python list练习题
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1.合并两个有序列表，并且保持合并后的列表有序
In [20]:
lst1 = [32, 42, 12, 5, 14, 4, 1]
In [21]:
lst2 = [199, 22, 324, 89, 2]
In [22]:
help(lst1.sort)
Help on built-in function sort:
sort(...) method of builtins.list instance
L.sort(key=None, reverse=False) -> None -- stable sort *IN PLACE*
In [23]:
help(lst1.extend)
Help on built-in function extend:
extend(...) method of builtins.list instance
L.extend(iterable) -> None -- extend list by appending elements from the iterable
In [27]:
#方法一：lst = []lst.extend(lst1)lst.extend(lst2)lst.sort()lst
Out[27]:
[1, 2, 4, 5, 12, 14, 22, 32, 42, 89, 199, 324]
In [38]:
lst1 = [32, 42, 12, 5, 14, 4, 1]lst2 = [199, 22, 324, 89, 2]
In [35]:
lst1.extend(lst2).sort() #lst1.extend(lst2) -> None.sort()lst1
---------------------------------------------------------------------------AttributeError Traceback (most recent call last)<ipython-input-35-e3d4966adf34> in <module>()----> 1 lst1.extend(lst2).sort() #lst1.extend(lst2) -> None.sort() 2 lst1AttributeError: ‘NoneType‘ object has no attribute ‘sort‘
In [39]:
方法二：lst3 = []lst3 = lst1 + lst2lst3.sort()lst3
Out[39]:
[1, 2, 4, 5, 12, 14, 22, 32, 42, 89, 199, 324]
In [68]:
lst1 = [32, 42, 12, 5, 14, 4, 1]lst2 = [199, 22, 324, 89, 2]
In [109]:
lst3 = [] #如果为空，enumerate出来的的count也是空，不会抛出异常
In [71]:
help(enumerate)
Help on class enumerate in module builtins:
class enumerate(object)
| enumerate(iterable[, start]) -> iterator for index, value of iterable
|
| Return an enumerate object. iterable must be another object that supports
| iteration. The enumerate object yields pairs containing a count (from
| start, which defaults to zero) and a value yielded by the iterable argument.
| enumerate is useful for obtaining an indexed list:
| (0, seq[0]), (1, seq[1]), (2, seq[2]), ...
|
| Methods defined here:
|
| __getattribute__(self, name, /)
| Return getattr(self, name).
|
| __iter__(self, /)
| Implement iter(self).
|
| __new__(*args, **kwargs) from builtins.type
| Create and return a new object. See help(type) for accurate signature.
|
| __next__(self, /)
| Implement next(self).
|
| __reduce__(...)
| Return state information for pickling.
In [72]:
for i,j in enumerate(lst1):
print(‘{}->{}‘.format(i,j))
0->32
1->42
2->12
3->5
4->14
5->4
6->1
In [74]:
for i,j in enumerate(lst1,start=3):
print(‘{}->{}‘.format(i,j))
3->32
4->42
5->12
6->5
7->14
8->4
9->1
In [84]:
en1 = enumerate(lst1,start=3)
In [85]:
en1
Out[85]:
<enumerate at 0x7fc6011736c0>
In [87]:
en1.__next__()
Out[87]:
(3, 32)
In [88]:
en1.__next__()
Out[88]:
(4, 42)
In [89]:
next(en1)
Out[89]:
(5, 12)
In [90]:
iter(en1)
Out[90]:
<enumerate at 0x7fc6011736c0>
In [96]:
en1.__reduce__()
Out[96]:
(enumerate, (<list_iterator at 0x7fc60116c390>, 6))
In [104]:
lsta = [5, 7, 9, 3, 5]
In [108]:
help(lsta.insert) #在指定的索引前插入指定的对象
Help on built-in function insert:
insert(...) method of builtins.list instance
L.insert(index, object) -- insert object before index
In [106]:
lsta.insert(3,88)
In [107]:
lsta
Out[107]:
[5, 7, 9, 88, 3, 5]
In [1]:
lst1 = [32, 42, 12, 5, 14, 4, 1]lst2 = [199, 22, 324, 89, 2]
In [2]:
lst3 = lst1 + lst2
In [3]:
lst3
Out[3]:
[32, 42, 12, 5, 14, 4, 1, 199, 22, 324, 89, 2]
In [4]:
lst4 = []for i in lst3:
for n, j in enumerate(lst4):
if i < j:
lst4.insert(n, i)
break
else:
lst4.append(i)
break
else:
lst4.append(i)
In [5]:
lst4 #看到问题所在了吗？ -> 从14开始出现了问题，原因是，每次循环只和第一个数做了比较
Out[5]:
[1, 4, 5, 12, 32, 42, 14, 199, 22, 324, 89, 2]
In [17]:
#修改之后，将值插入合适的位置#插入排序lst5 = []for i in lst3:
for n, j in enumerate(lst5):
if i < j:
lst5.insert(n, i)
break# else:# lst4.append(i)# break
else: #为了解决lst5为空的时候，添加第一个值和没有跳出循环（无break）-比lst5里面的数都大，就添加在末尾
lst5.append(i)
In [18]:
lst5 #成功了
Out[18]:
[1, 2, 4, 5, 12, 14, 22, 32, 42, 89, 199, 324]
2.按单词反转字符串 例如‘I love python’ 反转成 ‘python love I’
In [19]:
s1 = ‘I love python‘
In [24]:
help(reversed)
Help on class reversed in module builtins:
class reversed(object)
| reversed(sequence) -> reverse iterator over values of the sequence
|
| Return a reverse iterator
|
| Methods defined here:
|
| __getattribute__(self, name, /)
| Return getattr(self, name).
|
| __iter__(self, /)
| Implement iter(self).
|
| __length_hint__(...)
| Private method returning an estimate of len(list(it)).
|
| __new__(*args, **kwargs) from builtins.type
| Create and return a new object. See help(type) for accurate signature.
|
| __next__(self, /)
| Implement next(self).
|
| __reduce__(...)
| Return state information for pickling.
|
| __setstate__(...)
| Set state information for unpickling.
In [25]:
for i in reversed(s1): print(i)
n
o
h
t
y
p
e
v
o
l
I
In [30]:
s1[::-1]
Out[30]:
‘nohtyp evol I‘
In [36]:
l1 = s1.split()
In [37]:
l1
Out[37]:
[‘I‘, ‘love‘, ‘python‘]
In [40]:
l2 = l1[::-1]
In [41]:
l2
Out[41]:
[‘python‘, ‘love‘, ‘I‘]
In [46]:
s2 = ‘ ‘.join(l2)
In [47]:
s2
Out[47]:
‘python love I‘
In [48]:
s3 = ‘ ‘.join(s1.split()[::-1])
In [49]:
s3
Out[49]:
‘python love I‘
3.找出一个列表中，只出现了一次的数字，并且保持原来的次序，例如[1,2,1,3,2,5],结果为[3,5]
In [50]:
help(lst1.count)
Help on built-in function count:
count(...) method of builtins.list instance
L.count(value) -> integer -- return number of occurrences of value
In [76]:
lst1 = [1,2,1,3,2,5]
In [71]:
lst1.count(1)
Out[71]:
2
In [80]:
#方法一：效率不高，比如10万个元素的列表，其中9万个是重复的，每个都要count一次，count()了9w此，消耗非常大lst2 = []for x in lst1:
if lst1.count(x) == 1:
lst2.append(x)
In [82]:
lst2
Out[82]:
[3, 5]
In [93]:
lst1 = [1,2,1,3,2,5,1]
In [87]:
#方法二：这个是针对结果列表做操作的，但重复奇数次就不行了ret = []for x in lst1:
if x not in ret:
ret.append(x)
else:
ret.remove(x)
In [89]:
ret
Out[89]:
[3, 5, 1]
In [74]:
lst1
Out[74]:
[3, 2, 5]
In [75]:
3 not in lst1
Out[75]:
False
In [105]:
#方法二修订tmp = []ret = []for x in lst1:
if x not in tmp: #tmp存下去重之后的元素
tmp.append(x)
ret.append(x)
else: #只要在tmp中出现过一次，再一次出现，全部删除，tmp像个
if x in ret:
ret.remove(x)
In [99]:
ret
Out[99]:
[3, 5]
[1,1,1,2,..]
|
|
_ | _
tmp[1,2]
|
ret[2,]
4.查找一个列表中的最大值
In [100]:
lst6 = [128, 99, 100, 23, 54, 443, 118]
In [106]:
#方法一max(lst6)
Out[106]:
443
In [107]:
#方法二：排序之后求值|ret = []for x in lst6:
for k, y in enumerate(ret):
if x > y:
ret.insert(k,x)
break
else:
ret.append(x)
In [108]:
ret
Out[108]:
[443, 128, 118, 100, 99, 54, 23]
In [109]:
ret[0]
Out[109]:
443
In [110]:
#方法三：m = lst6[0]for x in lst6:
if x > m:
m = x
In [111]:
m
Out[111]:
443
5.写一个程序，把字符串转化为数字，例如‘123’转化为123，‘0.254’转化为0.254，不允许使用int函数，float函数，eval函数，不允许导入任何模块
In [112]:
s1 = ‘123.456‘
In [117]:
l2 = s1.split(‘.‘)
In [118]:
l2
Out[118]:
[‘123‘, ‘456‘]
In [123]:
len(l2[0])
Out[123]:
3
In [124]:
x, y = [‘1‘,‘3‘]
In [125]:
x
Out[125]:
‘1‘
In [126]:
y
Out[126]:
‘3‘
In [127]:
integer, decimal = s1.split(‘.‘)
In [128]:
integer
Out[128]:
‘123‘
In [129]:
decimal
Out[129]:
‘456‘
In [130]:
length_integer = len(integer)
In [131]:
length_decimal = len(decimal)
In [144]:
ret1 = 0for k, x in enumerate(integer):
if x == ‘0‘:
ret1 += 0 * 10 ** (length_integer - k -1)
if x == ‘1‘:
ret1 += 1 * 10 ** (length_integer - k -1)
if x == ‘2‘:
ret1 += 2 * 10 ** (length_integer - k -1)
if x == ‘3‘:
ret1 += 3 * 10 ** (length_integer - k -1)
if x == ‘4‘:
ret1 += 4 * 10 ** (length_integer - k -1)
if x == ‘5‘:
ret1 += 5 * 10 ** (length_integer - k -1)
if x == ‘6‘:
ret1 += 6 * 10 ** (length_integer - k -1)
if x == ‘7‘:
ret1 += 7 * 10 ** (length_integer - k -1)
if x == ‘8‘:
ret1 += 8 * 10 ** (length_integer - k -1)
if x == ‘9‘:
ret1 += 9 * 10 ** (length_integer - k -1)
In [145]:
ret1
Out[145]:
123
In [146]:
ret2 = 0for k, x in enumerate(decimal):
if x == ‘0‘:
ret2 += 0 * 10 ** (-1 - k)
if x == ‘1‘:
ret2 += 1 * 10 ** (-1 - k)
if x == ‘2‘:
ret2 += 2 * 10 ** (-1 - k)
if x == ‘3‘:
ret2 += 3 * 10 ** (-1 - k)
if x == ‘4‘:
ret2 += 4 * 10 ** (-1 - k)
if x == ‘5‘:
ret2 += 5 * 10 ** (-1 - k)
if x == ‘6‘:
ret2 += 6 * 10 ** (-1 - k)
if x == ‘7‘:
ret2 += 7 * 10 ** (-1 - k)
if x == ‘8‘:
ret2 += 8 * 10 ** (-1 - k)
if x == ‘9‘:
ret2 += 9 * 10 ** (-1 - k)
In [147]:
ret2
Out[147]:
0.456
In [148]:
ret1 + ret2
Out[148]:
123.456
In [150]:
def str2int(strx):
integer, decimal = strx.split(‘.‘)
length_integer = len(integer)
length_decimal = len(decimal)
ret1 = 0
for k, x in enumerate(integer):
if x == ‘0‘:
ret1 += 0 * 10 ** (length_integer - k -1)
if x == ‘1‘:
ret1 += 1 * 10 ** (length_integer - k -1)
if x == ‘2‘:
ret1 += 2 * 10 ** (length_integer - k -1)
if x == ‘3‘:
ret1 += 3 * 10 ** (length_integer - k -1)
if x == ‘4‘:
ret1 += 4 * 10 ** (length_integer - k -1)
if x == ‘5‘:
ret1 += 5 * 10 ** (length_integer - k -1)
if x == ‘6‘:
ret1 += 6 * 10 ** (length_integer - k -1)
if x == ‘7‘:
ret1 += 7 * 10 ** (length_integer - k -1)
if x == ‘8‘:
ret1 += 8 * 10 ** (length_integer - k -1)
if x == ‘9‘:
ret1 += 9 * 10 ** (length_integer - k -1)
ret2 = 0
for k, x in enumerate(decimal):
if x == ‘0‘:
ret2 += 0 * 10 ** (-1 - k)
if x == ‘1‘:
ret2 += 1 * 10 ** (-1 - k)
if x == ‘2‘:
ret2 += 2 * 10 ** (-1 - k)
if x == ‘3‘:
ret2 += 3 * 10 ** (-1 - k)
if x == ‘4‘:
ret2 += 4 * 10 ** (-1 - k)
if x == ‘5‘:
ret2 += 5 * 10 ** (-1 - k)
if x == ‘6‘:
ret2 += 6 * 10 ** (-1 - k)
if x == ‘7‘:
ret2 += 7 * 10 ** (-1 - k)
if x == ‘8‘:
ret2 += 8 * 10 ** (-1 - k)
if x == ‘9‘:
ret2 += 9 * 10 ** (-1 - k)
return ret1 + ret2
In [151]:
str3 = ‘839.3452‘
In [153]:
type(str2int(str3))
Out[153]:
float
In [154]:
str2int(str3)
Out[154]:
839.3452
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