Python pI';我在想我做错了什么?
pI'mPython pI';我在想我做错了什么?,python,list,function,Python,List,Function,pI'm [2, 4, 2300, 4],[3, 6, 7500, 3],[2, 4, 4100, 7],[2, 4, 2800, 5],[2, 4, 5800, 2], [2, 4, 4800, 8], [2, 4, 2750, 4]] cars=[[2,4,2300,4],[3,6,7500,3],[2,4,4100,7],[2,4,2800,5],[2,4,5800,2],[2,4,4800,8],[2,4,2750,4]。返回语句缩进错误,因此函数结束得太快: def divisor(a
[2, 4, 2300, 4],[3, 6, 7500, 3],[2, 4, 4100, 7],[2, 4, 2800, 5],[2, 4, 5800, 2], [2, 4, 4800, 8], [2, 4, 2750, 4]]
cars=[[2,4,2300,4],[3,6,7500,3],[2,4,4100,7],[2,4,2800,5],[2,4,5800,2],[2,4,4800,8],[2,4,2750,4]。返回语句缩进错误,因此函数结束得太快:
def divisor(a, c):
list = []
for i in range(a, c):
if i % 3 == 0 or i % 7 == 0:
list.append(i)
return list
b = 1234
d = 1422
chamador = divisor(b, d)
print(chamador)
输出:
您的return语句缩进错误,因此函数结束得太快:
def divisor(a, c):
list = []
for i in range(a, c):
if i % 3 == 0 or i % 7 == 0:
list.append(i)
return list
b = 1234
d = 1422
chamador = divisor(b, d)
print(chamador)
输出:
您也可以使用简单的列表理解来完成此操作:
a = 1234
b = 1422
>>> [i for i in range(a, b) if not i % 3 or not i % 7]
[1236,
1239,
1242,
1245,
...]
您也可以使用简单的列表理解来完成此操作:
a = 1234
b = 1422
>>> [i for i in range(a, b) if not i % 3 or not i % 7]
[1236,
1239,
1242,
1245,
...]
在函数中定义
list1=[]
,并将return
函数取消缩进到与范围(a,c)中i的相同的级别。
您应该将list1
添加到函数中,并且它应该与for循环具有相同的缩进级别。definelist1=[]
在函数中,取消缩进将函数返回到与范围(a,c)中i的相同的缩进级别您应该将列表1
添加到函数中,并且它应该与for循环具有相同的缩进级别。