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)));
(Im using python on Jupiter Notebook 5.7.8)
I have a project in which are 3 lists, and a list(list_of_lists) that refer to those 3.
I want my program to receive an input, compare this input to the content of my "list_of_lists" and if find a match I want to store the match in another variable for later use.
Im just learning, so here is the code I wrote:
first = ["item1", "item2","item3"]
second = ["item4","item5","item6"]
list1 = [first,second]
list2 = ["asd","asd","asd"]
list_of_lists = [list1,list2]
x = input("Which list are you going to use?: ")
for item in list_of_lists:
if item == x:
match = item
print(match)
print('There was a match')
else:
print('didnt match')
I expect a match but it always output "the didnt match",
I assume it fail to compare the contect of the input with the list inside the list_of lists. The question is also why and how to do it properly(if possible), thanks.
input in python3 returns a string. if you want to convert it into a list, use ast.literal_eval or json.loads or your own parsing method.
list_str = input("Which list are you going to use?: ")
user_list = ast.literal_eval(list_str)
assert isinstance(user_list, list)
...
# do your thing...
So here i tried this code, and it does what i desire, I dont know if its too rudimentary and if there is another way to achieve this.
Here I use a second list to catch the moment when there is a match, after I give to that list the value of my true list and from there print it to be used.
I was wondering if there is a way to take out of the ressults the symbols "[]" and the quotes '', so I can have a clean text format, thanks for the help
first = ["item1", "item2","item3"]
second = ["item4","item5","item6"]
list1 = [first,second]
list2 = ["asd","asd","asd"]
list3 = ["qwe","qwe","qwe"]
list_of_lists = [list1,list2,list3]
reference_list = ["list1","list2","list3"]
count = -1
x = input('Which list are you going to use? ')
for item in reference_list:
count += 1
if x == item:
reference_list = list_of_lists
print(reference_list[count])
Write a function called stop_at_z that iterates through a list of strings. Using a while loop, append each string to a new list until the string that appears is “z”. The function should return the new list.
def stop_at_z(str):
d = 0
x=[]
str1 = list(str)
while True :
if str1[d] != 'Z' :
x.append(str1[d])
d+=1
if str1[d] == 'Z' :
break
return x
Using a while loop, append each string to a new list until the string that appears is “z”. The function should return the new list.
You're getting this error because d keeps increasing infinitely if there is no uppercase 'Z' in the string. Instead, you should only stay in the while loop while the full length of the input string has not been reached:
def stop_at_z(inputstr):
d = 0
x=[]
str1 = list(inputstr)
while d<len(inputstr) :
if str1[d] == 'z' :
break
else:
x.append(str1[d])
d+=1
return x
Note that you can achieve the same thing using takewhile() from the itertools module:
from itertools import takewhile
def stop_at_z(inputstr):
return list(takewhile(lambda i: i != 'z', inputstr))
print(stop_at_z("hello wzrld"))
Output:
['h', 'e', 'l', 'l', 'o', ' ', 'w']
Is the the way you are doing it, searching for “z” is case-sensitive, try something like:
If str1[d].strip().lower() == “z”
It strips off leading and trailing white space and then converts the str1 element to lower case (both of these simply return the modified string, so the original is unchanged) and compares it to a lower case z
What if the string 'z' is never in the list?
Then it keeps on increasing the index and eventually runs into an error.
Just restricting the loop to the length of the list should help.
def stop_at_z(str):
d = 0
x=[]
str1 = list(str)
for d in range(0,len(str1)) :
print(d)
if str1[d] != 'Z' :
x.append(str1[d])
else:
break
return x
Basically, we needed to have a list that could have all the characters until we get "z". One way we could do that is we first convert the string into a list and iterate that list and add every character to a new list ls until we get "z". But the problem is we may get a string that doesn't have "z" so we need to iterate till the length of that list. I hope it is clear.
def stop_at_z(s):
ls = []
idx = 0
x = list(s)
while idx<len(x):
if x[idx]=="z":
break
ls.append(x[idx])
idx+=1
return ls
It's my first time posting here, but I use this while loop:
def stop_at_z(input_list):
print (input_list)
output_list=[]
index=0
while index< len(input_list):
if input_list[index] != "z":
output_list.append(input_list[index])
index+=1
else:
break
return output_list
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.
I wrote the simple amateurish code:
#!/usr/bin/env python2
# this code reads a string from a file
with open('text1.txt', 'r') as myfile:
data=myfile.read()
c = data[3]
if c == ["a"]:
print c
else:
print (c, 'is not a')
As I wrote the string in the text1.txt file, I knew for a fact that the character at position [3] is indeed ['a']. However, the code returns the impertinent answer:
('a', 'is not a')
I tried many things: converting the string into a list, using "is" instead of "==", adding or not adding citation marks to "a", but the result is always the same: the code fails to establish identity. What is going on here?
Any help is much appreciated.
You are comparing against a list in the if statement. It should be:
if c == "a":
print c
else:
print (c, 'is not a')