How do I take the first character from each string in a list, join them together, then the second character from each string, join them together, and so on - and eventually create one combined string?
eg. if I have strings like these:
homanif
eiesdnt
ltiwege
lsworar
I want the end result to be helloitsmeiwaswonderingafter
I put together a very hackneyed version of this which does the job but produces an extra line of gibberish. Considering this is prone to index going out of range, I don't think this is a good approach:
final_c = ['homanif', 'eiesdnt', 'ltiwege', 'lsworar']
final_message = ""
current_char = 0
for i in range(len(final_c[1])):
for c in final_c:
final_message += c[current_char]
current_char += 1
final_message += final_c[0][:-1]
print(final_message)
gives me helloitsmeiwaswonderingafterhomani when it should simply stop at helloitsmeiwaswonderingafter.
How do I improve this?
Problems related to iterating in some convoluted order can often be solved elegantly with itertools.
Using zip
You can use zip and itertools.chain together.
from itertools import chain
final_c = ['homanif', 'eiesdnt', 'ltiwege', 'lsworar']
final_message = ''.join(chain.from_iterable(zip(*final_c))) # 'helloitsmeiwaswonderingafter'
In the event you needed the strings in final_c to be of different lengths, you could tweak your code a bit by using itertools.zip_longest.
final_message = ''.join(filter(None, chain.from_iterable(zip_longest(*final_c))))
Using cycle
The fun part with itertools is that it offers plenty of clever short solutions for iterating over objects. Here is another using itertools.cycle.
from itertools import cycle
final_c = ['homanif', 'eiesdnt', 'ltiwege', 'lsworara']
final_message = ''.join(next(w) for w in cycle(iter(w) for w in final_c))
You can use a nested comprehension:
x = ["homanif",
"eiesdnt",
"ltiwege",
"lsworar"]
y = "".join(x[i][j]
for j in range(len(x[0]))
for i in range(len(x)))
or use nested joins and zip
y = "".join("".join(y) for y in zip(*x))
Here is a code that works for me :
final_c = ["homanif", "eiesdnt", "ltiwege", "lsworar"]
final_message = ""
current_char = 0
for i in range(len(final_c[1])):
for c in final_c:
final_message += c[current_char]
current_char += 1
# final_message += final_c[0][:-1]
print(final_message)
I hope it helps
I don't understand what you are expecting with the line
final_message += final_c[0][:-1]
The code works just fine without that. Either remove that line or go with something like list comprehensions :
final_message = "".join(final_c[i][j] for j in range(len(final_c[0])) for i in range(len(final_c)))
This gives the expected output:
helloitsmeiwaswonderingafter
looks like you can have a matrix of form nxm where n is the number of words and m is the number of character in a word (the following code will work if all your words have the same length)
import numpy as np
n = len(final_c) # number of words in your list
m = len(final_c[0]) # number of character in a word
array = np_array(''.join([list(w) for w in ''.join(final_c)])
# reshape the array
matrix = array.reshape(n,1,m )
''.join(matrix.transpose().flatten())
Related
I'm trying to get how many any character repeats in a word. The repetitions must be sequential.
For example, the method with input "loooooveee" should return 6 (4 times 'o', 2 times 'e').
I'm trying to implement string level functions and I can do it this way but, is there an easy way to do this? Regex, or some other sort of things?
Original question: order of repetition does not matter
You can subtract the number of unique letters by the number of total letters. set applied to a string will return a unique collection of letters.
x = "loooooveee"
res = len(x) - len(set(x)) # 6
Or you can use collections.Counter, subtract 1 from each value, then sum:
from collections import Counter
c = Counter("loooooveee")
res = sum(i-1 for i in c.values()) # 6
New question: repetitions must be sequential
You can use itertools.groupby to group sequential identical characters:
from itertools import groupby
g = groupby("aooooaooaoo")
res = sum(sum(1 for _ in j) - 1 for i, j in g) # 5
To avoid the nested sum calls, you can use itertools.islice:
from itertools import groupby, islice
g = groupby("aooooaooaoo")
res = sum(1 for _, j in g for _ in islice(j, 1, None)) # 5
You could use a regular expression if you want:
import re
rx = re.compile(r'(\w)\1+')
repeating = sum(x[1] - x[0] - 1
for m in rx.finditer("loooooveee")
for x in [m.span()])
print(repeating)
This correctly yields 6 and makes use of the .span() function.
The expression is
(\w)\1+
which captures a word character (one of a-zA-Z0-9_) and tries to repeat it as often as possible.
See a demo on regex101.com for the repeating pattern.
If you want to match any character (that is, not only word characters), change your expression to:
(.)\1+
See another demo on regex101.com.
try this:
word=input('something:')
sum = 0
chars=set(list(word)) #get the set of unique characters
for item in chars: #iterate over the set and output the count for each item
if word.count(char)>1:
sum+=word.count(char)
print('{}|{}'.format(item,str(word.count(char)))
print('Total:'+str(sum))
EDIT:
added total count of repetitions
Since it doesn't matter where the repetition is occurring or which characters are being repeated, you can make use of the set data structure provided in Python. It will discard the duplicate occurrences of any character or an object.
Therefore, the solution would look something like this:
def measure_normalized_emphasis(text):
return len(text) - len(set(text))
This will give you the exact result.
Also, make sure to look out for some edge cases, which you should as it is a good practice.
I think your code is comparing the wrong things
You start by finding the last character:
char = text[-1]
Then you compare this to itself:
for i in range(1, len(text)):
if text[-i] == char: #<-- surely this is test[-1] to begin with?
Why not just run through the characters:
def measure_normalized_emphasis(text):
char = text[0]
emphasis_size = 0
for i in range(1, len(text)):
if text[i] == char:
emphasis_size += 1
else:
char = text[i]
return emphasis_size
This seems to work.
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 am trying to create a loop where I can generate string using loop. What I am trying to achieve is that I want to create a small collection of strings starting from 1 character to up to 5 characters.
So, starting from sting 1, I want to go to 55555 but this is number so it seems easy if I just add them, but when it comes to alpha numeric, it gets tricky.
Here is explanation,
I have collection of alpha-numeric chars as string s = "123ABC" and what I want to do is that I want to create all possible 1 character string out of it, so I will have 1,2,3,A,B,C and after that I want to add one more digit in length of string so I can get 11, 12, 13 and so on until I get all possible combination out of it up to CA, CB, CC and I want to get it up to CCCCCC. I am confused in loop because I can get it to generate a temp sting but looping inside to rotate characters is tricky,
this is what I have done so far,
i = 0
strr = "123ABC"
while i < len(strr):
t = strr[0] * (i+1)
for q in range(0, len(t)):
# Here I need help to rotate more
pass
i += 1
Can anyone explain me or point me to resource where I can find solution for it?
You may want to use itertools.permutations function:
import itertools
chars = '123ABC'
for i in xrange(1, len(chars)+1):
print list(itertools.permutations(chars, i))
EDIT:
To get a list of strings, try this:
import itertools
chars = '123ABC'
strings = []
for i in xrange(1, len(chars)+1):
strings.extend(''.join(x) for x in itertools.permutations(chars, i))
This is a nested loop. Different depths of recursion produce all possible combinations.
strr = "123ABC"
def prod(items, level):
if level == 0:
yield []
else:
for first in items:
for rest in prod(items, level-1):
yield [first] + rest
for ln in range(1, len(strr)+1):
print("length:", ln)
for s in prod(strr, ln):
print(''.join(s))
It is also called cartesian product and there is a corresponding function in itertools.
we've started doing Lists in our class and I'm a bit confused thus coming here since previous questions/answers have helped me in the past.
The first question was to sum up all negative numbers in a list, I think I got it right but just want to double check.
import random
def sumNegative(lst):
sum = 0
for e in lst:
if e < 0:
sum = sum + e
return sum
lst = []
for i in range(100):
lst.append(random.randrange(-1000, 1000))
print(sumNegative(lst))
For the 2nd question, I'm a bit stuck on how to write it. The question was:
Count how many words occur in a list up to and including the first occurrence of the word “sap”. I'm assuming it's a random list but wasn't given much info so just going off that.
I know the ending would be similar but no idea how the initial part would be since it's string opposed to numbers.
I wrote a code for a in-class problem which was to count how many odd numbers are on a list(It was random list here, so assuming it's random for that question as well) and got:
import random
def countOdd(lst):
odd = 0
for e in lst:
if e % 2 = 0:
odd = odd + 1
return odd
lst = []
for i in range(100):
lst.append(random.randint(0, 1000))
print(countOdd(lst))
How exactly would I change this to fit the criteria for the 2nd question? I'm just confused on that part. Thanks.
The code to sum -ve numbers looks fine! I might suggest testing it on a list that you can manually check, such as:
print(sumNegative([1, -1, -2]))
The same logic would apply to your random list.
A note about your countOdd function, it appears that you are missing an = (== checks for equality, = is for assignment) and the code seems to count even numbers, not odd. The code should be:
def countOdd(lst):
odd = 0
for e in lst:
if e%2 == 1: # Odd%2 == 1
odd = odd + 1
return odd
As for your second question, you can use a very similar function:
def countWordsBeforeSap(inputList):
numWords = 0
for word in inputList:
if word.lower() != "sap":
numWords = numWords + 1
else:
return numWords
inputList = ["trees", "produce", "sap"]
print(countWordsBeforeSap(inputList))
To explain the above, the countWordsBeforeSap function:
Starts iterating through the words.
If the word is anything other than "sap" it increments the counter and continues
If the word IS "sap" then it returns early from the function
The function could be more general by passing in the word that you wanted to check for:
def countWordsBefore(inputList, wordToCheckFor):
numWords = 0
for word in inputList:
if word.lower() != wordToCheckFor:
numWords = numWords + 1
else:
return numWords
inputList = ["trees", "produce", "sap"]
print(countWordsBeforeSap(inputList, "sap"))
If the words that you are checking come from a single string then you would initially need to split the string into individual words like so:
inputString = "Trees produce sap"
inputList = inputString.split(" ")
Which splits the initial string into words that are separated by spaces.
Hope this helps!
Tom
def count_words(lst, end="sap"):
"""Note that I added an extra input parameter.
This input parameter has a default value of "sap" which is the actual question.
However you can change this input parameter to any other word if you want to by
just doing "count_words(lst, "another_word".
"""
words = []
# First we need to loop through each item in the list.
for item in lst:
# We append the item to our "words" list first thing in this loop,
# as this will make sure we will count up to and INCLUDING.
words.append(item)
# Now check if we have reached the 'end' word.
if item == end:
# Break out of the loop prematurely, as we have reached the end.
break
# Our 'words' list now has all the words up to and including the 'end' variable.
# 'len' will return how many items there are in the list.
return len(words)
lst = ["something", "another", "woo", "sap", "this_wont_be_counted"]
print(count_words(lst))
Hope this helps you understand lists better!
You can make effective use of list/generator comprehensions. Below are fast and memory efficient.
1. Sum of negatives:
print(sum( i<0 for i in lst))
2. Count of words before sap: Like you sample list, it assumes no numbers are there in list.
print(lst.index('sap'))
If it's a random list. Filter strings. Find Index for sap
l = ['a','b',1,2,'sap',3,'d']
l = filter(lambda x: type(x)==str, l)
print(l.index('sap'))
3. Count of odd numbers:
print(sum(i%2 != 0 for i in lst))
In a numerical sequence (e.g. one-dimensional array) I want to find different patterns of numbers and count each finding separately. However, the numbers can occur repeatedly but only the basic pattern is important.
# Example signal (1d array)
a = np.array([1,1,2,2,2,2,1,1,1,2,1,1,2,3,3,3,3,3,2,2,1,1,1])
# Search for these exact following "patterns": [1,2,1], [1,2,3], [3,2,1]
# Count the number of pattern occurrences
# [1,2,1] = 2 (occurs 2 times)
# [1,2,3] = 1
# [3,2,1] = 1
I have come up with the Knuth-Morris-Pratt string matching (http://code.activestate.com/recipes/117214/), which gives me the index of the searched pattern.
for s in KnuthMorrisPratt(list(a), [1,2,1]):
print('s')
The problem is, I don't know how to find the case, where the pattern [1,2,1] "hides" in the sequence [1,2,2,2,1]. I need to find a way to reduce this sequence of repeated numbers in order to get to [1,2,1]. Any ideas?
I don't use NumPy and I am quite new to Python, so there might be a better and more efficient solution.
I would write a function like this:
def dac(data, pattern):
count = 0
for i in range(len(data)-len(pattern)+1):
tmp = data[i:(i+len(pattern))]
if tmp == pattern:
count +=1
return count
If you want to ignore repeated numbers in the middle of your pattern:
def dac(data, pattern):
count = 0
for i in range(len(data)-len(pattern)+1):
tmp = [data[i], data [i+1]]
try:
for j in range(len(data)-i):
print(i, i+j)
if tmp[-1] != data[i+j+1]:
tmp.append(data[i+j+1])
if len(tmp) == len(pattern):
print(tmp)
break
except:
pass
if tmp == pattern:
count +=1
return count
Hope that might help.
Here's a one-liner that will do it
import numpy as np
a = np.array([1,1,2,2,2,2,1,1,1,2,1,1,2,3,3,3,3,3,2,2,1,1,1])
p = np.array([1,2,1])
num = sum(1 for k in
[a[j:j+len(p)] for j in range(len(a) - len(p) + 1)]
if np.array_equal(k, p))
The innermost part is a list comprehension that generates all pieces of the array that are the same length as the pattern. The outer part sums 1 for every element of this list which matches the pattern.
The only way I could think of solving your problem with the
subpatterns matching was to use regex.
The following is a demonstration for findind for example the sequence [1,2,1] in list1:
import re
list1 = [1,1,2,2,2,2,1,1,1,2,1,1,2,3,3,3,3,3,2,2,1,1,1]
str_list = ''.join(str(i) for i in list1)
print re.findall(r'1+2+1', str_list)
This will give you as a result:
>>> print re.findall(r'1+2+1', str_list)
['1122221', '1121']