I have a list of values, some are numeric only, others made up of words, others a mix of the two.
I would like to select only those items composed by the combination number, single letter, number.
let me explain, this is my list of values
l = ['980X2350', 'DO_UN_HPL_Glas_Links', 'DO_UN_HPL_Glas_Rechts',
'930x2115', 'DO_UN_HPL_Links', 'DO_UN_HPL_Rechts', '830X2115',
'Deuropening', 'BF_32_Tourniquets_dubbeledeur_Aluminium']
I'd like to just get back:
['980X2350', '930x2115', '830X2115']
There is no need of importing re for such trivial matter.
Here is an approach that is more efficient than the regex based one:
allowed = '0123456786x'
def filter_str(lst):
output = []
for s in lst:
c = s.lower().strip()
if all(i in allowed for i in c) and c.count('x') == 1:
output.append(s)
return output
If the strings must contain two numeric fields:
allowed = '0123456786x'
def filter_str(lst):
output = []
for s in lst:
c = s.lower().strip()
n = len(c) - 1
if all(i in allowed for i in c) and c.count('x') == 1 and c.index('x') not in (0, n):
output.append(s)
return output
all function short-circuits (i.e. it stops checking as soon as Falsy value is registered), all Python logical operators also short-circuit, for the and operator, the right-hand operand won't be executed if the left-hand operand is Falsy, so my code does look it's longer than the regex based one, but it actually executes faster because regex checks whole strings and does not short-circuit.
Assuming a list of strings as input, you can use a regex and a list comprehension:
l = ['980X2350', 'DO_UN_HPL_Glas_Links', 'DO_UN_HPL_Glas_Rechts',
'930x2115', 'DO_UN_HPL_Links', 'DO_UN_HPL_Rechts', '830X2115',
'Deuropening', 'BF_32_Tourniquets_dubbeledeur_Aluminium']
import re
regex = re.compile('\d+x\d+', flags=re.I)
out = [s for s in l if regex.match(s.strip())]
output:
['980X2350', '930x2115', '830X2115']
Assuming a list of strings :
you can store in a counter the number of letters encountered, if this number is exactly equal to 1 and you have encountered some numbers then you can store it to your output list :
a = ['980X2350', 'DO_UN_HPL_Glas_Links', 'DO_UN_HPL_Glas_Rechts', '930x2115', 'DO_UN_HPL_Links',
'DO_UN_HPL_Rechts', '830X2115', 'Deuropening' ]
alphabet = 'abcdefghijklmnopqrstuvwxyz'
alphabet+= alphabet.upper()
numeric = '0123456789'
numeric_flag = False
output = []
for item in a:
alphabet_count = 0
for char in item:
if char in alphabet:
alphabet_count += 1
if char in numeric:
numeric_flag = True
if alphabet_count == 1 and numeric_flag:
output.append(item)
print(output)
# ['980X2350', '930x2115', '830X2115']
i want to write a function in which it receives a string (s) and a single letter (s). the function needs to return the length of the longest substring of this letter. i dont know why the function i wrote doesn't work
for exmaple: print(count_longest_repetition('eabbaaaacccaaddd', 'a') supposed to return '4'
def count_longest_repetition(s, c):
n= len(s)
lst=[]
length_charachter=0
for i in range(n-1):
if s[i]==c and s[i+1]==c:
if s[i] in lst:
lst.append(s[i])
length_charachter= len(lst)
return length_charachter
Due to the condition if s[i] in lst, nothing will be appended to 'lst' as originally 'lst' is empty and the if condition will never be satisfied. Also, to traverse through the entire string you need to use range(n) as it generates numbers from 0 to n-1. This should work -
def count_longest_repetition(s, c):
n= len(s)
length_charachter=0
max_length = 0
for i in range(n):
if s[i] == c:
length_charachter += 1
else:
length_charachter = 0
max_length = max(max_length, length_charachter)
return max_length
I might suggest using a regex approach here with re.findall:
def count_longest_repetition(s, c):
matches = re.findall(r'' + c + '+', s)
matches = sorted(matches, key=len, reverse=True)
return len(matches[0])
cnt = count_longest_repetition('eabbaaaacccaaddd', 'a')
print(cnt)
This prints: 4
To better explain the above, given the inputs shown, the regex used is a+, that is, find groups of one or more a characters. The sorted list result from the call to re.findall is:
['aaaa', 'aa', 'a']
By sorting descending by string length, we push the longest match to the front of the list. Then, we return this length from the function.
Your function doesn't work because if s[i] in lst: will initially return false and never gets to add anything to to the lst list (so it will remain false throughout the loop).
You should look into regular expressions for this kind of string processing/search:
import re
def count_longest_repetition(s, c):
return max((0,*map(len,re.findall(f"{re.escape(c)}+",s))))
If you're not allowed to use libraries, you could compute repetitions without using a list by adding matches to a counter that you reset on every mismatch:
def count_longest_repetition(s, c):
maxCount = count = 0
for b in s:
count = (count+1)*(b==c)
maxCount = max(count,maxCount)
return maxCount
This can also be done by groupby
from itertools import groupby
def count_longest_repetition(text,let):
return max([len(list(group)) for key, group in groupby(list(text)) if key==let])
count_longest_repetition("eabbaaaacccaaddd",'a')
#returns 4
there was a similar question asked on here but they wanted the remaining letters returned if one word was longer. I'm trying to return the same number of characters for both strings.
Here's my code:
def one_each(st, dum):
total = ""
for i in (st, dm):
total += i
return total
x = one_each("bofa", "BOFAAAA")
print(x)
It doesn't work but I'm trying to get this desired output:
>>>bBoOfFaA
How would I go about solving this? Thank you!
str.join with zip is possible, since zip only iterates pairwise up to the shortest iterable. You can combine with itertools.chain to flatten an iterable of tuples:
from itertools import chain
def one_each(st, dum):
return ''.join(chain.from_iterable(zip(st, dum)))
x = one_each("bofa", "BOFAAAA")
print(x)
bBoOfFaA
I'd probably do something like this
s1 = "abc"
s2 = "123"
ret = "".join(a+b for a,b in zip(s1, s2))
print (ret)
Here's a short way of doing it.
def one_each(short, long):
if len(short) > len(long):
short, long = long, short # Swap if the input is in incorrect order
index = 0
new_string = ""
for character in short:
new_string += character + long[index]
index += 1
return new_string
x = one_each("bofa", "BOFAAAA") # returns bBoOfFaA
print(x)
It might show wrong results when you enter x = one_each("abcdefghij", "ABCD") i.e when the small letters are longer than capital letters, but that can be easily fixed if you alter the case of each letter of the output.
From any *.fasta DNA sequence (only 'ACTG' characters) I must find all sequences which contain at least one repetition of each letter.
For examle from sequence 'AAGTCCTAG' I should be able to find: 'AAGTC', 'AGTC', 'GTCCTA', 'TCCTAG', 'CCTAG' and 'CTAG' (iteration on each letter).
I have no clue how to do that in pyhton 2.7. I was trying with regular expressions but it was not searching for every variants.
How can I achive that?
You could find all substrings of length 4+, and then down select from those to find only the shortest possible combinations that contain one of each letter:
s = 'AAGTCCTAG'
def get_shortest(s):
l, b = len(s), set('ATCG')
options = [s[i:j+1] for i in range(l) for j in range(i,l) if (j+1)-i > 3]
return [i for i in options if len(set(i) & b) == 4 and (set(i) != set(i[:-1]))]
print(get_shortest(s))
Output:
['AAGTC', 'AGTC', 'GTCCTA', 'TCCTAG', 'CCTAG', 'CTAG']
This is another way you can do it. Maybe not as fast and nice as chrisz answere. But maybe a little simpler to read and understand for beginners.
DNA='AAGTCCTAG'
toSave=[]
for i in range(len(DNA)):
letters=['A','G','T','C']
j=i
seq=[]
while len(letters)>0 and j<(len(DNA)):
seq.append(DNA[j])
try:
letters.remove(DNA[j])
except:
pass
j+=1
if len(letters)==0:
toSave.append(seq)
print(toSave)
Since the substring you are looking for may be of about any length, a LIFO queue seems to work. Append each letter at a time, check if there are at least one of each letters. If found return it. Then remove letters at the front and keep checking until no longer valid.
def find_agtc_seq(seq_in):
chars = 'AGTC'
cur_str = []
for ch in seq_in:
cur_str.append(ch)
while all(map(cur_str.count,chars)):
yield("".join(cur_str))
cur_str.pop(0)
seq = 'AAGTCCTAG'
for substr in find_agtc_seq(seq):
print(substr)
That seems to result in the substrings you are looking for:
AAGTC
AGTC
GTCCTA
TCCTAG
CCTAG
CTAG
I really wanted to create a short answer for this, so this is what I came up with!
See code in use here
s = 'AAGTCCTAG'
d = 'ACGT'
c = len(d)
while c <= len(s):
x,c = s[:c],c+1
if all(l in x for l in d):
print(x)
s,c = s[1:],len(d)
It works as follows:
c is set to the length of the string of characters we are ensuring exist in the string (d = ACGT)
The while loop iterates over each possible substring of s such that c is smaller than the length of s.
This works by increasing c by 1 upon each iteration of the while loop.
If every character in our string d (ACGT) exist in the substring, we print the result, reset c to its default value and slice the string by 1 character from the start.
The loop continues until the string s is shorter than d
Result:
AAGTC
AGTC
GTCCTA
TCCTAG
CCTAG
CTAG
To get the output in a list instead (see code in use here):
s = 'AAGTCCTAG'
d = 'ACGT'
c,r = len(d),[]
while c <= len(s):
x,c = s[:c],c+1
if all(l in x for l in d):
r.append(x)
s,c = s[1:],len(d)
print(r)
Result:
['AAGTC', 'AGTC', 'GTCCTA', 'TCCTAG', 'CCTAG', 'CTAG']
If you can break the sequence into a list, e.g. of 5-letter sequences, you could then use this function to find repeated sequences.
from itertools import groupby
import numpy as np
def find_repeats(input_list, n_repeats):
flagged_items = []
for item in input_list:
# Create itertools.groupby object
groups = groupby(str(item))
# Create list of tuples: (digit, number of repeats)
result = [(label, sum(1 for _ in group)) for label, group in groups]
# Extract just number of repeats
char_lens = np.array([x[1] for x in result])
# Append to flagged items
if any(char_lens >= n_repeats):
flagged_items.append(item)
# Return flagged items
return flagged_items
#--------------------------------------
test_list = ['aatcg', 'ctagg', 'catcg']
find_repeats(test_list, n_repeats=2) # Returns ['aatcg', 'ctagg']
I came across a strange Codecademy exercise that required a function that would take a string as input and return it in reverse order. The only problem was you could not use the reversed method or the common answer here on stackoverflow, [::-1].
Obviously in the real world of programming, one would most likely go with the extended slice method, or even using the reversed function but perhaps there is some case where this would not work?
I present a solution below in Q&A style, in case it is helpful for people in the future.
You can also do it with recursion:
def reverse(text):
if len(text) <= 1:
return text
return reverse(text[1:]) + text[0]
And a simple example for the string hello:
reverse(hello)
= reverse(ello) + h # The recursive step
= reverse(llo) + e + h
= reverse(lo) + l + e + h
= reverse(o) + l + l + e + h # Base case
= o + l + l + e + h
= olleh
Just another option:
from collections import deque
def reverse(iterable):
d = deque()
d.extendleft(iterable)
return ''.join(d)
Use reversed range:
def reverse(strs):
for i in xrange(len(strs)-1, -1, -1):
yield strs[i]
...
>>> ''.join(reverse('hello'))
'olleh'
xrange or range with -1 step would return items in reversed order, so we need to iterate from len(string)-1 to -1(exclusive) and fetch items from the string one by one.
>>> list(xrange(len(strs) -1, -1 , -1))
[4, 3, 2, 1, 0] #iterate over these indexes and fetch the items from the string
One-liner:
def reverse(strs):
return ''.join([strs[i] for i in xrange(len(strs)-1, -1, -1)])
...
>>> reverse('hello')
'olleh'
EDIT
Recent activity on this question caused me to look back and change my solution to a quick one-liner using a generator:
rev = ''.join([text[len(text) - count] for count in xrange(1,len(text)+1)])
Although obviously there are some better answers here like a negative step in the range or xrange function. The following is my original solution:
Here is my solution, I'll explain it step by step
def reverse(text):
lst = []
count = 1
for i in range(0,len(text)):
lst.append(text[len(text)-count])
count += 1
lst = ''.join(lst)
return lst
print reverse('hello')
First, we have to pass a parameter to the function, in this case text.
Next, I set an empty list, named lst to use later. (I actually didn't know I'd need the list until I got to the for loop, you'll see why it's necessary in a second.)
The count variable will make sense once I get into the for loop
So let's take a look at a basic version of what we are trying to accomplish:
It makes sense that appending the last character to the list would start the reverse order. For example:
>>lst = []
>>word = 'foo'
>>lst.append(word[2])
>>print lst
['o']
But in order to continue reversing the order, we need to then append word[1] and then word[0]:
>>lst.append(word[2])
>>lst.append(word[1])
>>lst.append(word[0])
>>print lst
['o','o','f']
This is great, we now have a list that has our original word in reverse order and it can be converted back into a string by using .join(). But there's a problem. This works for the word foo, it even works for any word that has a length of 3 characters. But what about a word with 5 characters? Or 10 characters? Now it won't work. What if there was a way we could dynamically change the index we append so that any word will be returned in reverse order?
Enter for loop.
for i in range(0,len(text)):
lst.append(text[len(text)-count])
count += 1
First off, it is necessary to use in range() rather than just in, because we need to iterate through the characters in the word, but we also need to pull the index value of the word so that we change the order.
The first part of the body of our for loop should look familiar. Its very similar to
>>lst.append(word[..index..])
In fact, the base concept of it is exactly the same:
>>lst.append(text[..index..])
So what's all the stuff in the middle doing?
Well, we need to first append the index of the last letter to our list, which is the length of the word, text, -1. From now on we'll refer to it as l(t) -1
>>lst.append(text[len(text)-1])
That alone will always get the last letter of our word, and append it to lst, regardless of the length of the word. But now that we have the last letter, which is l(t) - 1, we need the second to last letter, which is l(t) - 2, and so on, until there are no more characters to append to the list. Remember our count variable from above? That will come in handy. By using a for loop, we can increment the value of count by 1 through each iteration, so that the value we subtract by increases, until the for loop has iterated through the entire word:
>>for i in range(0,len(text)):
..
.. lst.append(text[len(text)-count])
.. count += 1
Now that we have the heart of our function, let's look at what we have so far:
def reverse(text):
lst = []
count = 1
for i in range(0,len(text)):
lst.append(text[len(text)-count])
count += 1
We're almost done! Right now, if we were to call our function with the word 'hello', we would get a list that looks like:
['o','l','l','e','h']
We don't want a list, we want a string. We can use .join for that:
def reverse(text):
lst = []
count = 1
for i in range(0,len(text)):
lst.append(text[len(text)-count])
count += 1
lst = ''.join(lst) # join the letters together without a space
return lst
And that's it. If we call the word 'hello' on reverse(), we'd get this:
>>print reverse('hello')
olleh
Obviously, this is way more code than is necessary in a real life situation. Using the reversed function or extended slice would be the optimal way to accomplish this task, but maybe there is some instance when it would not work, and you would need this. Either way, I figured I'd share it for anyone who would be interested.
If you guys have any other ideas, I'd love to hear them!
Only been coding Python for a few days, but I feel like this was a fairly clean solution. Create an empty list, loop through each letter in the string and append it to the front of the list, return the joined list as a string.
def reverse(text):
backwardstext = []
for letter in text:
backwardstext.insert(0, letter)
return ''.join(backwardstext)
I used this:
def reverse(text):
s=""
l=len(text)
for i in range(l):
s+=text[l-1-i]
return s
Inspired by Jon's answer, how about this one
word = 'hello'
q = deque(word)
''.join(q.pop() for _ in range(len(word)))
This is a very interesting question, I will like to offer a simple one
liner answer:
>>> S='abcdefg'
>>> ''.join(item[1] for item in sorted(enumerate(S), reverse=True))
'gfedcba'
Brief explanation:
enumerate() returns [(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'), (6, 'g')]. The indices and the values.
To reverse the values, just reverse sort it by sorted().
Finally, just put it together back to a str
I created different versions of how to reverse a string in python in my repo:
https://github.com/fedmich/Python-Codes/tree/master/Reverse%20a%20String
You can do it by using list-comprehension or lambda technique:
# Reverse a string without using reverse() function
s = 'Federico';
li = list( s ) #convert string to list
ret = [ li[i-1] for i in xrange(len(li),0,-1) ] #1 liner lambda
print ( "".join( ret ) )
or by doing a backward for loop
# Reverse a string without using reverse() function
s = 'Federico';
r = []
length = len(s)
for i in xrange(length,0,-1):
r.append( s[ i - 1] )
print ( "".join(r) )
reduce(lambda x, y : y + x, "hello world")
A golfed version: r=lambda x:"".join(x[i] for i in range(len(x-1),-1,-1)).
i just solved this in code academy and was checking my answers and ran across this list. so with a very limited understanding of python i just did this and it seamed to work.
def reverse(s):
i = len(s) - 1
sNew = ''
while i >= 0:
sNew = sNew + str(s[i])
i = i -1
return sNew
def reverse(s):
return "".join(s[i] for i in range(len(s)-1, -1, -1))
Blender's answer is lovely, but for a very long string, it will result in a whopping RuntimeError: maximum recursion depth exceeded. One might refactor the same code into a while loop, as one frequently must do with recursion in python. Obviously still bad due to time and memory issues, but at least will not error.
def reverse(text):
answer = ""
while text:
answer = text[0] + answer
text = text[1:]
return answer
Today I was asked this same exercise on pen&paper, so I come up with this function for lists:
def rev(s):
l = len(s)
for i,j in zip(range(l-1, 0, -1), range(l//2)):
s[i], s[j] = s[j], s[i]
return s
which can be used with strings with "".join(rev(list("hello")))
This is a way to do it with a while loop:
def reverse(s):
t = -1
s2 = ''
while abs(t) < len(s) + 1:
s2 = s2 + s[t]
t = t - 1
return s2
I have also just solved the coresponding exercise on codeacademy and wanted to compare my approach to others. I have not found the solution I used so far, so I thought that I sign up here and provide my solution to others. And maybe I get a suggestion or a helpful comment on how to improve the code.
Ok here it goes, I did not use any list to store the string, instead I have just accessed the string index. It took me a bit at first to deal with the len() and index number, but in the end it worked :).
def reverse(x):
reversestring = ""
for n in range(len(str(x))-1,-1, -1):
reversestring += x[n]
return reversestring
I am still wondering if the reversestring = "" could be solved in a more elegant way, or if it is "bad style" even, but i couldn't find an answer so far.
def reverse(text):
a=""
l=len(text)
while(l>=1):
a+=text[l-1]
l-=1
return a
i just concatenated the string a with highest indexes of text (which keeps on decrementing by 1 each loop).
All I did to achieve a reverse string is use the xrange function with the length of the string in a for loop and step back per the following:
myString = "ABC"
for index in xrange(len(myString),-1):
print index
My output is "CBA"
You can simply reverse iterate your string starting from the last character. With python you can use list comprehension to construct the list of characters in reverse order and then join them to get the reversed string in a one-liner:
def reverse(s):
return "".join([s[-i-1] for i in xrange(len(s))])
if you are not allowed to even use negative indexing you should replace s[-i-1] with s[len(s)-i-1]
You've received a lot of alternative answers, but just to add another simple solution -- the first thing that came to mind something like this:
def reverse(text):
reversed_text = ""
for n in range(len(text)):
reversed_text += text[-1 - n]
return reversed_text
It's not as fast as some of the other options people have mentioned(or built in methods), but easy to follow as we're simply using the length of the text string to concatenate one character at a time by slicing from the end toward the front.
def reverseThatString(theString):
reversedString = ""
lenOfString = len(theString)
for i,j in enumerate(theString):
lenOfString -= 1
reversedString += theString[lenOfString]
return reversedString
This is my solution using the for i in range loop:
def reverse(string):
tmp = ""
for i in range(1,len(string)+1):
tmp += string[len(string)-i]
return tmp
It's pretty easy to understand. I start from 1 to avoid index out of bound.
Here's my contribution:
def rev(test):
test = list(test)
i = len(test)-1
result = []
print test
while i >= 0:
result.append(test.pop(i))
i -= 1
return "".join(result)
You can do simply like this
def rev(str):
rev = ""
for i in range(0,len(str)):
rev = rev + str[(len(str)-1)-i]
return rev
Here is one using a list as a stack:
def reverse(s):
rev = [_t for _t in s]
t = ''
while len(rev) != 0:
t+=rev.pop()
return t
Try this simple and elegant code.
my_string= "sentence"
new_str = ""
for i in my_string:
new_str = i + new_str
print(new_str)
you have got enough answer.
Just want to share another way.
you can write a two small function for reverse and compare the function output with the given string
var = ''
def reverse(data):
for i in data:
var = i + var
return var
if not var == data :
print "No palindrome"
else :
print "Palindrome"
Not very clever, but tricky solution
def reverse(t):
for j in range(len(t) // 2):
t = t[:j] + t[- j - 1] + t[j + 1:- j - 1] + t[j] + t[len(t) - j:]
return t
Pointfree:
from functools import partial
from operator import add
flip = lambda f: lambda x, y: f(y, x)
rev = partial(reduce, flip(add))
Test:
>>> rev('hello')
'olleh'