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<pre><code>def sem1Sort1(semester1, selectionSEM1):
list = []
for period in semester1:
if period == 1:
for index in semester1[period]:
if index in selectionSEM1:
list.append(index)
return list
def sem1Sort2(semester1, selectionSEM1):
list = []
for period in semester1:
if semester1 == 2:
for index in semester1[period]:
if index in selectionSEM1:
list.append(index)
return list
def main():
selectionSEM1 = ["a", "b", "c", "d", "e", "f", "g", "h"]
selectionSEM2 = []
semester1 = {
1: ["e", "f", "g", "h"], 2: ["a", "b", "c", "d"] ,
}
SEM1period1 = sem1Sort1(semester1, selectionSEM1)
SEM1period2 = sem1Sort2(semester1, selectionSEM1)
print SEM1period1
print SEM1period2
main()
</code></pre>
<p>当我运行这段代码时,它会输出SEM1period1 fine,如[“e”,“f”,“g”,“h”],但是第二个方法sem1Sort2似乎不会将任何内容保存到SEM1period2中-因为print语句会输出[]</p>
<p>更新:</p>
<pre><code>def sem1Sort1(semester1, selectionSEM1):
list = []
for period in semester1:
if period == 1:
for index in semester1[period]:
if index in selectionSEM1:
list.append(index)
return list
def sem1Sort2(semester1, selectionSEM1):
list = []
for period in semester1:
if period == 2:
for index in semester1[period]:
if index in selectionSEM1:
list.append(index)
return list
def sem1Sort3(semester1, selectionSEM1):
list = []
for period in semester1:
if period == 3:
for index in semester1[period]:
if index in selectionSEM1:
list.append(index)
def main():
selectionSEM1 = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l"]
selectionSEM2 = []
semester1 = {
1: ["e", "f", "g", "h"], 2: ["a", "b", "c", "d"] ,
3: ["i", "j", "k", "l"]
}
SEM1period1 = sem1Sort1(semester1, selectionSEM1)
SEM1period2 = sem1Sort2(semester1, selectionSEM1)
SEM1period3 = sem1Sort3(semester1, selectionSEM1)
print SEM1period1
print SEM1period2
print SEM1period3
main()
</code></pre>
<p>为什么print SEM1period3不返回none?你知道吗</p>