題主讓我有點不知道是不頭暈感覺，if判斷句應該在循環體內，險答案外，以下代碼可能符合題主想法

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>'

你的代碼中，

代碼塊1實現了把生成了一個列表對象tds, 具體內容如下：

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

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

代碼塊2，我實在不知道有什么用，if語句中判斷的score有具體指向嗎？

score首先不是一個對象，不能直接對其判斷，其次，score在這個題目中是一個用在for循環中的臨時抽象的名稱，不能不直接判斷吧。

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

代碼塊3 ?join()函數的功能，是把一個list對象拼接成一個字符串對象

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

正確的代碼的邏輯應該是這樣的： 在通過for循環生成dts列表對象的時候，使對應分數是59的那個列表元素就變成這樣一個字符串?'<tr><td>%Bart</td><td style="color:red">%59</td></tr>' ?

generate_tr ? 在這個函數里面去實現 ?效果 如 不及格 數字顯示紅色

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>'