CODEWITHSUNDEEP
pythonarraysloops
Mr Keystrokes
Mr Keystrokes

Reputation: 81

How do you check for the presence of a string in a text file based on the elements of an array?

I have an array containing strings. I have a text file. I want to loop through the text file line by line. And check whether each element of my array is present or not. (they must be whole words and not substrings) I am stuck because my script only checks for the presence of the first array element. However, I would like it to return results with each array element and a note as to whether this array element is present in the entire file or not. 

#!/usr/bin/python



with open("/home/all_genera.txt") as file:

    generaA=[]

    for line in file:
        line=line.strip('\n')
        generaA.append(line)


with open("/home/config/config2.cnf") as config_file:
    counter = 0
    for line in config_file:
        line=line.strip('\n')

        for part in line .split():
            if generaA[counter]in part:
                print (generaA[counter], "is -----> PRESENT")
            else:
                continue
    counter += 1

Upvotes: 0

Views: 121

Answers (3)

bruno desthuilliers
bruno desthuilliers

Reputation: 77912

If I understand correctly, you want a sequence of words that are in both files. If yes, set is your friend: 

def parse(f):
    return set(word for line in f for word in line.strip().split())

with open("path/to/genera/file") as f:
    source = parse(f)
with open("path/to/conf/file" as f:
    conf = parse(f)

# elements that are common to both sets
common = conf & source
print(common)

# elements that are in `source` but not in `conf`
print(source - conf)

# elements that are in `conf` but not in `source`
print(conf - source)

So to answer "I would like it to return results with each array element and a note as to whether this array element is present in the entire file or not", you can use either common elements or the source - conf difference to annotate your source list:

# using common elements
common = conf & source
result = [(word, word in common) for word in source]
print(result)

# using difference
diff = source - conf
result = [(word, word not in diff) for word in source]

Both will yeld the same result and since set lookup is O(1) perfs should be similar too, so I suggest the first solution (positive assertions are easier to the brain than negative ones).

You can of course apply further cleaning / normalisation when building the sets, ie if you want case insensitive search:

def parse(f):
    return set(word.lower() for line in f for word in line.strip().split())

Upvotes: 1

Gsk
Gsk

Reputation: 2945

the counter is not increasing because it's outside the for loops.

with open("/home/all_genera.txt") as myfile: # don't use 'file' as variable, is a reserved word! use myfile instead

    generaA=[]

    for line in myfile: # use .readlines() if you want a list of lines!
        generaA.append(line)

# if you just need to know if string are present in your file, you can use .read():
with open("/home/config/config2.cnf") as config_file:
    mytext = config_file.read()
    for mystring in generaA:
        if mystring in mytext:
            print mystring, "is -----> PRESENT"

# if you want to check if your string in line N is present in your file in the same line, you can go with:
with open("/home/config/config2.cnf") as config_file:
    for N, line in enumerate(config):
        if generaA[N] in line:
            print "{0} is -----> PRESENT in line {1}".format(generaA[N], N)

I hope that everything is clear.

This code could be improved in many ways, but i tried to have it as similar as yours so it will be easier to understand

Upvotes: -1

Joran Beasley
Joran Beasley

Reputation: 113988

from collection import Counter
import re

#first normalize the text (lowercase everything and remove puncuation(anything not alphanumeric)
normalized_text = re.sub("[^a-z0-9 ]","",open("some.txt","rb").read().lower())
# note that this normalization is subject to the rules of the language/alphabet/dialect you are using, and english ascii may not cover it

#counter will collect all the words into a dictionary of [word]:count
words = Counter(normalized_text.split())

# create a new set of all the words in both the text and our word_list_array
set(my_word_list_array).intersection(words.keys())

Upvotes: 1

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