列表元素就变成这样一个字符串 '<tr><td>%Bart</td><td style="color:red">%59</td></tr>'  

generate_tr   在这个函数里面去实现  效果 如 不及格 数字显示红色

d = { 'Adam': 95, 'Lisa': 85, 'Bart': 59 }

def generate_tr(name, score):

    return '<tr><td>%s</td><td>%s</td></tr>' % (name, score)

tds = [generate_tr(name, score) for name, score in d.iteritems()]

print '<table border="1">'

print '\n'.join(tds)

print '</table>'