0. Try it yourself, and analysis in this case, the data should be added to the list which method is better?
Suppose you are given in the following list:
Member = [ 'Turtle small', 'night', 'lost', 'Yi Jing', 'setting sun Qiuwu']
required to modify the list:
Member = [ 'small turtle', 88 ' night ', 90,' lost ', 85,' Yi Jing ', 90' Qiuwu sun ', 88]
错误做法：
member = ['小甲鱼','黑夜','迷途','易经','斜阳']
member.insert(1,88)
member.insert(2,90)
member.insert(3,85)
member.insert(4,90)
member.insert(5,88)
print(member)
#没有考虑到每个字符串就是一位小甲鱼为0，后面为2，黑夜为3以此类推，因此正确的做法应该是：
member = ['小甲鱼','黑夜','迷途','易经','斜阳']
member.insert(1,88)
member.insert(3,90)
member.insert(5,85)
member.insert(7,90)
member.insert(9,88)
print(member)
1. The use of the print cycle for each of the content list upper member
member = ['小甲鱼',88,'黑夜',90,'迷途',85,'易经',90,'斜阳',88]
for i in member:
print(i)
2. On a question of style print is not very good, you can modify the code printed pattern in the figure below it? [Please use at least two methods] (the style is probably the figure below, there is no online map)
小甲鱼 88
黑夜 90
迷途 85
易经 90
斜阳 88
**法一：由于不知道怎么直接打印列表当中的东西，因此必须通过for+in循环来将列表当中的字符串提取出来打印**、
member = ['小甲鱼',88,'黑夜',90,'迷途',85,'易经',90,'斜阳',88]
count = 2
for i in member:
if count%2 == 0:
print(i,end=" ")
else:
print(i)
count += 1
**法一改进：知道了直接打印列表中字符串打印方式后**
#首先法一改进是在知道可以用print(member[count],member[count+1]))基础上，这样可以直接输入列表的值
#于是可以将列表数字化，在引入一个conut进行循环变量
member = ['小甲鱼',88,'黑夜',90,'迷途',85,'易经',90,'斜阳',88]
lengenty= len(member)
count = 0
while count < lengenty:
print(member[count],member[count+1])
count += 2
再次改进的法二：
member = ['小甲鱼',88,'黑夜',90,'迷途',85,'易经',90,'斜阳',88]
for i in range(len(member)):
if i%2 == 0:
print (member [i], member [i+1])
#for 和 range 的日常狼狈为奸。其明显节约了定义一个count和while循环的时间，直接把表变成数值变量并循环，然后在满足数值要求的情况下输出列表的值
Summary: 1. list append, extend, insert use
2.range for the benefits of working hand in glove
