I'm trying to compare each character of the string num with each digit of the strings contained in computer_guesses.
I must use the numbers as strings.
I tried to run this code but the output was:
0
0
0
0
0
0
0
0
0
0
Process finished with exit code 0
Probably i'm just missing the syntax, i'm sure that it can be made without using the function split(word) .
def split(word):
return [char for char in word]
computer_guesses = ["12345", "67890"]
num = "23648"
correct = 0
for x in range(len(computer_guesses)):
for y in range(len(computer_guesses[x])):
if num[y] == split(computer_guesses[x]):
correct += 1
print(correct)
else:
print(correct)
You can use enumerate for iterating on iterable with index counting.
computer_guesses = ["12345", "67890"]
num = "23648"
correct = 0
for guess in computer_guesses:
for idx, digit in enumerate(guess):
if num[idx] == digit:
correct += 1
print(correct)
else:
print(correct)
This can be one approach for your problem:
def split(num,word):
num = num
default = [char for char in num]
wd = [char for char in word]
correct = 0
for m in range(len(wd)):
if default[m]==wd[m]:
correct += 1
return correct
computer_guesses = ["12345", "67890"]
num = "23648"
for x in range(len(computer_guesses)):
print(split(num, computer_guesses[x]))
from my vision the wrong part in code it's 'if condition' side
if num[y] == split(computer_guesses[x]):
i switch it to :
if num[y] == computer_guesses[x][y]:
and the code working , i hope that what you looking for
This question already has answers here:
Run length encoding in Python
(10 answers)
Closed 2 years ago.
How the input should be like:
aaabbcca
How the output should be like:
a3b2c2a1
My attempt:
string = input()
ans = ""
i = 0
j=0
while i < len(string):
num=1
ans += string[i]
j = i + 1
if j >= len(string): break
while j < len(string):
if string[i] == string[j]:
num += 1
else:
ans += str(num)
i = j
break
print(ans)
# i = 0
# for i in range(len(string)):
# num=0
# ans += string[i]
# for j in range(i,len(string)):
# if string[i] == string[j]:
# num +=1
# i = j
# else: break
# ans += str(num)
# print(ans)
My problem: nothing is printed
how can i get this code right?
You might want to explore groupby from itertools.
For example:
def encode(data: str) -> str:
return "".join(f"{x}{sum(1 for _ in y)}" for x, y in groupby(data))
print(encode("aaabbcca"))
Output:
a3b2c2a1
Second loop is redundant for this case. You can simply take first char as a base case and with one loop you can do all you need.
raw_str = 'aaabbcca'
last_char = raw_str[0]
res = ''
count = 0
for i in range(len(raw_str)):
if raw_str[i] == last_char:
count +=1
else:
res += raw_str[i - 1] + str(count)
last_char = raw_str[i]
count = 1
# If there is only 1 different char at the end of the str, it will not call if
# so you need to check explicitly if the last char is equal to your lastly added one
if last_char != raw_str[-2]:
res += last_char + str(count)
I want to count the number of occurrences of the substring "bob" within the string s. I do this exercise for an edX Course.
s = 'azcbobobegghakl'
counter = 0
numofiterations = len(s)
position = 0
#loop that goes through the string char by char
for iteration in range(numofiterations):
if s[position] == "b": # search pos. for starting point
if s[position+1:position+2] == "ob": # check if complete
counter += 1
position +=1
print("Number of times bob occurs is: " + str(counter))
However it seems that the s[position+1:position+2] statement is not working properly. How do i adress the two chars behind a "b"?
The second slice index isn't included. It means that s[position+1:position+2] is a single character at position position + 1, and this substring cannot be equal to ob. See a related answer. You need [:position + 3]:
s = 'azcbobobegghakl'
counter = 0
numofiterations = len(s)
position = 0
#loop that goes through the string char by char
for iteration in range(numofiterations - 2):
if s[position] == "b": # search pos. for starting point
if s[position+1:position+3] == "ob": # check if complete
counter += 1
position +=1
print("Number of times bob occurs is: " + str(counter))
# 2
You could use .find with an index:
s = 'azcbobobegghakl'
needle = 'bob'
idx = -1; cnt = 0
while True:
idx = s.find(needle, idx+1)
if idx >= 0:
cnt += 1
else:
break
print("{} was found {} times.".format(needle, cnt))
# bob was found 2 times.
Eric's answer explains perfectly why your approach didn't work (slicing in Python is end-exclusive), but let me propose another option:
s = 'azcbobobegghakl'
substrings = [s[i:] for i in range(0, len(s))]
filtered_s = filter(substrings, lambda s: s.startswith("bob"))
result = len(filtered_s)
or simply
s = 'azcbobobegghakl'
result = sum(1 for ss in [s[i:] for i in range(0, len(s))] if ss.startswith("bob"))
I have this code that I found on another topic, but it sorts the substring by contiguous characters and not by alphabetical order. How do I correct it for alphabetical order? It prints out lk, and I want to print ccl. Thanks
ps: I'm a beginner in python
s = 'cyqfjhcclkbxpbojgkar'
from itertools import count
def long_alphabet(input_string):
maxsubstr = input_string[0:0] # empty slice (to accept subclasses of str)
for start in range(len(input_string)): # O(n)
for end in count(start + len(maxsubstr) + 1): # O(m)
substr = input_string[start:end] # O(m)
if len(set(substr)) != (end - start): # found duplicates or EOS
break
if (ord(max(sorted(substr))) - ord(min(sorted(substr))) + 1) == len(substr):
maxsubstr = substr
return maxsubstr
bla = (long_alphabet(s))
print "Longest substring in alphabetical order is: %s" %bla
s = 'cyqfjhcclkbxpbojgkar'
r = ''
c = ''
for char in s:
if (c == ''):
c = char
elif (c[-1] <= char):
c += char
elif (c[-1] > char):
if (len(r) < len(c)):
r = c
c = char
else:
c = char
if (len(c) > len(r)):
r = c
print(r)
Try changing this:
if len(set(substr)) != (end - start): # found duplicates or EOS
break
if (ord(max(sorted(substr))) - ord(min(sorted(substr))) + 1) == len(substr):
to this:
if len(substr) != (end - start): # found duplicates or EOS
break
if sorted(substr) == list(substr):
That will display ccl for your example input string. The code is simpler because you're trying to solve a simpler problem :-)
You can improve your algorithm by noticing that the string can be broken into runs of ordered substrings of maximal length. Any ordered substring must be contained in one of these runs
This allows you to just iterate once through the string O(n)
def longest_substring(string):
curr, subs = '', ''
for char in string:
if not curr or char >= curr[-1]:
curr += char
else:
curr, subs = '', max(curr, subs, key=len)
return max(curr, subs, key=len)
s = 'cyqfjhcclkbxpbojgkar'
longest = ""
max =""
for i in range(len(s) -1):
if(s[i] <= s[i+1] ):
longest = longest + s[i]
if(i==len(s) -2):
longest = longest + s[i+1]
else:
longest = longest + s[i]
if(len(longest) > len(max)):
max = longest
longest = ""
if(len(s) == 1):
longest = s
if(len(longest) > len(max)):
print("Longest substring in alphabetical order is: " + longest)
else:
print("Longest substring in alphabetical order is: " + max)
In a recursive way, you can import count from itertools
Or define a same method:
def loops( I=0, S=1 ):
n = I
while True:
yield n
n += S
With this method, you can obtain the value of an endpoint, when you create any substring in your anallitic process.
Now looks the anallize method (based on spacegame issue and Mr. Tim Petters suggestion)
def anallize(inStr):
# empty slice (maxStr) to implement
# str native methods
# in the anallize search execution
maxStr = inStr[0:0]
# loop to read the input string (inStr)
for i in range(len(inStr)):
# loop to sort and compare each new substring
# the loop uses the loops method of past
# I = sum of:
# (i) current read index
# (len(maxStr)) current answer length
# and 1
for o in loops(i + len(maxStr) + 1):
# create a new substring (newStr)
# the substring is taked:
# from: index of read loop (i)
# to: index of sort and compare loop (o)
newStr = inStr[i:o]
if len(newStr) != (o - i):# detect and found duplicates
break
if sorted(newStr) == list(newStr):# compares if sorted string is equal to listed string
# if success, the substring of sort and compare is assigned as answer
maxStr = newStr
# return the string recovered as longest substring
return maxStr
Finally, for test or execution pourposes:
# for execution pourposes of the exercise:
s = "azcbobobegghakl"
print "Longest substring in alphabetical order is: " + anallize( s )
The great piece of this job started by: spacegame and attended by Mr. Tim Petters, is in the use of the native str methods and the reusability of the code.
The answer is:
Longest substring in alphabetical order is: ccl
In Python character comparison is easy compared to java script where the ASCII values have to be compared. According to python
a>b gives a Boolean False and b>a gives a Boolean True
Using this the longest sub string in alphabetical order can be found by using the following algorithm :
def comp(a,b):
if a<=b:
return True
else:
return False
s = raw_input("Enter the required sting: ")
final = []
nIndex = 0
temp = []
for i in range(nIndex, len(s)-1):
res = comp(s[i], s[i+1])
if res == True:
if temp == []:
#print i
temp.append(s[i])
temp.append(s[i+1])
else:
temp.append(s[i+1])
final.append(temp)
else:
if temp == []:
#print i
temp.append(s[i])
final.append(temp)
temp = []
lengths = []
for el in final:
lengths.append(len(el))
print lengths
print final
lngStr = ''.join(final[lengths.index(max(lengths))])
print "Longest substring in alphabetical order is: " + lngStr
Use list and max function to reduce the code drastically.
actual_string = 'azcbobobegghakl'
strlist = []
i = 0
while i < len(actual_string)-1:
substr = ''
while actial_string[i + 1] > actual_string[i] :
substr += actual_string[i]
i += 1
if i > len(actual_string)-2:
break
substr += actual-string[i]
i += 1
strlist.append(subst)
print(max(strlist, key=len))
Wow, some really impressing code snippets here...
I want to add my solution, as I think it's quite clean:
s = 'cyqfjhcclkbxpbojgkar'
res = ''
tmp = ''
for i in range(len(s)):
tmp += s[i]
if len(tmp) > len(res):
res = tmp
if i > len(s)-2:
break
if s[i] > s[i+1]:
tmp = ''
print("Longest substring in alphabetical order is: {}".format(res))
Without using a library, but using a function ord() which returns ascii value for a character.
Assumption: input will be in lowercase, and no special characters are used
s = 'azcbobobegghakl'
longest = ''
for i in range(len(s)):
temp_longest=s[i]
for j in range(i+1,len(s)):
if ord(s[i])<=ord(s[j]):
temp_longest+=s[j]
i+=1
else:
break
if len(temp_longest)>len(longest):
longest = temp_longest
print(longest)
Slightly different implementation, building up a list of all substrings in alphabetical order and returning the longest one:
def longest_substring(s):
in_orders = ['' for i in range(len(s))]
index = 0
for i in range(len(s)):
if (i == len(s) - 1 and s[i] >= s[i - 1]) or s[i] <= s[i + 1]:
in_orders[index] += s[i]
else:
in_orders[index] += s[i]
index += 1
return max(in_orders, key=len)
s = "azcbobobegghakl"
ls = ""
for i in range(0, len(s)-1):
b = ""
ss = ""
j = 2
while j < len(s):
ss = s[i:i+j]
b = sorted(ss)
str1 = ''.join(b)
j += 1
if str1 == ss:
ks = ss
else:
break
if len(ks) > len(ls):
ls = ks
print("The Longest substring in alphabetical order is "+ls)
This worked for me
s = 'cyqfjhcclkbxpbojgkar'
lstring = s[0]
slen = 1
for i in range(len(s)):
for j in range(i,len(s)-1):
if s[j+1] >= s[j]:
if (j+1)-i+1 > slen:
lstring = s[i:(j+1)+1]
slen = (j+1)-i+1
else:
break
print("Longest substring in alphabetical order is: " + lstring)
Output: Longest substring in alphabetical order is: ccl
input_str = "cyqfjhcclkbxpbojgkar"
length = len(input_str) # length of the input string
iter = 0
result_str = '' # contains latest processed sub string
longest = '' # contains longest sub string alphabetic order
while length > 1: # loop till all char processed from string
count = 1
key = input_str[iter] #set last checked char as key
result_str += key # start of the new sub string
for i in range(iter+1, len(input_str)): # discard processed char to set new range
length -= 1
if(key <= input_str[i]): # check the char is in alphabetic order
key = input_str[i]
result_str += key # concatenate the char to result_str
count += 1
else:
if(len(longest) < len(result_str)): # check result and longest str length
longest = result_str # if yes set longest to result
result_str = '' # re initiate result_str for new sub string
iter += count # update iter value to point the index of last processed char
break
if length is 1: # check for the last iteration of while loop
if(len(longest) < len(result_str)):
longest = result_str
print(longest);
finding the longest substring in alphabetical order in Python
in python shell 'a' < 'b' or 'a' <= 'a' is True
result = ''
temp = ''
for char in s:
if (not temp or temp[-1] <= char):
temp += char
elif (temp[-1] > char):
if (len(result) < len(temp)):
result = temp
temp = char
if (len(temp) > len(result)):
result = temp
print('Longest substring in alphabetical order is:', result)
s=input()
temp=s[0]
output=s[0]
for i in range(len(s)-1):
if s[i]<=s[i+1]:
temp=temp+s[i+1]
if len(temp)>len(output):
output=temp
else:
temp=s[i+1]
print('Longest substring in alphabetic order is:' + output)
I had similar question on one of the tests on EDX online something. Spent 20 minutes brainstorming and couldn't find solution. But the answer got to me. And it is very simple. The thing that stopped me on other solutions - the cursor should not stop or have unique value so to say if we have the edx string s = 'azcbobobegghakl' it should output - 'beggh' not 'begh'(unique set) or 'kl'(as per the longest identical to alphabet string). Here is my answer and it works
n=0
for i in range(1,len(s)):
if s[n:i]==''.join(sorted(s[n:i])):
longest_try=s[n:i]
else:
n+=1
In some cases, input is in mixed characters like "Hello" or "HelloWorld"
**Condition 1:**order determination is case insensitive, i.e. the string "Ab" is considered to be in alphabetical order.
**Condition 2:**You can assume that the input will not have a string where the number of possible consecutive sub-strings in alphabetical order is 0. i.e. the input will not have a string like " zxec ".
string ="HelloWorld"
s=string.lower()
r = ''
c = ''
last=''
for char in s:
if (c == ''):
c = char
elif (c[-1] <= char):
c += char
elif (c[-1] > char):
if (len(r) < len(c)):
r = c
c = char
else:
c = char
if (len(c) > len(r)):
r = c
for i in r:
if i in string:
last=last+i
else:
last=last+i.upper()
if len(r)==1:
print(0)
else:
print(last)
Out:elloW
```python
s = "cyqfjhcclkbxpbojgkar" # change this to any word
word, temp = "", s[0] # temp = s[0] for fence post problem
for i in range(1, len(s)): # starting from 1 not zero because we already add first char
x = temp[-1] # last word in var temp
y = s[i] # index in for-loop
if x <= y:
temp += s[i]
elif x > y:
if len(temp) > len(word): #storing longest substring so we can use temp for make another substring
word = temp
temp = s[i] #reseting var temp with last char in loop
if len(temp) > len(word):
word = temp
print("Longest substring in alphabetical order is:", word)
```
My code store longest substring at the moment in variable temp, then compare every string index in for-loop with last char in temp (temp[-1]) if index higher or same with (temp[-1]) then add that char from index in temp. If index lower than (temp[-1]) checking variable word and temp which one have longest substring, after that reset variable temp so we can make another substring until last char in strings.
s = 'cyqfjhcclkbxpbojgkar'
long_sub = '' #longest substring
temp = '' # temporarily hold current substr
if len(s) == 1: # if only one character
long_sub = s
else:
for i in range(len(s) - 1):
index = i
temp = s[index]
while index < len(s) - 1:
if s[index] <= s[index + 1]:
temp += s[index + 1]
else:
break
index += 1
if len(temp) > len(long_sub):
long_sub = temp
temp = ''
print(long_sub)
For comprehensibility, I also add this code snippet based on regular expressions. It's hard-coded and seems clunky. On the other hand, it seems to be the shortest and easiest answer to this problem. And it's also among the most efficient in terms of runtime complexity (see graph).
import re
def longest_substring(s):
substrings = re.findall('a*b*c*d*e*f*g*h*i*j*k*l*m*n*o*p*q*r*s*t*u*v*w*x*y*z*', s)
return max(substrings, key=len)
(Unfortunately, I'm not allowed to paste a graph here as a "newbie".)
Source + Explanation + Graph: https://blog.finxter.com/python-how-to-find-the-longest-substring-in-alphabetical-order/
Another way:
s = input("Please enter a sentence: ")
count = 0
maxcount = 0
result = 0
for char in range(len(s)-1):
if(s[char]<=s[char+1]):
count += 1
if(count > maxcount):
maxcount = count
result = char + 1
else:
count = 0
startposition = result - maxcount
print("Longest substring in alphabetical order is: ", s[startposition:result+1])
My intent is to have the program list all strings in a text file that have 3 sets of double letters. Here is the function that is supposed to return True if 3 or more double letter sets are found:
def three_double(s):
doubnum = 0
i=0
while i < len(s)-1:
if s[i] == s[i+1]:
doubnum += 1
elif doubnum >= 3:
return True
else:
i += 1
return False
I'm not sure why it doesn't print anything. Here is the rest of the program.
# Function to read and apply the three_double test to each string in
# an input file. It counts the number of results.
def find_three_double(fin):
count = 0
for w in fin:
w = w.strip()
if three_double(w):
print w
count = count + 1
if count == 0:
print '<None found>'
else:
print count, 'found'
# Bring in a package to access files over the web
import urllib
# Access the file containing the valid letters
words_url = "http://thinkpython.com/code/words.txt"
words_file = urllib.urlopen(words_url)
# Apply the actual test
find_three_double(words_file)
I didn't read your code carefully at first, turns out it isn't related to read() or readlines() as you are iterating in find_three_doubles() function.
In your three_double() function:
while i < len(s)-1:
if s[i] == s[i+1]:
doubnum += 1
elif doubnum >= 3:
return True
else:
i += 1
return False
There are two problems:
You need to increment i by 1 otherwise the while loop will never stop if there is a "double".
You also need to change elif to if here because otherwise some qualified words will not be selected.
Fixed Code:
def three_double(s):
doubnum = 0
i=0
while i < len(s)-1:
if s[i] == s[i+1]:
doubnum += 1
if doubnum >= 3:
return True
i += 1
return False
# Function to read and apply the three_double test to each string in
# an input file. It counts the number of results.
def find_three_double(fin):
count = 0
for w in fin:
w = w.strip()
if three_double(w):
print w
count = count + 1
if count == 0:
print '<None found>'
else:
print count, 'found'
# Bring in a package to access files over the web
import urllib
# Access the file containing the valid letters
words_url = "http://thinkpython.com/code/words.txt"
words_file = urllib.urlopen(words_url)
# Apply the actual test
find_three_double(words_file)
Results:
aggressiveness
aggressivenesses
allottee
allottees
appellee
appellees
barrenness
barrennesses
bookkeeper
bookkeepers
bookkeeping
bookkeepings
cheerlessness
cheerlessnesses
committee
committees
greenness
greennesses
heedlessness
heedlessnesses
heelless
hyperaggressiveness
hyperaggressivenesses
keelless
keenness
keennesses
masslessness
masslessnesses
possessiveness
possessivenesses
rottenness
rottennesses
sleeplessness
stubbornness
stubbornnesses
successfully
suddenness
suddennesses
sullenness
sullennesses
toolless
wheelless
whippoorwill
whippoorwills
woodenness
woodennesses
46 found
itertools.groupby can greatly simplify your program (= less bugs)
from itertools import groupby
import urllib
def find_three_double(words_file):
for word in words_file:
word = word.strip()
if sum(sum(1 for i in g) == 2 for k,g in groupby(word)) == 3:
print word
# Access the file containing the valid letters
words_url = "http://thinkpython.com/code/words.txt"
words_file = urllib.urlopen(words_url)
# Apply the actual test
find_three_double(words_file)
Explanation:
inside the generator expression we see groupby(word). This scans the word and gathers the double letters together.
sum(1 for i in g) is applied to each group. It is equivalent to finding the length of the group. If the length is 2, then this is a double letter so sum(1 for i in g) == 2 evaluates to True
The outer sum() adds up all the True and False values, True is added as 1 and False is added as 0. If there are exactly 3 True values, the word is printed
while i < len(s)-1:
if s[i] == s[i+1]:
doubnum += 1
elif doubnum >= 3:
return True
else:
i += 1
If your first check (s[i] == s[i+1]) is True, then you'll never increment i so the loop continues forever.