输入到列表并找到相同输入python的最长条纹

Question

我正在编写一个程序，其中用户将值输入到列表中，直到想要结束它，程序会告诉用户他们输入的最长连续数字。例如，用户输入 7,7,7,6,6,4,end 会得到输出：你最长的连续是 3。因为 7 连续输入了 3 次。

到目前为止，我有这个，它似乎不想结束当前的运行，所以如果我输入 7、7、7、6、6、6、6、5、4，它会告诉我最长的连胜是 7就像它正在继续从 7 进入的连胜纪录。这就是我所拥有的：

mylist = []

run = 1

currentrun = 1

number = input('enter a number: ')
mylist.append(number)



while number != 'end' :
    number = input ('enter a number: ')
    mylist.append(number)
for i in range (len(mylist)):

    if mylist[i] == mylist[i-1] and mylist[i] == mylist[i+1] :
        currentrun = currentrun + 1
    else:
        currentrun = 0

    print (currentrun)
    if currentrun > run:
        run = currentrun

print (mylist)

print ('Your longest run was' ,run)

非常感谢任何帮助。

Answer 1

试试这个

mylist = []
while True:
    mylist.append(int(raw_input("enter number:")))
streak = 0
cur, last = 0, None
for num in mylist:
    if num == last:
        curr += 1
    else:
        streak = max(streak, cur)
        last = num
        cur = 0
print("longest run was ",streak)

Answer 2

假设您有一个诸如[7, 7, 7, 6, 6, 6, 6, 5, 4, 6, 7] 之类的列表，您可以使用groupby() 函数计算重复次数，然后在末尾打印最大次数。

from itertools import groupby
a = [7, 7, 7, 6, 6, 6, 6, 5, 4, 6, 7]
lst = []
for n,c in groupby(a):
   num,count = n,sum(1 for i in c)
   lst.append((num,count))

maxx = max([y for x,y in lst])
print 'Your longest run was {}'.format(maxx)

在这种情况下，它返回 4，因为数字 6 连续重复了 4 次

Answer 3

这是您所描述的如何完成的冗长版本。中途我意识到我正在使用 Python 16 运行它，所以它是向后兼容的！

a = None # stores the last number seen
b = 0 # stores the count of the last number
result = [0, 0] # number, count result array
for c in "7,7,7,6,6,6,6,5,4".split(','): # split string into array of
                                         # just our numbers
    c = int(c) # change the character into a bast ten int
    if c != a: # check current number against last
        a = c # if different than last, remember current number as last
        b = 1 # start over counting at one
    else: # if current number is same as last
        b = b + 1 # increment counter
        if b > result[1]: result = a, b # if counter higher than highest
                                        # previous count, store the
                                        # current number and count
print(("value: %i, count: %i" % result)) # print resulting number, count

输出：

value: 6, count 4

如果您有任何问题，请随时发表评论。

Answer 4

这将跟踪条纹的值。如果另一个条纹更长，它将用新的值和长度替换最后一个条纹。

我对输入使用了异常处理。如果您输入一个非数字，它将忽略它并再次要求输入一个数字。如果您不输入任何内容，它将结束输入循环。

请注意，我使用的是raw_input 而不是输入。

while True:
    number = raw_input('enter a number: ')
    if number == '':
      break

    try:
        number = int(number)
    except ValueError:
        print ("'" + number + "' is not a number.")
        continue

    mylist.append(number)

if len(mylist) > 0:
    #print (mylist)

    # Chain is the current run we're tracking. Longest is the longest chain we've tracked so far.
    chain = longest = 1
    # Current is the value of the chain we're tracking now. Value is the value of the longest chain we've tracked so far. We use the first value of the list.
    current = value = mylist[0]

    # Starting with the second number in the list and iterating to the end we compare values.
    for number in mylist[1:]:
        # Did we find another in our current chain?
        if number == current:
            chain += 1
        else:
            chain = 1
            current = number

        # This will require chains to exceed the previous longest chain to be chosen as the longest. Change this to >= to track the last chain (in the case of a tie).
        if chain > longest:
            longest = chain
            value = current

    print ('Your longest run was', longest)

Answer 5

>>> from itertools import groupby
>>> input_iter = iter(lambda: input('enter a number: '), 'end')
>>> max(sum(1 for x in v) for k,v in groupby(input_iter))
enter a number: 7
enter a number: 7
enter a number: 7
enter a number: 6
enter a number: 6
enter a number: 4
enter a number: end
3