How can I check if any of the strings in an array exists in another string?
For example:
a = ['a', 'b', 'c']
s = "a123"
if a in s:
print("some of the strings found in s")
else:
print("no strings found in s")
How can I replace the if a in s: line to get the appropriate result?
You can use any:
a_string = "A string is more than its parts!"
matches = ["more", "wholesome", "milk"]
if any([x in a_string for x in matches]):
Similarly to check if all the strings from the list are found, use all instead of any.
any() is by far the best approach if all you want is True or False, but if you want to know specifically which string/strings match, you can use a couple things.
If you want the first match (with False as a default):
match = next((x for x in a if x in str), False)
If you want to get all matches (including duplicates):
matches = [x for x in a if x in str]
If you want to get all non-duplicate matches (disregarding order):
matches = {x for x in a if x in str}
If you want to get all non-duplicate matches in the right order:
matches = []
for x in a:
if x in str and x not in matches:
matches.append(x)
You should be careful if the strings in a or str gets longer. The straightforward solutions take O(S*(A^2)), where S is the length of str and A is the sum of the lenghts of all strings in a. For a faster solution, look at Aho-Corasick algorithm for string matching, which runs in linear time O(S+A).
Just to add some diversity with regex:
import re
if any(re.findall(r'a|b|c', str, re.IGNORECASE)):
print 'possible matches thanks to regex'
else:
print 'no matches'
or if your list is too long - any(re.findall(r'|'.join(a), str, re.IGNORECASE))
A surprisingly fast approach is to use set:
a = ['a', 'b', 'c']
str = "a123"
if set(a) & set(str):
print("some of the strings found in str")
else:
print("no strings found in str")
This works if a does not contain any multiple-character values (in which case use any as listed above). If so, it's simpler to specify a as a string: a = 'abc'.
You need to iterate on the elements of a.
a = ['a', 'b', 'c']
str = "a123"
found_a_string = False
for item in a:
if item in str:
found_a_string = True
if found_a_string:
print "found a match"
else:
print "no match found"
a = ['a', 'b', 'c']
str = "a123"
a_match = [True for match in a if match in str]
if True in a_match:
print "some of the strings found in str"
else:
print "no strings found in str"
jbernadas already mentioned the Aho-Corasick-Algorithm in order to reduce complexity.
Here is one way to use it in Python:
Download aho_corasick.py from here
Put it in the same directory as your main Python file and name it aho_corasick.py
Try the alrorithm with the following code:
from aho_corasick import aho_corasick #(string, keywords)
print(aho_corasick(string, ["keyword1", "keyword2"]))
Note that the search is case-sensitive
The regex module recommended in python docs, supports this
words = {'he', 'or', 'low'}
p = regex.compile(r"\L<name>", name=words)
m = p.findall('helloworld')
print(m)
output:
['he', 'low', 'or']
Some details on implementation: link
A compact way to find multiple strings in another list of strings is to use set.intersection. This executes much faster than list comprehension in large sets or lists.
>>> astring = ['abc','def','ghi','jkl','mno']
>>> bstring = ['def', 'jkl']
>>> a_set = set(astring) # convert list to set
>>> b_set = set(bstring)
>>> matches = a_set.intersection(b_set)
>>> matches
{'def', 'jkl'}
>>> list(matches) # if you want a list instead of a set
['def', 'jkl']
>>>
Just some more info on how to get all list elements availlable in String
a = ['a', 'b', 'c']
str = "a123"
list(filter(lambda x: x in str, a))
It depends on the context
suppose if you want to check single literal like(any single word a,e,w,..etc) in is enough
original_word ="hackerearcth"
for 'h' in original_word:
print("YES")
if you want to check any of the character among the original_word:
make use of
if any(your_required in yourinput for your_required in original_word ):
if you want all the input you want in that original_word,make use of all
simple
original_word = ['h', 'a', 'c', 'k', 'e', 'r', 'e', 'a', 'r', 't', 'h']
yourinput = str(input()).lower()
if all(requested_word in yourinput for requested_word in original_word):
print("yes")
flog = open('test.txt', 'r')
flogLines = flog.readlines()
strlist = ['SUCCESS', 'Done','SUCCESSFUL']
res = False
for line in flogLines:
for fstr in strlist:
if line.find(fstr) != -1:
print('found')
res = True
if res:
print('res true')
else:
print('res false')
I would use this kind of function for speed:
def check_string(string, substring_list):
for substring in substring_list:
if substring in string:
return True
return False
Yet another solution with set. using set.intersection. For a one-liner.
subset = {"some" ,"words"}
text = "some words to be searched here"
if len(subset & set(text.split())) == len(subset):
print("All values present in text")
if subset & set(text.split()):
print("Atleast one values present in text")
If you want exact matches of words then consider word tokenizing the target string. I use the recommended word_tokenize from nltk:
from nltk.tokenize import word_tokenize
Here is the tokenized string from the accepted answer:
a_string = "A string is more than its parts!"
tokens = word_tokenize(a_string)
tokens
Out[46]: ['A', 'string', 'is', 'more', 'than', 'its', 'parts', '!']
The accepted answer gets modified as follows:
matches_1 = ["more", "wholesome", "milk"]
[x in tokens for x in matches_1]
Out[42]: [True, False, False]
As in the accepted answer, the word "more" is still matched. If "mo" becomes a match string, however, the accepted answer still finds a match. That is a behavior I did not want.
matches_2 = ["mo", "wholesome", "milk"]
[x in a_string for x in matches_1]
Out[43]: [True, False, False]
Using word tokenization, "mo" is no longer matched:
[x in tokens for x in matches_2]
Out[44]: [False, False, False]
That is the additional behavior that I wanted. This answer also responds to the duplicate question here.
data = "firstName and favoriteFood"
mandatory_fields = ['firstName', 'lastName', 'age']
# for each
for field in mandatory_fields:
if field not in data:
print("Error, missing req field {0}".format(field));
# still fine, multiple if statements
if ('firstName' not in data or
'lastName' not in data or
'age' not in data):
print("Error, missing a req field");
# not very readable, list comprehension
missing_fields = [x for x in mandatory_fields if x not in data]
if (len(missing_fields)>0):
print("Error, missing fields {0}".format(", ".join(missing_fields)));
I am trying to build a function which returns the most occurred character in a given string and it's working pretty nicely, but how do I return None if the characters have same frequency?
Like for input: 'abac'
Expected output is: 'a'
and for input: 'abab'
Expected output is: None
I have tried using a dictionary to store character frequency and then returning the element with largest value.
def most_occuring_char(str1):
count = {}
max = 0
c = ''
for char in str1:
if char in count.keys():
count[char]+=1
else:
count[char] = 1
for char in str1:
if max < count[char]:
max = count[char]
c = char
return c
I don't know how to check whether the count dictionary elements have same frequency.
You can do that counting with the dict using collections.Counter.
You basically only have to add a check to see if the maximum count is unique (if so, return the char with maximum number of occurrences) or not (if so, return None):
from collections import Counter
def most_occurring_char(string):
counter = Counter(string)
max_char_count = max(counter.values())
is_unique = len([char_count for char_count in counter.values() if char_count == max_char_count]) == 1
if is_unique:
char = [char for char, count in counter.items() if count == max_char_count][0]
return char
return None
# Tests
assert most_occurring_char('abac') == 'a'
assert most_occurring_char('abab') is None
Once you have a dictionary containing the counts of every character (after your first for loop), you can inspect this to determine whether certain counts are the same or not.
If you wish to return None only when all the character counts are the same, you could extract the values (i.e. the character counts) from your dictionary, sort them so they are in numerical order, and compare the first and last values. Since they are sorted, if the first and last values are the same, so are all the intervening values. This can be done using the following code:
count_values = sorted(count.values())
if count_values[0] == count_values[-1]: return None
If you wish to return None whenever there is no single most frequent character, you could instead compare the last value of the sorted list to the second last. If these are equal, there are two or more characters that occur most frequently. The code for this is very similar to the code above.
count_values = sorted(count.values())
if count_values[-1] == count_values[-2]: return None
Another possibility:
def most_occuring_char(s):
from collections import Counter
d = Counter(s)
k = sorted(d, key=lambda x:d[x], reverse=True)
if len(k) == 1: return k[0]
return None if len(k) == 0 or d[k[0]] == d[k[1]] else k[0]
#Test
print(most_occuring_char('abac')) #a
print(most_occuring_char('abab')) #None (same frequencies)
print(most_occuring_char('x')) #x
print(most_occuring_char('abcccba')) #c
print(most_occuring_char('')) #None (empty string)
I have a simple for loop iterating through a user input string. If the it encounters a certain value, I want to be able to look ahead at the next value in order to determine what to do next. What is the best way to accomplish this? For example in my code below, if it encounters a "1" in the string, it should look at the next character in the string. If it is also a 1, print 1. Otherwise do nothing.
UserInput = input("Enter calculator expression:")
for x in UserInput:
...
...
if x == 1:
# if the next value in the string is 1:
# print 1
# else:
# do nothing
Use range and len in the loop to get position address
UserInput = input("Enter calculator expression")
for i in range(len(UserInput)-1):
if x[i]=='1':
if x[i+1]=='1':
print "something"
else:
pass
You can zip a string with itself, but offset with a slice.
>>> from itertools import zip_longest
>>> the_input = 'foobar' # example
>>> the_input[1:] # slice off the first character
'oobar'
>>> for c, lookahead in zip_longest(the_input, the_input[1:]):
... print(c, lookahead)
...
f o
o o
o b
b a
a r
r None
This is more Pythonic than using an index like you would in a language like C.
Zipping pairs things elementwise, like the teeth of a zipper.
You can user enumerate() function.
something like this,
#!/user/bin/python3
for i,x in enumerate(UserInput):
if x=='1':
if UserInput[i+1] == '1':
print('1')
Firstly, python convention is snake-case (user_input instead)
code below is my solution
from itertools import islice
def window(seq, n=2):
it = iter(seq)
result = tuple(islice(it, n))
if len(result) == n:
yield result
for elem in it:
result = result[1:] + (elem,)
yield result
pair = list(window(user_input))
comparison_list = ["1"]
result = [x for x,y in pair if x in comparison_list and x == y]
You don't need to look ahead. At least it is not what we usually do when we tackle this kind of problem. Instead, you should look backward - by storing the information. The following example stores the information in a variable state and you can decide the next action in the subsequent iteration.
UserInput = input("Enter calculator expression:")
state = '0'
for x in UserInput:
if x == '1':
if state == '1':
print('1')
else:
state = '1'
else:
state = '0'
A different (probably more pythonic) way of doing this with the zip function:
for char, next_char in zip(UserInput, UserInput[1:]):
if char == "1" and char == next_char:
print(1)
prev_char = None
for char in input("Enter calculator expression:"):
pass # Here you can check the current character and the previous character
if 1 == prev_char and 1 == char:
print(1)
prev_char = char
I actually need help evaluating what is going on with the code which I wrote.
It is meant to function like this:
input: remove_duple('WubbaLubbaDubDub')
output: 'WubaLubaDubDub'
another example:
input: remove_duple('aabbccdd')
output: 'abcd'
I am still a beginner and I would like to know both what is wrong with my code and an easier way to do it. (There are some lines in the code which were part of my efforts to visualize what was happening and debug it)
def remove_duple(string):
to_test = list(string)
print (to_test)
icount = 0
dcount = icount + 1
for char in to_test:
if to_test[icount] == to_test[dcount]:
del to_test[dcount]
print ('duplicate deleted')
print (to_test)
icount += 1
elif to_test[icount] != to_test[dcount]:
print ('no duplicated deleted')
print (to_test)
icount += 1
print ("".join(to_test))
Don't modify a list (e.g. del to_test[dcount]) that you are iterating over. Your iterator will get screwed up. The appropriate way to deal with this would be to create a new list with only the values you want.
A fix for your code could look like:
In []:
def remove_duple(s):
new_list = []
for i in range(len(s)-1): # one less than length to avoid IndexError
if s[i] != s[i+1]:
new_list.append(s[i])
if s: # handle passing in an empty string
new_list.append(s[-1]) # need to add the last character
return "".join(new_list) # return it (print it outside the function)
remove_duple('WubbaLubbaDubDub')
Out[]:
WubaLubaDubDub
As you are looking to step through the string, sliding 2 characters at a time, you can do that simply by ziping the string with itself shifted one, and adding the first character if the 2 characters are not equal, e.g.:
In []:
import itertools as it
def remove_duple(s):
return ''.join(x for x, y in it.zip_longest(s, s[1:]) if x != y)
remove_duple('WubbaLubbaDubDub')
Out[]:
'WubaLubaDubDub'
In []:
remove_duple('aabbccdd')
Out[]:
'abcd'
Note: you need itertools.zip_longest() or you will drop the last character. The default fillvalue of None is fine for a string.
Okay so, my code:
def isPalindrome():
string = requestString("give me a Palendrom!, add spaces between each letter")
list = string.split()
print list
reverseList = list.reverse()
print reverseList
this is unfinished, but the idea is to detect Palindromes, the user is to input a word and, what I want to be able to do is say.
if list = reverseList:
print "yes"
else:
print "no!"
But unfortunately the return from what I have is:
======= Loading Progam =======
>>> isPalindrome()
['r', 'a', 'd', 'a', 'r']
None
>>>
My class mates are taking a different approach to this question, but I have a reputation for 'unique' code, so I was hoping this would work.
My question is
1 is this even possible?
2 is there a better approach to this problem?
Side note, I am very new to this, I am using JES, Jython and this is my first question on stackoverflow, be kind :D
Edit:
def isPalindrome2():
string = requestString("give me a Palindrome, make sure the letters are spaced")
print string
reversedString = string[::-1]
print reversedString
if string == reversedString:
print ("this is a Palindrome")
else:
print ("this is not a Palindrome")
OutPut:
>>> isPalindrome2()
r a d a r
r a d a r
this is a Palindrome
string[::-1]
It should return the string reversed.