Python determines whether there are consecutive numbers

s_list=['25','24','23','02','04','06','08','22','23','24','25','02','04','06','08','23','24','25'If the character number to be judged is purely a number, just delete the int.

end list= List of results.

t_len=len(s_list)

z=0for i in range(1,t_len,1):

if (int(s_list[i])-int(s_list[i-1]))==1:

if z<=0:

if z<-1:

argtemp=

for j in range(i z-1,i,1):

end_z=1else:

z=z 1elif (int(s_list[i])-int(s_list[i-1]))==-1:

if z>=0:

if z>1:

argtemp=

for j in range(i-z-1,i,1):

end_z=-1else:

z=z-1else:

z=abs(z)

if z>1:

argtemp=

for j in range(i-z-1,i,1):

end_z=0z=abs(z)

if z>1:

argtemp=

for j in range(t_len-z-1,t_len,1):

end_for argtemp in end_list:

for t in argtemp[:-1]:

print t, print argtemp[-1] results. ###

Just look at the ascending order.

How many in a row?

"Qianlong Manuscript One Hundred and Twenty Returns to the Dream of the Red Chamber", in the Xianfeng period, the source title is called "The Dream of the Red Chamber", Yang Jizhen's old collection, also known as "Yang Collection", "Fat Manuscript". Article 78

Use it first to group each successive number.

import itertools

num_times = [(k, len(list(v)))for k, v in

The num times thus obtained is the number of consecutive occurrences of each number, num times = [

Then for each number, count the maximum number of occurrences, this step is relatively simple, just think about it yourself.

mylist = [1,1,0,1,1,1,0,0,0,0,1,1,1,1,0,1,0,11,0]

result = {}

tmp = none

for i in mylist:

if not :

The new value is 1

result[i] =

else:if i == tmp :

Same as last time, the tmpcount number is added by one, and the maxcount is updated at the same time

result[tmp]['tmpcount'] = result[tmp]['tmpcount'] +1

if result[tmp]['maxcount'] result[tmp]['maxcount'] = result[tmp]['tmpcount']

else: If it is different, the tmpcount of the last digit is zeroed, and the tmpcount of the digit this time is normalized.

result[i]['tmpcount'] = 1

result[tmp]['tmpcount'] = 0

tmp = i

for j,k in :

print 'Figures' + str(j) +'The maximum number of consecutive occurrences is' + str(k['maxcount'])

Try this, it's a bit of a hassle but it should fix the problem.

In terms of regular expressions, the regular expression can be written like [0-9], which means that as long as the numbers 0-9 are matched, it means that there are numbers in the string.

Judgment of strings:

isdigit()

true: Unicode digits, byte digits (single bytes), full-width digits (double bytes), Roman numerals.

false: Chinese character number.

error: None.

isdecimal()

true: unicode digits, full-width digits (double bytes) false: roman numerals, Chinese numerals.

error: byte number (single byte).

isnumeric()

true: Unicode digits, full-width digits (double bytes), Roman numerals, Chinese numerals.

false: None.

error: byte number (single byte).

Such as s="12335"