I'm writing a function that when given a list of lists it can determine if a specific first character is followed by a specific second character (either once or repeated) or is followed by another random string. When another random string is present the function should return False but when no random string is present the function should return True.
Basically if any other character other then . comes after B in its current list or any following lists then it should return False but if only . is present then it should return True.
For example, if the first character was B and the second character was . and the list of lists was [['.','.','B','.'],['.','.','.','.']] then it should return True but if the list of lists was [['a','c','B','r'],['.','s','g','h']] it should return False since a series of random strings follows B.
Any tips or help would be appreciated this is the code I have so far:
def free_of_obstacles(lst):
A = 'B'
B = '.'
for i, v in enumerate(lst):
if A in v:
continue
if B in v:
continue
return True
else:
return False
You could join all the chars in each list and join all the joined strings into a single string and then apply a regex to check if there is a match or not:
>>> lst=[['.','.','B','.'],['.','.','.','.']]
>>> import re
>>> bool(re.search(r'B(\.+)$', ''.join(''.join(i) for i in lst)))
True
>>> lst=[['a','c','B','r'],['.','s','g','h']]
>>> bool(re.search(r'B(\.+)$', ''.join(''.join(i) for i in lst)))
False
>>>
EDIT 1 ----> Above solution as a function returning True or False:
>>> import re
>>> def free_of_obstacles(lst):
... return bool(re.search(r'B(\.+)$', ''.join(''.join(i) for i in lst)))
...
>>> lst=[['a','c','B','r'],['.','s','g','h']]
>>> free_of_obstacles(lst)
False
>>> lst=[['.','.','B','.'],['.','.','.','.']]
>>> free_of_obstacles(lst)
True
Without using any imported modules:
Sample run 1
>>> lst=[['.','.','B','.'],['.','.','.','.']]
>>> newlst=[j for i in lst for j in i]
>>> newlst=newlst[newlst.index('B')+1:]
>>> newlst
['.', '.', '.', '.', '.']
>>> list(map(lambda x:x=='.', newlst))
[True, True, True, True, True]
>>> all(list(map(lambda x:x=='.', newlst)))
True
Sample run 2
>>> lst=[['a','c','B','r'],['.','s','g','h']]
>>> newlst=[j for i in lst for j in i]
>>> newlst=newlst[newlst.index('B')+1:]
>>> newlst
['r', '.', 's', 'g', 'h']
>>> list(map(lambda x:x=='.', newlst))
[False, True, False, False, False]
>>> all(list(map(lambda x:x=='.', newlst)))
False
EDIT 2 ----> Above solution as a function returning True or False:
>>> def free_of_obstacles(lst):
... newlst=[j for i in lst for j in i]
... newlst=newlst[newlst.index('B')+1:]
... return all(list(map(lambda x:x=='.', newlst)))
...
>>> lst=[['.','.','B','.'],['.','.','.','.']]
>>> free_of_obstacles(lst)
True
>>> lst=[['a','c','B','r'],['.','s','g','h']]
>>> free_of_obstacles(lst)
False
>>>
Related
I want to compare 2 lists, say A and B.
A = [1,2,3]
B = [3,1,2]
I want to write a function to check if all the items in A are present in B or not.
i.e. the function should return True for the example above.
Note: The items of my lists are non-hashable objects.
For this, I wrote this code:
for elem in A:
if elem not in B:
return False
return True
The code works fine.
But I want a more efficient and a more pythonic way to do this.
You may use all
all(i in B for i in A)
Example:
>>> A = [1,2,3]
>>> B = [3,1,2]
>>> all(i in B for i in A)
True
>>> B = [3,1,4]
>>> all(i in B for i in A)
False
>>>
You could use Counter, which runs in O(n) time.
from collections import Counter
A = [1,2,3]
B = [3,1,2]
print(Counter(A) == Counter(B)) # Prints True
I think using set is a good solution for this task :
In [2]: A = [1,2,3]
In [3]: B = [3,1,2]
In [5]: set(A)==set(B)
Out[5]: True
This code will have O(n) complexity for more information you can check this link
If you don't want to use set you can try to sort lists :
def cmp(A,B):
A1 = sorted(A)
B1 = sorted(B)
for item in A1:
if binary_search(B1,item)==-1:
return False
for item in B1:
if binary_search(A1,item)==-1:
return False
return True
I create a list of dictionaries like this:
list = []
for i in xrange(4):
list.append({})
list[i]['a'] = False
Now after a while, I want to (using a single line of code) see if any of the 'a' values are True.
I have tried:
anyTrue = True in list # always returns false
anyTrue = True in list[:]['a']
Is there such a convenient way of doing this?
Thanks!
Using any with generator expression:
>>> lst = []
>>>
>>> for i in xrange(4):
... lst.append({})
... lst[i]['a'] = False
...
>>> any(d['a'] for d in lst)
False
>>> lst[1]['a'] = True
>>> any(d['a'] for d in lst)
True
BTW, don't use the list as a variable name. It shadows builtin function/type list.
You can use any and a generator expression:
if any(x['a'] for x in list):
# Do stuff
See a demonstration below:
>>> lst = []
>>> for i in xrange(4):
... lst.append({})
... lst[i]['a'] = False
...
>>> any(x['a'] for x in lst)
False
>>> lst[2]['a'] = True # Set an 'a' value to True
>>> any(x['a'] for x in lst)
True
>>>
Also, you should refrain from naming a variable list. Doing so overshadows the built-in.
let two strings
s='chayote'
d='aceihkjouty'
the characters in string s is present in d Is there any built-in python function to accomplish this ?
Thanks In advance
Using sets:
>>> set("chayote").issubset("aceihkjouty")
True
Or, equivalently:
>>> set("chayote") <= set("aceihkjouty")
True
I believe you are looking for all and a generator expression:
>>> s='chayote'
>>> d='aceihkjouty'
>>> all(x in d for x in s)
True
>>>
The code will return True if all characters in string s can be found in string d.
Also, if string s contains duplicate characters, it would be more efficient to make it a set using set:
>>> s='chayote'
>>> d='aceihkjouty'
>>> all(x in d for x in set(s))
True
>>>
Try this
for i in s:
if i in d:
print i
Could someone please help me strip characters from a string to leave me with just the characters held within '[....]'?
For example:
a = newyork_74[mylocation]
b = # strip the frist characters until you reach the first bracket [
c = [mylocation]
Something like this:
>>> import re
>>> strs = "newyork_74[mylocation]"
>>> re.sub(r'(.*)?(\[)','\g<2>',strs)
'[mylocation]'
Assuming no nested structures, one way would be using itertools.dropwhile,
>>> from itertools import dropwhile
>>> b = ''.join(dropwhile(lambda c: c != '[', a))
>>> b
'[mylocation]'
Another would be to use regexs,
>>> import re
>>> pat = re.compile(r'\[.*\]')
>>> b = pat.search(a).group(0)
>>> b
'[mylocation]'
I need to search a string for a list of several different matches, let's say I have this list:
['this', 'is', 'a', 'regex', 'test']
I want to see if any of those items is within a string, either using regex or any other method in Python.
I tried first just doing string in list, but that proved to be insufficient, so I tried concatenating the conditions in a regex like:
(this|is)(a|regex)(text)
But that tries to match several of the items as if they were concatenated.
You can use the built-in function any():
In [1]: strs="I am a string"
In [2]: lis=['this', 'is', 'a', 'regex', 'test']
In [3]: any(x in strs for x in lis)
Out[3]: True
This will return True for something like "thisisafoobar" as well.
But if you want to match the exact word, then try re.search() or str.split():
In [4]: import re
In [5]: any(re.search(r"\b{0}\b".format(x),strs) for x in lis)
Out[5]: True
In [6]: strs="foo bar"
In [7]: any(re.search(r"\b{0}\b".format(x),strs) for x in lis)
Out[7]: False
Using str.split():
In [12]: strs="I am a string"
In [13]: spl=strs.split() #use set(strs.split()) if the list returned is huge
In [14]: any(x in spl for x in lis)
Out[14]: True
In [15]: strs="Iamastring"
In [16]: spl=strs.split()
In [17]: any(x in spl for x in lis)
Out[17]: False
>>> l = ['this', 'is', 'a', 'regex', 'test']
>>> s = 'this is a test string'
>>> def check(elements, string):
... for element in elements:
... if element in string:
... return True
... return False
...
>>> check(l, s)
True
Apparently this function has better performance than any()
import time
def main():
# Making a huge list
l = ['this', 'is', 'a', 'regex', 'test'] * 10000
s = 'this is a test string'
def check(elements, string):
for element in elements:
if element in string:
return True
return False
def test_a(elements, string):
"""Testing check()"""
start = time.time()
check(elements, string)
end = time.time()
return end - start
def test_b(elements, string):
"""Testing any()"""
start = time.time()
any(element in string for element in elements)
end = time.time()
return end - start
print 'Using check(): %s' % test_a(l, s)
print 'Using any(): %s' % test_b(l, s)
if __name__ == '__main__':
main()
Results:
pearl:~ pato$ python test.py
Using check(): 3.09944152832e-06
Using any(): 5.96046447754e-06
pearl:~ pato$ python test.py
Using check(): 1.90734863281e-06
Using any(): 7.15255737305e-06
pearl:~ pato$ python test.py
Using check(): 2.86102294922e-06
Using any(): 6.91413879395e-06
But if you combine any() with map() in something like any(map(lambda element: element in string, elements)), these are the results:
pearl:~ pato$ python test.py
Using check(): 3.09944152832e-06
Using any(): 0.00903916358948
pearl:~ pato$ python test.py
Using check(): 2.86102294922e-06
Using any(): 0.00799989700317
pearl:~ pato$ python test.py
Using check(): 3.09944152832e-06
Using any(): 0.00829982757568
You could do:
if any(test in your_string for test in tests):
...