This question already has answers here:
Understanding slicing
(38 answers)
Closed 7 years ago.
I'm a new programmer and I'm having a lot of trouble understanding for loops and while loops. In what situations would I know to use a for loop and in what situations would I know to use a while loop?
Also, could you explain to me what these 2 codes mean? I have a a lot of confusion.
1 function:
def every_nth_character(s, n):
""" (str, int) -> str
Precondition: n > 0
Return a string that contains every nth character from s, starting at index 0.
>>> every_nth_character('Computer Science', 3)
'CpeSee'
"""
result = ''
i = 0
while i < len(s):
result = result + s[i]
i = i + n
return result
****What does s[i] mean?****
2nd function:
def find_letter_n_times(s, letter, n):
""" (str, str, int) -> str
Precondition: letter occurs at least n times in s
Return the smallest substring of s starting from index 0 that contains
n occurrences of letter.
>>> find_letter_n_times('Computer Science', 'e', 2)
'Computer Scie'
"""
i = 0
count = 0
while count < n:
if s[i] == letter:
count = count + 1
i = i + 1
return s[:i]
what does s[i] and s[:i] mean??
S is a list of characters 'Computer Science'["C","o","m","p"...], and i is the indexposition for each item/character in the list S, so in your case you've stated that your loop counts each third(3) item in S as long as there are items in S, that is, S[i] = [C], S[i]=[p], S=[e], S[i]= C, S[i]=p, where i is each third element in S.
In the second case you've defined i as a variable with value 0, after each loop i increases with +1, i = i + 1, and [:i] means return elements in S up to the latest looped slice, for instance "Computer Scie" + one additional loop would give you "Computer Scien" (i = 9 (the current range of S/number looped characters in S) -> i+1 (increased by +1) -> i=10 (i = 10, S[i]=10 means the first 10 indexpositions/charachters in S]
Your first question about differencies in while and for loops is completely answered here.
Strings and indexing:
Variable s holds a string value. As you may have noticed, it has been submitted as an argument for every_nth_character(s, n) function.
Now every letter in a string is in some position and that position is called index. indexing starts from 0. So if we have a string s containing value 'foo123', it's first character s[0] is 'f' and the last character s[5] = 3.
String can be cutted and sliced using ':' in the index field. Referring to the previous example, we have determined string s. Now you can take only first three characters of that string by using s[:3] and get 'foo' as a result. Read more about it here
Loops:
While and for loops start over and over again until they reach the limit you have determined.
For example:
x = 0
while x < 5:
print x
x = x + 1
prints numbers from 0 to 4. Variable x increases +1 at every single run and the loop ends when x reaches value 5.
Get familiar with Python documentation page, it will help you a lot in future especially in basic things. Google search: Python (your-python-version) documentation
Related
I'm working through the Codility problems and I have gotten the first one almost correct. The task is to write a function which returns the longest binary gap (a sequence of 0s in between 1s) in a binary number. I have gotten all of the test numbers correct apart from 9, which should be 2 (its binary representation is 1001) but which my function returns as 0. I can't seem to figure out why.
My function is as follows:
def Solution(N):
x = bin(N)[2:]
x_string = str(x)
y = (len(x_string))
count = 0
max = 0
for index, item in enumerate(x_string):
if item == "1":
count = 0
elif item == "0" and x_string[index + 1:y-1] != "0"*(y -1 - (index + 1)):
count = count + 1
if count > max:
max = count
print(max)
The complicated indexing and second condition in the elif statement is so that when a 0 is not contained between two 1s then it isn't recognised as the beginning of a binary gap e.g. when the for loop looks at the second character in bin(16) = 10000, it doesn't set count to 1 because all of the remaining characters in the string are also zero.
Simple solution
x_string[index + 1:y-1] != "0"
this bit wants to take a look at the whole string that's left, but the end argument isn't inclusive,it's exluded, so if string length = 4; string[0:4] is the whole string.
source: https://docs.python.org/3/tutorial/introduction.html
-Sam
This question already has answers here:
How to count consecutive repetitions of a substring in a string?
(4 answers)
Closed 1 year ago.
I'm working on a cs50/pset6/dna project. I'm struggling with finding a way to analyze a sequence of strings, and gather the maximum number of times a certain sequence of characters repeats consecutively. Here is an example:
String: JOKHCNHBVDBVDBVDJHGSBVDBVD
Sequence of characters I should look for: BVD
Result: My function should be able to return 3, because in one point the characters BVD repeat three times consecutively, and even though it repeats again two times, I should look for the time that it repeats the most number of times.
It's a bit lame, but one "brute-force"ish way would be to just check for the presence of the longest substring possible. As soon as a substring is found, break out of the loop:
EDIT - Using a function might be more straight forward:
def get_longest_repeating_pattern(string, pattern):
if not pattern:
return ""
for i in range(len(string)//len(pattern), 0, -1):
current_pattern = pattern * i
if current_pattern in string:
return current_pattern
return ""
string = "JOKHCNHBVDBVDBVDJHGSBVDBVD"
pattern = "BVD"
longest_repeating_pattern = get_longest_repeating_pattern(string, pattern)
print(len(longest_repeating_pattern))
EDIT - explanation:
First, just a simple for-loop that starts at a larger number and goes down to a smaller number. For example, we start at 5 and go down to 0 (but not including 0), with a step size of -1:
>>> for i in range(5, 0, -1):
print(i)
5
4
3
2
1
>>>
if string = "JOKHCNHBVDBVDBVDJHGSBVDBVD", then len(string) would be 26, if pattern = "BVD", then len(pattern) is 3.
Back to my original code:
for i in range(len(string)//len(pattern), 0, -1):
Plugging in the numbers:
for i in range(26//3, 0, -1):
26//3 is an integer division which yields 8, so this becomes:
for i in range(8, 0, -1):
So, it's a for-loop that goes from 8 to 1 (remember, it doesn't go down to 0). i takes on the new value for each iteration, first 8 , then 7, etc.
In Python, you can "multiply" strings, like so:
>>> pattern = "BVD"
>>> pattern * 1
'BVD'
>>> pattern * 2
'BVDBVD'
>>> pattern * 3
'BVDBVDBVD'
>>>
A slightly less bruteforcey solution:
string = 'JOKHCNHBVDBVDBVDJHGSBVDBVD'
key = 'BVD'
len_k = len(key)
max_l = 0
passes = 0
curr_len=0
for i in range(len(string) - len_k + 1): # split the string into substrings of same len as key
if passes > 0: # If key was found in previous sequences, pass ()this way, if key is 'BVD', we will ignore 'VD.' and 'D..'
passes-=1
continue
s = string[i:i+len_k]
if s == key:
curr_len+=1
if curr_len > max_l:
max_l=curr_len
passes = len(key)-1
if prev_s == key:
if curr_len > max_l:
max_l=curr_len
else:
curr_len=0
prev_s = s
print(max_l)
You can do that very easily, elegantly and efficiently using a regex.
We look for all sequences of at least one repetition of your search string. Then, we just need to take the maximum length of these sequences, and divide by the length of the search string.
The regex we use is '(:?<your_sequence>)+': at least one repetition (the +) of the group (<your_sequence>). The :? is just here to make the group non capturing, so that findall returns the whole match, and not just the group.
In case there is no match, we use the default parameter of the max function to return 0.
The code is very short, then:
import re
def max_consecutive_repetitions(search, data):
search_re = re.compile('(?:' + search + ')+')
return max((len(seq) for seq in search_re.findall(data)), default=0) // len(search)
Sample run:
print(max_consecutive_repetitions("BVD", "JOKHCNHBVDBVDBVDJHGSBVDBVD"))
# 3
This is my contribution, I'm not a professional but it worked for me (sorry for bad English)
results = {}
# Loops through all the STRs
for i in range(1, len(reader.fieldnames)):
STR = reader.fieldnames[i]
j = 0
s=0
pre_s = 0
# Loops through all the characters in sequence.txt
while j < (len(sequence) - len(STR)):
# checks if the character we are currently looping is the same than the first STR character
if STR[0] == sequence[j]:
# while the sub-string since j to j - STR lenght is the same than STR, I called this a streak
while sequence[j:(j + len(STR))] == STR:
# j skips to the end of sub-string
j += len(STR)
# streaks counter
s += 1
# if s > 0 means that that the whole STR and sequence coincided at least once
if s > 0:
# save the largest streak as pre_s
if s > pre_s:
pre_s = s
# restarts the streak counter to continue exploring the sequence
s=0
j += 1
# assigns pre_s value to a dictionary with the current STR as key
results[STR] = pre_s
print(results)
Similar to this and many other questions, I have many nested loops (up to 16) of the same structure.
Problem: I have 4-letter alphabet and want to get all possible words of length 16. I need to filter those words. These are DNA sequences (hence 4 letter: ATGC), filtering rules are quite simple:
no XXXX substrings (i.e. can't have same letter in a row more than 3 times, ATGCATGGGGCTA is "bad")
specific GC content, that is number of Gs + number of Cs should be in specific range (40-50%). ATATATATATATA and GCGCGCGCGCGC are bad words
itertools.product will work for that, but data structure here gonna be giant (4^16 = 4*10^9 words)
More importantly, if I do use product, then I still have to go through each element to filter it out. Thus I will have 4 billion steps times 2
My current solution is nested for loops
alphabet = ['a','t','g','c']
for p1 in alphabet:
for p2 in alphabet:
for p3 in alphabet:
...skip...
for p16 in alphabet:
word = p1+p2+p3+...+p16
if word_is_good(word):
good_words.append(word)
counter+=1
Is there good pattern to program that without 16 nested loops? Is there a way to parallelize it efficiently (on multi-core or multiple EC2 nodes)
Also with that pattern i can plug word_is_good? check inside middle of the loops: word that starts badly is bad
...skip...
for p3 in alphabet:
word_3 = p1+p2+p3
if not word_is_good(word_3):
break
for p4 in alphabet:
...skip...
from itertools import product, islice
from time import time
length = 16
def generate(start, alphabet):
"""
A recursive generator function which works like itertools.product
but restricts the alphabet as it goes based on the letters accumulated so far.
"""
if len(start) == length:
yield start
return
gcs = start.count('g') + start.count('c')
if gcs >= length * 0.5:
alphabet = 'at'
# consider the maximum number of Gs and Cs we can have in the end
# if we add one more A/T now
elif length - len(start) - 1 + gcs < length * 0.4:
alphabet = 'gc'
for c in alphabet:
if start.endswith(c * 3):
continue
for string in generate(start + c, alphabet):
yield string
def brute_force():
""" Straightforward method for comparison """
lower = length * 0.4
upper = length * 0.5
for s in product('atgc', repeat=length):
if lower <= s.count('g') + s.count('c') <= upper:
s = ''.join(s)
if not ('aaaa' in s or
'tttt' in s or
'cccc' in s or
'gggg' in s):
yield s
def main():
funcs = [
lambda: generate('', 'atgc'),
brute_force
]
# Testing performance
for func in funcs:
# This needs to be big to get an accurate measure,
# otherwise `brute_force` seems slower than it really is.
# This is probably because of how `itertools.product`
# is implemented.
count = 100000000
start = time()
for _ in islice(func(), count):
pass
print(time() - start)
# Testing correctness
global length
length = 12
for x, y in zip(*[func() for func in funcs]):
assert x == y, (x, y)
main()
On my machine, generate was just a bit faster than brute_force, at about 390 seconds vs 425. This was pretty much as fast as I could make them. I think the full thing would take about 2 hours. Of course, actually processing them will take much longer. The problem is that your constraints don't reduce the full set much.
Here's an example of how to use this in parallel across 16 processes:
from multiprocessing.pool import Pool
alpha = 'atgc'
def generate_worker(start):
start = ''.join(start)
for s in generate(start, alpha):
print(s)
Pool(16).map(generate_worker, product(alpha, repeat=2))
Since you happen to have an alphabet of length 4 (or any "power of 2 integer"), the idea of using and integer ID and bit-wise operations comes to mind instead of checking for consecutive characters in strings. We can assign an integer value to each of the characters in alphabet, for simplicity lets use the index corresponding to each letter.
Example:
6546354310 = 33212321033134 = 'aaaddcbcdcbaddbd'
The following function converts from a base 10 integer to a word using alphabet.
def id_to_word(word_id, word_len):
word = ''
while word_id:
rem = word_id & 0x3 # 2 bits pet letter
word = ALPHABET[rem] + word
word_id >>= 2 # Bit shift to the next letter
return '{2:{0}>{1}}'.format(ALPHABET[0], word_len, word)
Now for a function to check whether a word is "good" based on its integer ID. The following method is of a similar format to id_to_word, except a counter is used to keep track of consecutive characters. The function will return False if the maximum number of identical consecutive characters is exceeded, otherwise it returns True.
def check_word(word_id, max_consecutive):
consecutive = 0
previous = None
while word_id:
rem = word_id & 0x3
if rem != previous:
consecutive = 0
consecutive += 1
if consecutive == max_consecutive + 1:
return False
word_id >>= 2
previous = rem
return True
We're effectively thinking of each word as an integer with base 4. If the Alphabet length was not a "power of 2" value, then modulo % alpha_len and integer division // alpha_len could be used in place of & log2(alpha_len) and >> log2(alpha_len) respectively, although it would take much longer.
Finally, finding all the good words for a given word_len. The advantage of using a range of integer values is that you can reduce the number of for-loops in your code from word_len to 2, albeit the outer loop is very large. This may allow for more friendly multiprocessing of your good word finding task. I have also added in a quick calculation to determine the smallest and largest IDs corresponding to good words, which helps significantly narrow down the search for good words
ALPHABET = ('a', 'b', 'c', 'd')
def find_good_words(word_len):
max_consecutive = 3
alpha_len = len(ALPHABET)
# Determine the words corresponding to the smallest and largest ids
smallest_word = '' # aaabaaabaaabaaab
largest_word = '' # dddcdddcdddcdddc
for i in range(word_len):
if (i + 1) % (max_consecutive + 1):
smallest_word = ALPHABET[0] + smallest_word
largest_word = ALPHABET[-1] + largest_word
else:
smallest_word = ALPHABET[1] + smallest_word
largest_word = ALPHABET[-2] + largest_word
# Determine the integer ids of said words
trans_table = str.maketrans({c: str(i) for i, c in enumerate(ALPHABET)})
smallest_id = int(smallest_word.translate(trans_table), alpha_len) # 1077952576
largest_id = int(largest_word.translate(trans_table), alpha_len) # 3217014720
# Find and store the id's of "good" words
counter = 0
goodies = []
for i in range(smallest_id, largest_id + 1):
if check_word(i, max_consecutive):
goodies.append(i)
counter += 1
In this loop I have specifically stored the word's ID as opposed to the actual word itself incase you are going to use the words for further processing. However, if you are just after the words then change the second to last line to read goodies.append(id_to_word(i, word_len)).
NOTE: I receive a MemoryError when attempting to store all good IDs for word_len >= 14. I suggest writing these IDs/words to a file of some sort!
The wording of my question was not polite to the search feature on the site, so I apologize should someone feel this is a duplicate question, but I must ask anyway.
Working in Python 3.6.1, my goal is to find a substring of letters in a string that are in alphabetical order and if that substring of letters is the longest substring of letters in alphabetical order (aa would be considered alphabetical order), then print out the string. I have not gotten entirely close to the solution but I'm making progress; however, this came up and I'm confounded by it being completely new to Python. My question is, why is this valid:
s = 'hijkkpdgijklmnopqqrs'
n = len(s)
i = 0
a = 0
for i in range(n-2):
if s[i] <= s[i+1]:
a = s[i+1]
i = s[i+2]
a = i + a
print(a)
And yet this is not:
s = 'hijkkpdgijklmnopqqrs'
n = len(s)
i = 0
a = 0
b = ''
for i in range(n-2):
if s[i] <= s[i+1]:
b = a + i
a = s[i+1]
i = s[i+2]
a = a + i
print(b)
When the latter code is run, I receive the error:
Traceback (most recent call last):
File "C:\Users\spect\Desktop\newjackcity.py", line 14, in <module>
b = a + i
TypeError: must be str, not int
What I am ultimately trying to do is to 'index in' to the string s, compare the zeroth element to the zeroth+1 element and if s[I] < s[I+1], I want to concatenate the two into my variable a for later printing. Because when I do this, a only prints out two letters in the string. I thought, well initialize the variable first so that a and i can be incremented, then added into a for comparison purposes, and b for printing.
I see now that I'm only going through n-2 iterations (in order to compare the second to last letter to n-1 so the logic is flawed, but I still don't understand the error of why all of a sudden binding a+i to a variable b will produce a str/int error? In my view saying s[i]; etc. is pulling out the elements as a string and this to me is proven in the fact if I run the first set of code, I get the output:
sr
>>>
In both for loops, you use i as the loop variable, so it starts as an int.
In the first version, you reassign i to a string, then add.
for i in range(n-2):
# here i is an int, something between 0 and n-2
if s[i] <= s[i+1]:
a = s[i+1] # a is a string...
i = s[i+2] # now you change i to a string
a = i + a # string + string: OK!
In the second version you try to add i first:
for i in range(n-2):
# here i is an int, something between 0 and n-2
if s[i] <= s[i+1]:
b = a + i # string + int, can't do it...
a = s[i+1]
i = s[i+2]
a = a + i
You will have an easier time debugging your code if you pick more meaningful names.
edit: here is my cleaned up version of your code:
s = 'hijkkpdgijklmnopqqrs'
# i = 0 isn't needed, range starts at 0
# the first character is always 'alphabetical'
alph_substr = s[0]
# range(1,n) is [1,2, ..., n-1]
for i in range(1, len(s)):
if s[i-1] <= s[i]:
alph_substr = alph_substr + s[i]
else:
# we have to start over, since we're not alphabetical anymore
print(alph_substr)
alph_substr = s[i]
print(alph_substr)
I have tried plenty of different methods to achieve this, and I don't know what I'm doing wrong.
reps=[]
len_charac=0
def longest_charac(strng)
for i in range(len(strng)):
if strng[i] == strng[i+1]:
if strng[i] in reps:
reps.append(strng[i])
len_charac=len(reps)
return len_charac
Remember in Python counting loops and indexing strings aren't usually needed. There is also a builtin max function:
def longest(s):
maximum = count = 0
current = ''
for c in s:
if c == current:
count += 1
else:
count = 1
current = c
maximum = max(count,maximum)
return maximum
Output:
>>> longest('')
0
>>> longest('aab')
2
>>> longest('a')
1
>>> longest('abb')
2
>>> longest('aabccdddeffh')
3
>>> longest('aaabcaaddddefgh')
4
Simple solution:
def longest_substring(strng):
len_substring=0
longest=0
for i in range(len(strng)):
if i > 0:
if strng[i] != strng[i-1]:
len_substring = 0
len_substring += 1
if len_substring > longest:
longest = len_substring
return longest
Iterates through the characters in the string and checks against the previous one. If they are different then the count of repeating characters is reset to zero, then the count is incremented. If the current count beats the current record (stored in longest) then it becomes the new longest.
Compare two things and there is one relation between them:
'a' == 'a'
True
Compare three things, and there are two relations:
'a' == 'a' == 'b'
True False
Combine these ideas - repeatedly compare things with the things next to them, and the chain gets shorter each time:
'a' == 'a' == 'b'
True == False
False
It takes one reduction for the 'b' comparison to be False, because there was one 'b'; two reductions for the 'a' comparison to be False because there were two 'a'. Keep repeating until the relations are all all False, and that is how many consecutive equal characters there were.
def f(s):
repetitions = 0
while any(s):
repetitions += 1
s = [ s[i] and s[i] == s[i+1] for i in range(len(s)-1) ]
return repetitions
>>> f('aaabcaaddddefgh')
4
NB. matching characters at the start become True, only care about comparing the Trues with anything, and stop when all the Trues are gone and the list is all Falses.
It can also be squished into a recursive version, passing the depth in as an optional parameter:
def f(s, depth=1):
s = [ s[i] and s[i]==s[i+1] for i in range(len(s)-1) ]
return f(s, depth+1) if any(s) else depth
>>> f('aaabcaaddddefgh')
4
I stumbled on this while trying for something else, but it's quite pleasing.
You can use itertools.groupby to solve this pretty quickly, it will group characters together, and then you can sort the resulting list by length and get the last entry in the list as follows:
from itertools import groupby
print(sorted([list(g) for k, g in groupby('aaabcaaddddefgh')],key=len)[-1])
This should give you:
['d', 'd', 'd', 'd']
This works:
def longestRun(s):
if len(s) == 0: return 0
runs = ''.join('*' if x == y else ' ' for x,y in zip(s,s[1:]))
starStrings = runs.split()
if len(starStrings) == 0: return 1
return 1 + max(len(stars) for stars in starStrings)
Output:
>>> longestRun("aaabcaaddddefgh")
4
First off, Python is not my primary language, but I can still try to help.
1) you look like you are exceeding the bounds of the array. On the last iteration, you check the last character against the character beyond the last character. This normally leads to undefined behavior.
2) you start off with an empty reps[] array and compare every character to see if it's in it. Clearly, that check will fail every time and your append is within that if statement.
def longest_charac(string):
longest = 0
if string:
flag = string[0]
tmp_len = 0
for item in string:
if item == flag:
tmp_len += 1
else:
flag = item
tmp_len = 1
if tmp_len > longest:
longest = tmp_len
return longest
This is my solution. Maybe it will help you.
Just for context, here is a recursive approach that avoids dealing with loops:
def max_rep(prev, text, reps, rep=1):
"""Recursively consume all characters in text and find longest repetition.
Args
prev: string of previous character
text: string of remaining text
reps: list of ints of all reptitions observed
rep: int of current repetition observed
"""
if text == '': return max(reps)
if prev == text[0]:
rep += 1
else:
rep = 1
return max_rep(text[0], text[1:], reps + [rep], rep)
Tests:
>>> max_rep('', 'aaabcaaddddefgh', [])
4
>>> max_rep('', 'aaaaaabcaadddddefggghhhhhhh', [])
7