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面试真题

面试真题

作者: Ugfly | 来源:发表于2018-02-14 00:43 被阅读0次
""" 1 :对列表去重,并且保持原来顺序。"""
# l = [1,2,3,4,4,3,2,1,34,4,5,6,54,4,6,7,67,878,6,567,45,64,345]
# # ls = set(l)
# # print(ls)# {64, 1, 2, 3, 4, 34, 5, 6, 7, 67, 45, 878, 54, 567, 345}
# # ll = list(ls)
# # ll.sort(key=l.index)
# # print(ll)
# ls = list(set(l))
# ls.sort(key=l.index)
# print(ls)
""" 2. 现有两元组(('a'),('b')),(('c'),('d'))。请使用python中匿名函数生成列表[{'a':'c'},{'b':'d'}]"""
# t = (('a'),('b')),(('c'),('d'))
# d = {}
# for i in t :
#     print(i)
#     d[i[0]] = i[1]
#     print(d)
    # ('a', 'b')
    # {'a': 'b'}
    # ('c', 'd')
    # {'a': 'b', 'c': 'd'}
""" 3. 请给出二分查找的python示例代码。"""
# def search(num,l,start=None,end=None):
#     start = start if start else 0
#     end = end if end else len(l) - 1
#     mid = (end - start)//2 + start
#     if start > end:
#         return None
#     elif l[mid] > num :
#         return search(num,l,start,mid-1)
#     elif l[mid] < num:
#         return search(num,l,mid+1,end)
#     elif l[mid] == num:
#         return mid
# l = [1,2,3,4,4,3,2,1,34,4,5,6,54,4,6,7,67,878,6,567,45,64,345]
# num = 878
# print(search(num,l)) # 17

''' 4. 在python字符串格式化中,%和.format的主要区别是什么?'''

# %s普通版字符串格式化,还有%d 是十进制数。
# .format是格式化字符串。 可以按照顺序传入值,也可以根据关键字传值。

''' 5. *args 和 kwargs 在上面情况下会使用到?给出使用**kwargs的示例代码。'''


# x= ' f'
# y=2
# # print(x*y)
#
# d= {'1':5,'2':22}
# dd = d
# d['1'] = 10
# print(dd)
# ddd = d['1']+dd['1']
# print(ddd)


# l = [1,2,3,4,5]
# l1=l[:]
# l.append(6)
# print(l1)  # [1, 2, 3, 4, 5]


# name1 = [1,2,3,4,5]
# name2 = name1
# name3 = name1[:]
# name2[0] = 'a'
# name3[1] = 'b'
# sum = 0
# for i in (name1,name2,name3):
#     if i[0] == 'a':
#         sum += 1
#     if i[1] == 'b':
#         sum += 10
# print(sum)
# 12


# a = lambda p:p*2
# b = lambda p:p*3
# x = 2
# x = a(x)  # 4
# x = b(x)  # 12
# x = a(x)  # 24
# print(x)



# x = 1
# y = 0
# z = 0
# if x or y and z:
#     print('6666')
# else:
#     print('false')
# and优先级高

# 在python里面如何实现tuple和list的切换
# t = (1,2,3,4)
# print(t,type(t))
# l = list(t)
# print(l,type(l))
# (1, 2, 3, 4) <class 'tuple'>
# [1, 2, 3, 4] <class 'list'>

# 请用python代码实现删除一个列表里面的重复元素
# l = [1,2,3,4,5,6,4,3,2,1,6]
# # ll = []
# # for i in l :
# #     if not i in ll :
# #         ll.append(i)
# # print(i)
# ll = list(set(l))
# ll.sort(key=l.index)
# print(ll)


# 如何得到一个list的交集或者差集
# l = [1,2,3,4,5]
# l1 = [1,2,3,4,6,7]
# # 转为set类型,
# # >> x & y  # 交集
# # >> > x | y  # 并集
# # >> > x - y  # 差集
# s = set(l)
# s1 = set(l1)
# print(list(s&s1))
# print(list(s-s1))
# print(list(s|s1))



# 第三页!!!!!!!!!!!!!!!!!!!!


# 产生数组对象
# 将时间字符串转化为时间对象
# import datetime
# date = datetime.datetime.strptime('2018-1-11','%Y-%m-%d')
# # print(date) # 2018-01-11 00:00:00
#
# # 灵活处理时间对象。
# # import dateutil
# # date = dateutil.parser.parse('2018-2-2')
# # print(date)
#
import pandas  # 你知道开始和结尾就可以知道中间的日子,可以自动识别闰年。
# # print(pandas.date_range('2015-12-1','2016-2-2'))
# # print(pandas.date_range('2016-1-1',periods=40,freq='B')) # periods 时间 长度  第三个参数默认D
# """
# start     开始时间
# end       结束时间
# periods       时间长度
# freq      时间频率,默认为'D',可选H(our),W(eek),B(usiness),S(emi-)M(onth),(min)T(es), S(econd), A(year),…
# """
#
# # #####文件操作
# df = pandas.read_csv('') # 逗号分隔
# df1 = pandas.read_csv('',sep='\s+') # 逗号分隔
# df2 = pandas.read_table('',) # \t 分隔 制表符


import matplotlib.pyplot as plt
import numpy as np

# plt.plot([0,1,2,3],[4,7,9,13],marker=',')
# plt.plot([2,3,32,3,31])
# plt.title = ('pipipi')
# plt.show()

# x = np.linspace(-100,100,100000)  # start end  切成多少块
# y1 = x
# plt.plot(x,y1,label='$y=x')  # 加了$,就是公式环境。
# plt.legend()


plt.pie([1,2,3],labels=['a','b','c'])
plt.axis('equal')
# plt.show()




import socket
# s = socket.gethostbyname('baidu.com')
# print(s)

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