Python: 每日一题 34

475123763

jerryxjr1220 发表于 2017-4-27 13:39
def get_longest(lst,k):
  nlst = [len(i) for i in lst]
  xlst = [sum(nlst) for i in range(len(lst) ...

xlst = [sum(nlst[i:i+k]) for i in range(len(lst)-k)]
  maxi = xlst.index(max(xlst))
大哥能把这个按照初学者能看懂的方式写下吗

475123763

str = "Office workers who would normally step into a pub or gym to cope with the"
lst = str.split(" ")
nlst = [len(i) for i in lst]
t=max(nlst)
d=nlst.index(t)
c = lst[d]
del lst[d]
nlst = [len(i) for i in lst]
t=max(nlst)
d=nlst.index(t)
f=lst[d]
print(c+f)
初学者的献丑

475123763

g=['Office', 'workers', 'who','Office', 'woeqrkers', 'who', 'would', 'normally', 'step', 'into', 'a']

k=3
m=[]
def compare(list,k):
i = 0
while i<k:
      nlst = [len(i) for i in list]
      t = max(nlst)
      d = nlst.index(t)
      m.insert(0,list[d])
      del list[d]
      i+=1
print(  ''.join(m))
compare(g,k)

Akihisa

1. test.assert_equals(longest_consec(["it","wkppv","ixoyx", "3452", "zzzzzzzzzzzz"], 3), "ixoyx3452zzzzzzzzzzzz")
   
2. 为什么我的结果和楼主测试结果不一样
   
3. >>> longest_consec(["it","wkppv","ixoyx", "3452", "zzzzzzzzzzzz"], 3)
   
4. 'zzzzzzzzzzzzwkppvixoyx'

复制代码

Akihisa

1. def longest_consec(strarr,k):
   
2.     n = len(strarr)
   
3.     if n == 0 or k > n or k <= 0:
   
4.         return "无法执行"
   
5.     else:
   
6.         strarr.sort(key =lambda a:len(a),reverse = True)
   
7.         str1 =''
   
8.         for i in range(k):
   
9.             str1 += strarr[i]
   
10.         return str1

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jrro452

1. from heapq import nlargest
   
2. def longest_consec(lists,num):
   
3.     count_list = []
   
4.     dict_list = []
   
5.     result_list = []
   
6.     sorted_list = []
   
7.     fr = ''
   
8.     joint = ''
   
9.     if len(lists) == 0 or num>len(lists) or num<=0:
   
10.         return ""
    
11.     else:
    
12.         for each in lists:
    
13.             count_list.append(len(each))
    
14.         dict_list = zip(count_list,lists)
    
15.         result_list = nlargest(num,dict_list)
    
16.         for each in result_list:
    
17.             sorted_list.append(each[1])
    
18.         fr =joint.join(sorted_list)
    
19.     return fr

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做的好复杂，有2到结果和题主结果不一样。不知道是什么情况。

算幽幽

I Iove Fishc C

1. def longest_consec(list1, n):
   
2.     if n==0 or len(list1)<n or len(list1)<=0:
   
3.         return ''
   
4.     i = 0
   
5.     list_len = []
   
6.     while i + n <= len(list1):
   
7.         result = ""
   
8.         for m in range(i, i + n):
   
9.             result += list1[m]
   
10.         else:
    
11.             list_len.append(result)
    
12.         i+=1
    
13.     return sorted(list_len,key=lambda s:len(s))[-1]

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double-lee

看看...

3333332

谢谢

北城以北灬

shigure_takimi 发表于 2017-12-3 14:00

倒数第3个？结果是这样的？为什么？

捞鬼

捞

yu123py

貌似方法不太对：

1. def longest_consec(list, k):
   
2.     if len(list) == 0 or k > len(list) or k <= 0:
   
3.         return ''
   
4.     list.sort(reverse = True, key = len)
   
5.     s = ''
   
6.     for i in range(k):
   
7.         s += list[i]
   
8.     return s

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克里斯保罗

没看懂 = =

Jung

import time
def fun32():
    Tiangan = {3:"甲",4:"乙",5:"丙",6:"丁",7:"戊",8:"己",9:"庚",0:"辛",1:"壬",2:"癸"}
    DiZhi = {3:"子",4:"丑",5:"寅",6:"卯",7:"辰",8:"巳",9:"午",10:"未",11:"申",0:"酉",1:"戌",2:"亥"}
    ShengXiao = {"子":"鼠","丑":"牛","寅":"虎","卯":"兔","辰":"龙","巳":"蛇","午":"马","未":"羊","申":"猴","酉":"鸡","戌":"狗","亥":"猪"}
    Jiazibiao = {1:"甲子", 11:"甲戌", 21:"甲申",31:"甲午",41:"甲辰",51:"甲寅",
                 2:"乙丑",12:"乙亥",22:"乙酉",32:"乙未",42:"乙巳",52:"乙卯",
                 3:"丙寅",13:"丙子",23:"丙戌",33:"丙申",43:"丙午",53:"丙辰",
                 4:"丁卯",14:"丁丑",24:"丁亥",34:"丁酉",44:"丁未",54:"丁巳",
                 5:"戊辰",15:"戊寅",25:"戊子",35:"戊戌",45:"戊申",55:"戊午",
                 6:"己巳",16:"己卯",26:"己丑",36:"己亥",46:"己酉",56:"己未",
                 7:"庚午",17:"庚辰",27:"庚寅",37:"庚子",47:"庚戌",57:"庚申",
                 8:"辛未",18:"辛巳",28:"辛卯",38:"辛丑",48:"辛亥",58:"辛酉",
                 9:"壬申",19:"壬午",29:"壬辰",39:"壬寅",49:"壬子",59:"壬戌",
                 10:"癸酉",20:"癸未",30:"癸巳",40:"癸卯",50:"癸丑",60:"癸亥"}
    localtime = time.asctime( time.localtime(time.time()))
    LCurrentYear=localtime.split(" ")
    CurrentYear=LCurrentYear[len(LCurrentYear)-1]
    Year = str(input("请输入年份："))
    Tian = Tiangan[int(Year[len(Year)-1])]
    Di = DiZhi[int(Year) % 12]
    ShuXiang = ShengXiao[Di]
    print("公元%s是黄历的%s年。属相是%s" % (Year,Tian+Di,ShuXiang))
    GanZi = input("请输入干子：")
    if GanZi in Jiazibiao.values():
        for key in Jiazibiao.keys():
            if Jiazibiao[key]== GanZi:
                GanZiindex = key
                break
        
    else:
        print("输入干子不对")
        
    Year = CurrentYear
    Tian = Tiangan[int(Year[len(Year)-1])]
    Di = DiZhi[int(Year) % 12]
    CurrentYearGanZi=Tian+Di
    for key1 in Jiazibiao.keys():
        if Jiazibiao[key1]== CurrentYearGanZi:
            CurrentYearGanZiindex = key1
            break
    if CurrentYearGanZi==GanZi:
       print("上一个%s年是%d.下一个%s年时%d.本年%d也是%s年" % (GanZi,int(Year)-60,GanZi,int(Year)+60,int(CurrentYear),GanZi))
    else:
         print("上一个%s年是%d.下一个%s年时%d." % (GanZi,int(Year)-(CurrentYearGanZiindex-GanZiindex),GanZi,int(Year)-(CurrentYearGanZiindex-GanZiindex)+60))

Jung

Jung 发表于 2019-12-5 08:48
import time
def fun32():
    Tiangan = {3:"甲",4:"乙",5:"丙",6:"丁",7:"戊",8:"己",9:"庚",0:"辛",1: ...

发错地方了。

holiday_python

def longest_consec(strarr,k):
    n = len(strarr)
    chr_len = {}
    if n ==0 or k > n or k <= 0:
        return ''
    else:
        for idx in range(1,len(strarr)+1):
            length = len(strarr)
            chr_len[strarr[idx]] = length
        for k in chr_len.keys():
            if chr_len[k]
            #用字典排序


            return ''.join(strarr[-2:])

if __name__ == '__main__':
    strarr = list(input('请输入字符串元素(用逗号分隔): ').split(','))
    k = int(input('请输入需要取出的字符串个数: '))
    print(longest_consec(strarr, k))

nononoyes

666

nononoyes

'''

给你一个字符串组成列表strarr，给你一个整数k。
你的任务是找出由k个字符组成的最长的连续的字符串组合。
例子：
longest_consec(["zone", "abigail", "theta", "form", "libe", "zas", "theta", "abigail"], 2) --> "abigailtheta"
n是列表的长度，如果if n = 0 or k > n or k <= 0 return ""
'''

'''
解题思路：
1.新建一个列表存放原字符串列表中每个字符的长度
2.通过max函数和index函数找到最长字符串的索引值
3.同时删除两个列表最长字符串的索引值再次循环找到最长字符串的索引值
'''
#请将以下代码存为test.py用于检验答案。
def assert_equals(func, target, *args):
    if func == target:
        print('Success！')
    else:
        print('Fail!{0} not equals {1}'.format(func, target))
        print(*args)


def longest_consec(list,k):
    #1.新建一个列表存放原字符串列表中每个字符的长度
    listSort = []
    str=""
    if len(list) ==0 or k > len(list) or k <=0:
        return ""
    else:
        for i in list:
            #1.新建一个列表存放原字符串列表中每个字符的长度
            listSort.append(len(i))
        for j in range(k):
            #通过max函数和index函数找到最长字符串的索引值
            index = listSort.index(max(listSort))
            str += list[listSort.index(max(listSort))]
            #同时删除两个列表最长字符串的索引值对应的字符
            listSort.remove(max(listSort))
            list.remove(list[index])
    return str


assert_equals(longest_consec(["zone", "abigail", "theta", "form", "libe", "zas"], 2), "abigailtheta")
assert_equals(longest_consec(["ejjjjmmtthh", "zxxuueeg", "aanlljrrrxx", "dqqqaaabbb", "oocccffuucccjjjkkkjyyyeehh"], 1), "oocccffuucccjjjkkkjyyyeehh")
assert_equals(longest_consec([], 3), "")
assert_equals(longest_consec(["itvayloxrp","wkppqsztdkmvcuwvereiupccauycnjutlv","vweqilsfytihvrzlaodfixoyxvyuyvgpck"], 2), "wkppqsztdkmvcuwvereiupccauycnjutlvvweqilsfytihvrzlaodfixoyxvyuyvgpck")
assert_equals(longest_consec(["wlwsasphmxx","owiaxujylentrklctozmymu","wpgozvxxiu"], 2), "wlwsasphmxxowiaxujylentrklctozmymu")
assert_equals(longest_consec(["zone", "abigail", "theta", "form", "libe", "zas"], -2), "")
assert_equals(longest_consec(["it","wkppv","ixoyx", "3452", "zzzzzzzzzzzz"], 3), "ixoyx3452zzzzzzzzzzzz")
assert_equals(longest_consec(["it","wkppv","ixoyx", "3452", "zzzzzzzzzzzz"], 15), "")
assert_equals(longest_consec(["it","wkppv","ixoyx", "3452", "zzzzzzzzzzzz"], 0), "")


Success！
Success！
Success！
Success！
Fail!owiaxujylentrklctozmymuwlwsasphmxx not equals wlwsasphmxxowiaxujylentrklctozmymu

Success！
Fail!zzzzzzzzzzzzwkppvixoyx not equals ixoyx3452zzzzzzzzzzzz

Success！
Success！

wwwwwise

1