blocks|key|3300605|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3300606|mylist+=+[]
while+True:
++++mylist.append(int(raw_input("enter+number:")))
streak+=+0
cur,+last+=+0,+None
for+num+in+mylist:
++++if+num+==+last:
++++++++curr+%2B=+1
++++else:
++++++++streak+=+max(streak,+cur)
++++++++last+=+num
++++++++cur+=+0
print("longest+run+was+",streak)|code-block|syntax|javascript|3300607|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this

<pre><code>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)
</code></pre>

blocks|key|3300634|text|假设您有一个列表，比如[7,+7,+7,+6,+6,+6,+6,+5,+4,+6,+7]，您可以使用groupby()函数来计算重复次数，然后在最后打印最大次数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3300635|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)|code-block|syntax|javascript|3300636|在这种情况下，它返回4，因为数字6连续重复了4次|3300637|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/itertools.html#itertools.groupby^0|B|X|1E|9|1E|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Assuming you have a list such as <code>[7, 7, 7, 6, 6, 6, 6, 5, 4, 6, 7]</code> you can use the <a href="https://docs.python.org/2/library/itertools.html#itertools.groupby" rel="nofollow"><code>groupby()</code></a> function to count the number of repetitions and then print the max number at the end.

<pre><code>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)
</code></pre>

In this case it returns 4 since the number six was repeated 4 consecutive times

blocks|key|358339|text|这是你所描述的如何做到的一个冗长的版本。中途我意识到我是在Python16上运行它，所以它是向后兼容的！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|358340|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+%2B+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:+%25i,+count:+%25i"+%25+result))+#+print+resulting+number,+count|code-block|syntax|javascript|358341|输出：|358342|value:+6,+count+4|358343|如果您有任何问题，请随时发表意见。|358344|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This is a drawn-out version of how what you describe can be done. Half-way through I realized I was running it with Python 16, so it's backwards compatible!

<pre><code>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 &gt; 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
</code></pre>

Output:

<pre><code>value: 6, count 4
</code></pre>

If you have any questions, feel free to comment them.

blocks|key|2270646|text|这将跟踪streak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。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2270647|我对输入使用了异常处理。如果您输入了一个非数字，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。|2270648|请注意，我使用的是raw_input而不是input。|offset|length|2270649|while+True:
++++number+=+raw_input('enter+a+number:+')
++++if+number+==+'':
++++++break

++++try:
++++++++number+=+int(number)
++++except+ValueError:
++++++++print+("'"+%2B+number+%2B+"'+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+%2B=+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)|code-block|syntax|javascript|2270650|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/functions.html#raw_input^0|0|0|9|9|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@$F|X|G|Y|1|Z]]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

This will track the value of streak. If another streak is longer it will replace the last streak with the new value and length.

I used exception handling for input. If you enter a non-number it will ignore it and ask for a number again. If you enter nothing it will end the input loop.

Note that I'm using <a href="https://docs.python.org/2/library/functions.html#raw_input" rel="nofollow">raw_input</a> instead of input.

<pre><code>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) &gt; 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 &gt;= to track the last chain (in the case of a tie).
 if chain &gt; longest:
 longest = chain
 value = current

 print ('Your longest run was', longest)
</code></pre>

blocks|key|3300742|text|>>>+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|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3300743|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>&gt;&gt;&gt; from itertools import groupby
&gt;&gt;&gt; input_iter = iter(lambda: input('enter a number: '), 'end')
&gt;&gt;&gt; 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
</code></pre>

I am writing a program in which a user inputs values into a list until the want to end it and the program will tell the user the longest streak of numbers they entered. For example, is the user inputted 7,7,7,6,6,4,end the would get the output: your longest streak was 3. As 7 was entered 3 times in a row.

So far I have this and it seems to not want to ever end the current run so if i enter 7,7,7,6,6,6,6,5,4 it will tell me the longest streak is 7 like it is continuing the streak from the 7 being entered. This is what i have:

<pre><code>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 &gt; run:
 run = currentrun

print (mylist)

print ('Your longest run was' ,run)
</code></pre>

Any help is greatly appreciated.

inputting to a list and finding longest streak of the same input python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在写一个程序，在这个程序中，用户在列表中输入值，直到想要结束它，程序会告诉用户他们输入的最长的数字。例如，如果用户输入7，7，6，6，4，则最终将得到输出:您的最长连线是3。因为连续输入了3次7。到目前为止，我已经有了这个，它似乎不想结束当前的运行，所以如果我输入7，7，7，6，6，6，5，4，它会告诉我最长的连线...

问输入到列表并查找相同输入python的最长条纹
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