Given 2 strings, each containing a DNA sequence, the function returns a bool to show if a contiguous sub-fragment of length 5 or above exists in string1 that can pair w/ a fragment of str2.
Here is what I tried using the functions "complement" and "reverese_complement" that I created but it doesn't give the right result:
def complement5(sequence1,sequence2):
for i in range(len(sequence1) - 4):
substr1 = sequence1[i:i+5]
if complement(substr1) in sequence2 or reverese_complement(substr1) in sequence2:
print(f'{substr1, complement(substr1), reverese_complement(substr1)} in {sequence2}')
return True
else:
return False
Then when I try:
complement5('CAATTCC','CTTAAGG')
it gives False instead of True
I personally would try to identify the whole complement first, then try to use the python function find(). See the example below
s = "This be a string"
if s.find("is") == -1:
print("No 'is' here!")
else:
print("Found 'is' in the string.")
So... for your genetics problem here:
Given str1 = "AGAGAG..."
Given str2 = "TCTCTC..."
Iterate through str1 sub-chunks (of length 5?) with a for loop and identify possible complements.
str1-complement-chunk = "TCTCTC" #use a method here. Obviously i'm simplifiying.
if (str2.find(str1-complement-chunk ) == -1):
print("dang couldn't find it")
else:
print('There exists a complement at <index>')
Additionally, you may reverse the chunk if you need to do so. I believe it can be accessed with str1[:-1], or something like that idk tho.
I'm trying to write a while loop that goes through a certain list looking for certain substrings. It will find words with those substrings and print out those strings.
Here is a example that works perfectly but for only one word:
lista = ['applepie','appleseed','bananacake','bananabread']
i = 0
z = len(lista)
while i < z:
if ('pie' in lista[0+i]) == True:
print(lista[0+i])
break
i = i + 1
else:
print('Not There Yet')
This prints out applepie (which is what is desired!).
How do I go about fixing this while loop to add in multiple constraints?
I'm trying to do this:
lista = ['applepie','appleseed','bananacake','bananabread']
i = 0
z = len(lista)
while i < z:
if ('pie' in lista[0+i]) == True:
print(lista[0+i])
if ('cake' in lista[0+1]) == True:
print(lista[0+i])
i = i + 1
else:
print('Not There Yet')
This prints out:
applepie
Not There Yet
When I want this to print out:
applepie
bananacake
I used multiple 'if' statements, because I know if I want to use an 'elif', it will only run if the first 'if' statement is false.
Any help is appreciated!
You have two issues smallish issues and I think one larger. The first of the two small ones are what Nick and Brenden noted above. The second is your conditional. It should be <= as opposed to the < you used.
The larger seems that you're having a problem conceptualizing the actual workings. For that, let me suggest you step through it here
You can use any() to go through a list of conditions. It will evaluate to True when the condition is first met and False if it never happens. If you combine that with a regular python for loop, it will be nice a succinct:
lista = ['applepie','appleseed','bananacake','bananabread']
words = ['pie', 'cake']
for food in lista:
if any(word in food for word in words):
print(food)
It prints:
applepie
bananacake
You can also so the same thing as a list comprehension to get a list of words that match:
lista = ['applepie','appleseed','bananacake','bananabread']
words = ['pie', 'cake']
found = [food for food in lista if any(word in food for word in words)]
# ['applepie', 'bananacake']
Generally speaking, Python discourages you from using indices in loops unless you really need to. It tends to be error prone and harder to read.
def is_palindrome(x):
for i in range(0,len(x)):
x = x.lower()
if x[i::1] == x[::-1]:
return("%s Is a palindrome" %(x))
else:
return("%s Is not a palindrome" %(x))
print(is_palindrome('racecar'))
Here is my palindrome program I was just wondering is it right? I checked using the words racecar and madam and it says they are palindrome and I tested nonpalindrome words and it worked. Can someone tell me if it is wrong and then correct me on my mistakes or If its right but there is an easier way can you show me?
There is one major mistake there: you use a for loop which is always going to return in its first iteration. The if ... else ... statement always includes a return, therefore i will never reach anything else then 0 and you only check, if x[0::1] == x[::-1]. This might be desired behaviour, then however you should remove the for-loop.
just a advise, you don't need to use the for loop, as palindrome property say, reverse of the word is same as word, so you need to apply that way.
so your code be like this (one of all possible cases):
direct implementation
def is_palindrome(word):
return word==word[::-1]
this function will return True OR False based on if word is palindrome or not
implementation with the for loop ( check that first character is equal to last character , second to the second last character and so on)
def is_palindrome(word):
wlen = len(word)
for index in range(wlen//2):
if not word[i]==word[wlen-i-1]:
return False
return True
implementation
words = ['aa', 'racecar', 'hello']
for word in words:
if is_palindrome(word):
print("{} is palindrome".format(word))
else:
print("{} is not palindrome".format(word))
I'm supposed to write a function to determine if two strings are anagrams or not.
The codes I've written now, doesn't really work well.
For example, one of sample input is
Tom Cruise
So I'm cuter
output should be True, but my code keep says False.
For another example, when the input is
the eyes
they see
My code actually says True which is the right answer.
So I have no idea why my code only works for certain input.
Can anyone help?
def anagram(a, b):
if(sorted(a)==sorted(b)):
return True
else:
return False
You need to remove all the non alphabet characters and then convert all letters to lower case:
import re
regex = re.compile('[^a-zA-Z]')
return sorted(regex.sub('', a).lower()) == sorted(regex.sub('', b).lower())
As I mentioned in the comment section above, you need to remove any symbol such as ' and convert each letter to either uppercase or lowercase to avoid case mismatches. So, your code should look like this:
def anagram(a, b):
newA = ''.join(elem.lower() for elem in a if elem.isalpha())
newB = ''.join(elem.lower() for elem in b if elem.isalpha())
if(sorted(newA)==sorted(newB)):
return True
else:
return False
a = "Tom Cruise"
b = "So I'm cuter"
print(anagram(a,b))
This will give you:
True
Check this one...
s1=input("Enter first string:")
s2=input("Enter second string:")
a=''.join(t for t in s1 if t.isalnum())
b=''.join(t for t in s2 if t.isalnum())
a=a.lower()
b=b.lower()
if(sorted(a)==sorted(b)):
print("The strings are anagrams.")
else:
print("The strings aren't anagrams.")
def anagram(a, b):
a = ''.join(e for e in a if e.isalpha()).lower()
b = ''.join(e for e in b if e.isalpha()).lower()
return sorted(a)==sorted(b)
a = "Tom Cruise"
b = "So I'm cuter"
anagram(a,b)
You got to remove the none alpha characters in the string and convert the two string into the consistent case.
Most of the questions I've found are biased on the fact they're looking for letters in their numbers, whereas I'm looking for numbers in what I'd like to be a numberless string.
I need to enter a string and check to see if it contains any numbers and if it does reject it.
The function isdigit() only returns True if ALL of the characters are numbers. I just want to see if the user has entered a number so a sentence like "I own 1 dog" or something.
Any ideas?
You can use any function, with the str.isdigit function, like this
def has_numbers(inputString):
return any(char.isdigit() for char in inputString)
has_numbers("I own 1 dog")
# True
has_numbers("I own no dog")
# False
Alternatively you can use a Regular Expression, like this
import re
def has_numbers(inputString):
return bool(re.search(r'\d', inputString))
has_numbers("I own 1 dog")
# True
has_numbers("I own no dog")
# False
You can use a combination of any and str.isdigit:
def num_there(s):
return any(i.isdigit() for i in s)
The function will return True if a digit exists in the string, otherwise False.
Demo:
>>> king = 'I shall have 3 cakes'
>>> num_there(king)
True
>>> servant = 'I do not have any cakes'
>>> num_there(servant)
False
Use the Python method str.isalpha(). This function returns True if all characters in the string are alphabetic and there is at least one character; returns False otherwise.
Python Docs: https://docs.python.org/3/library/stdtypes.html#str.isalpha
https://docs.python.org/2/library/re.html
You should better use regular expression. It's much faster.
import re
def f1(string):
return any(i.isdigit() for i in string)
def f2(string):
return re.search('\d', string)
# if you compile the regex string first, it's even faster
RE_D = re.compile('\d')
def f3(string):
return RE_D.search(string)
# Output from iPython
# In [18]: %timeit f1('assdfgag123')
# 1000000 loops, best of 3: 1.18 µs per loop
# In [19]: %timeit f2('assdfgag123')
# 1000000 loops, best of 3: 923 ns per loop
# In [20]: %timeit f3('assdfgag123')
# 1000000 loops, best of 3: 384 ns per loop
You could apply the function isdigit() on every character in the String. Or you could use regular expressions.
Also I found How do I find one number in a string in Python? with very suitable ways to return numbers. The solution below is from the answer in that question.
number = re.search(r'\d+', yourString).group()
Alternatively:
number = filter(str.isdigit, yourString)
For further Information take a look at the regex docu: http://docs.python.org/2/library/re.html
Edit: This Returns the actual numbers, not a boolean value, so the answers above are more correct for your case
The first method will return the first digit and subsequent consecutive digits. Thus 1.56 will be returned as 1. 10,000 will be returned as 10. 0207-100-1000 will be returned as 0207.
The second method does not work.
To extract all digits, dots and commas, and not lose non-consecutive digits, use:
re.sub('[^\d.,]' , '', yourString)
I'm surprised that no-one mentionned this combination of any and map:
def contains_digit(s):
isdigit = str.isdigit
return any(map(isdigit,s))
in python 3 it's probably the fastest there (except maybe for regexes) is because it doesn't contain any loop (and aliasing the function avoids looking it up in str).
Don't use that in python 2 as map returns a list, which breaks any short-circuiting
You can accomplish this as follows:
if a_string.isdigit():
do_this()
else:
do_that()
https://docs.python.org/2/library/stdtypes.html#str.isdigit
Using .isdigit() also means not having to resort to exception handling (try/except) in cases where you need to use list comprehension (try/except is not possible inside a list comprehension).
You can use NLTK method for it.
This will find both '1' and 'One' in the text:
import nltk
def existence_of_numeric_data(text):
text=nltk.word_tokenize(text)
pos = nltk.pos_tag(text)
count = 0
for i in range(len(pos)):
word , pos_tag = pos[i]
if pos_tag == 'CD':
return True
return False
existence_of_numeric_data('We are going out. Just five you and me.')
You can use range with count to check how many times a number appears in the string by checking it against the range:
def count_digit(a):
sum = 0
for i in range(10):
sum += a.count(str(i))
return sum
ans = count_digit("apple3rh5")
print(ans)
#This print 2
import string
import random
n = 10
p = ''
while (string.ascii_uppercase not in p) and (string.ascii_lowercase not in p) and (string.digits not in p):
for _ in range(n):
state = random.randint(0, 2)
if state == 0:
p = p + chr(random.randint(97, 122))
elif state == 1:
p = p + chr(random.randint(65, 90))
else:
p = p + str(random.randint(0, 9))
break
print(p)
This code generates a sequence with size n which at least contain an uppercase, lowercase, and a digit. By using the while loop, we have guaranteed this event.
any and ord can be combined to serve the purpose as shown below.
>>> def hasDigits(s):
... return any( 48 <= ord(char) <= 57 for char in s)
...
>>> hasDigits('as1')
True
>>> hasDigits('as')
False
>>> hasDigits('as9')
True
>>> hasDigits('as_')
False
>>> hasDigits('1as')
True
>>>
A couple of points about this implementation.
any is better because it works like short circuit expression in C Language and will return result as soon as it can be determined i.e. in case of string 'a1bbbbbbc' 'b's and 'c's won't even be compared.
ord is better because it provides more flexibility like check numbers only between '0' and '5' or any other range. For example if you were to write a validator for Hexadecimal representation of numbers you would want string to have alphabets in the range 'A' to 'F' only.
What about this one?
import string
def containsNumber(line):
res = False
try:
for val in line.split():
if (float(val.strip(string.punctuation))):
res = True
break
except ValueError:
pass
return res
containsNumber('234.12 a22') # returns True
containsNumber('234.12L a22') # returns False
containsNumber('234.12, a22') # returns True
I'll make the #zyxue answer a bit more explicit:
RE_D = re.compile('\d')
def has_digits(string):
res = RE_D.search(string)
return res is not None
has_digits('asdf1')
Out: True
has_digits('asdf')
Out: False
which is the solution with the fastest benchmark from the solutions that #zyxue proposed on the answer.
Also, you could use regex findall. It's a more general solution since it adds more control over the length of the number. It could be helpful in cases where you require a number with minimal length.
s = '67389kjsdk'
contains_digit = len(re.findall('\d+', s)) > 0
Simpler way to solve is as
s = '1dfss3sw235fsf7s'
count = 0
temp = list(s)
for item in temp:
if(item.isdigit()):
count = count + 1
else:
pass
print count
alp_num = [x for x in string.split() if x.isalnum() and re.search(r'\d',x) and
re.search(r'[a-z]',x)]
print(alp_num)
This returns all the string that has both alphabets and numbers in it. isalpha() returns the string with all digits or all characters.
This too will work.
if any(i.isdigit() for i in s):
print("True")
You can also use set.intersection
It is quite fast, better than regex for small strings.
def contains_number(string):
return True if set(string).intersection('0123456789') else False
An iterator approach. It consumes all characters unless a digit is met. The second argument of next fix the default value to return when the iterator is "empty". In this case it set to False but also '' works since it is casted to a boolean value in the if.
def has_digit(string):
str_iter = iter(string)
while True:
char = next(str_iter, False)
# check if iterator is empty
if char:
if char.isdigit():
return True
else:
return False
or by looking only at the 1st term of a generator comprehension
def has_digit(string):
return next((True for char in string if char.isdigit()), False)
I'm surprised nobody has used the python operator in. Using this would work as follows:
foo = '1dfss3sw235fsf7s'
bar = 'lorem ipsum sit dolor amet'
def contains_number(string):
for i in range(10):
if str(i) in list(string):
return True
return False
print(contains_number(foo)) #True
print(contains_number(bar)) #False
Or we could use the function isdigit():
foo = '1dfss3sw235fsf7s'
bar = 'lorem ipsum sit dolor amet'
def contains_number(string):
for i in list(string):
if i.isdigit():
return True
return False
print(contains_number(foo)) #True
print(contains_number(bar)) #False
These functions basically just convert s into a list, and check whether the list contains a digit. If it does, it returns True, if not, it returns False.