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Python count words of split sentence?

Not sure how to remove the "\n" thing at the end of output
Basically, i have this txt file with sentences such as:
"What does Bessie say I have done?" I asked.
"Jane, I don't like cavillers or questioners; besides, there is something truly forbidding in a child
taking up her elders in that manner.
Be seated somewhere; and until you can speak pleasantly, remain silent."
I managed to split the sentences by semicolon with code:
import re
with open("testing.txt") as file:
read_file = file.readlines()
for i, word in enumerate(read_file):
low = word.lower()
re.split(';',low)
But not sure how to count the words of the split sentences as len() doesn't work:
The output of the sentences:
['"what does bessie say i have done?" i asked.\n']
['"jane, i don\'t like cavillers or questioners', ' besides, there is something truly forbidding in a
child taking up her elders in that manner.\n']
['be seated somewhere', ' and until you can speak pleasantly, remain silent."\n']
The third sentence for example, i am trying to count the 3 words at left and 8 words at right.
Thanks for reading!

The number of words is the number of spaces plus one:
e.g.
Two spaces, three words:
World is wonderful
Code:
import re
import string
lines = []
with open('file.txt', 'r') as f:
lines = f.readlines()
DELIMETER = ';'
word_count = []
for i, sentence in enumerate(lines):
# Remove empty sentance
if not sentence.strip():
continue
# Remove punctuation besides our delimiter ';'
sentence = sentence.translate(str.maketrans('', '', string.punctuation.replace(DELIMETER, '')))
# Split by our delimeter
splitted = re.split(DELIMETER, sentence)
# The number of words is the number of spaces plus one
word_count.append([1 + x.strip().count(' ') for x in splitted])
# [[9], [7, 9], [7], [3, 8]]
print(word_count)

Use str.rstrip('\n') to remove the \n at the end of each sentence.
To count the words in a sentence, you can use len(sentence.split(' '))
To transform a list of sentences into a list of counts, you can use the map function.
So here it is:
import re
with open("testing.txt") as file:
for i, line in enumerate(file.readlines()):
# Ignore empty lines
if line.strip(' ') != '\n':
line = line.lower()
# Split by semicolons
parts = re.split(';', line)
print("SENTENCES:", parts)
counts = list(map(lambda part: len(part.split()), parts))
print("COUNTS:", counts)
Outputs
SENTENCES: ['"what does bessie say i have done?" i asked.']
COUNTS: [9]
SENTENCES: ['"jane, i don\'t like cavillers or questioners', ' besides, there is something truly forbidding in a child ']
COUNTS: [7, 9]
SENTENCES: [' taking up her elders in that manner.']
COUNTS: [7]
SENTENCES: ['be seated somewhere', ' and until you can speak pleasantly, remain silent."']
COUNTS: [3, 8]

You'll need the library nltk
from nltk import sent_tokenize, word_tokenize
mytext = """I have a dog.
The dog is called Bob."""
for sent in sent_tokenize(mytext):
print(len(word_tokenize(sent)))
Output
5
6
Step by step explanation:
for sent in sent_tokenize(mytext):
print('Sentence >>>',sent)
print('List of words >>>',word_tokenize(sent))
print('Count words per sentence>>>', len(word_tokenize(sent)))
Output:
Sentence >>> I have a dog.
List of words >>> ['I', 'have', 'a', 'dog', '.']
Count words per sentence>>> 5
Sentence >>> The dog is called Bob.
List of words >>> ['The', 'dog', 'is', 'called', 'Bob', '.']
Count words per sentence>>> 6

`
import re
sentences = [] #empty list for storing result
with open('testtext.txt') as fileObj:
lines = [line.strip() for line in fileObj if line.strip()] #makin list of lines allready striped from '\n's
for line in lines:
sentences += re.split(';', line) #spliting lines by ';' and store result in sentences
for sentence in sentences:
print(sentence +' ' + str(len(sentence.split()))) #out

try this one:
import re
with open("testing.txt") as file:
read_file = file.readlines()
for i, word in enumerate(read_file):
low = word.lower()
low = low.strip()
low = low.replace('\n', '')
re.split(';',low)

What function should I use to filter a text file for any element in array?

I have a list of words that I'm trying to filter from a text file stored in an array, but I'm not sure what function to use. Here's my code:
words = ["liverpool","arsenal","chelsea","manutd","mancity"]
test = ["LiverpoolFan","ArsenalFan"]
test2 = []
for i in range (len(test)):
test2[i] = test[i].lower()
if *any word in words* in test2[i]:
print("True")
I've used a test array to simulate reading from a text file.
I'm not sure what to use inbetween the two **

You can use builtin any:
>>> test2 = [team for team in words
if any(team.lower() in fanbase.lower() for fanbase in test)]
>>> test2
['liverpool', 'arsenal']
Or any with filter:
>>> def check_match(team):
return any(team.lower() in fanbase.lower() for fanbase in test)
>>> test2 = list(filter(check_match, words))
>>> test2
['liverpool', 'arsenal']
Or you could use str.join with a separator that is not in your words list, such as ',':
>>> all_fans = ','.join(test).lower()
>>> test2 = [team for team in words if team in all_fans]
>>> test2
['liverpool', 'arsenal']

You can do something like this
words = ["liverpool","arsenal","chelsea","manutd","mancity"]
fans = ["LiverpoolFan","ArsenalFan"]
for fan in fans:
for word in words:
if word.lower() in fan.lower():
print(f"{word} in {fan}", True)

brute force approach. Checks for every item if any word from words contains it
matches = []
for item in test:
for word in words:
# word.lower() thanks to #sushanth
if item.lower() in word.lower():
matches.append(item)
print("True")

Here is simple attempt, you can simply remove Fan from the test in order to check if there is team in words match that in test
import re
words = ["liverpool","arsenal","chelsea","manutd","mancity"]
test = ["LiverpoolFan","ArsenalFan"]
purified_test = [re.sub('Fan','', i) for i in test]
print(test)
for i in words:
if i.title() in test:
print('True');
or make it lower() while removing Fan
purified_test = [re.sub('Fan','', i).lower() for i in test]
for i in words:
if i in purified_test:
print('True');
or you can append it to test_2 and get the array like the following
import re
words = ["liverpool","arsenal","chelsea","manutd","mancity"]
test = ["LiverpoolFan","ArsenalFan"]
test2 = []
purified_test = [re.sub('Fan','', i).lower() for i in test]
for i in words:
if i in purified_test:
test2.append(i)
print(test2)
output
['liverpool', 'arsenal']
if all files don't all end in fan you can simple make the character set of all words that ends it like the following
import re
regex = re.compile('(Fan|FC|etc)')

How to print out print out only elements of a list containing certain letters?

I am working on a project and I want to write a code, that would find words containing only certain letters in a sentence and then return them (print them out).
sentence = "I am asking a question on Stack Overflow"
lst = []
# this gives me a list of all words in a sentence
change = sentence.split()
# NOTE: I know this isn't correct syntax, but that's basically what I want to do.
lst.append(only words containing "a")
print(lst)
Now the part I am struggeling with is, how do I append only words containig letter "a" for example?

you can act like this:
words = sentence.split()
lst = [word for word in words if 'a' in word]
print(lst)
# ['am', 'asking', 'a', 'Stack']

Try this! I hope it's well understood!
sentence = "I am asking a question on Stack Overflow"
lst = []
change = sentence.split()
#we are going to check in every word of the sentence, if letter 'a' is in it.
for a in change:
if 'a' in a:
print(a+" has an a! ")
lst.append(a)
print(lst)
This will output:
['am', 'asking', 'a', 'Stack']

How to get my definite loop to print one per line

I'm trying to process a list of words and return a new list
containing only unique word. My definite loop works, however it will only print the words all together, instead of one per line. Can anyone help me out? This is probably a simple question but I am very new to Python. Thank you!
uniqueWords = [ ]
for word in allWords:
if word not in uniqueWords:
uniqueWords.append(word)
else:
uniqueWords.remove(word)
return uniqueWords

You can use str.join:
>>> all_words = ['two', 'two', 'one', 'uno']
>>> print('\n'.join(get_unique_words(all_words)))
one
uno
Or plain for loop:
>>> for word in get_unique_words(all_words):
... print(word)
...
one
uno
However, your method won't work for odd counts:
>>> get_unique_words(['three', 'three', 'three'])
['three']
If your goal is to get all words that appear exactly once, here's a shorter method that works using collections.Counter:
from collections import Counter
def get_unique_words(all_words):
return [word for word, count in Counter(all_words).items() if count == 1]

This code may help, it prints unique words line by line, is what I understood in your question:
allWords = ['hola', 'hello', 'distance', 'hello', 'hola', 'yes']
uniqueWords = [ ]
for word in allWords:
if word not in uniqueWords:
uniqueWords.append(word)
else:
uniqueWords.remove(word)
for i in uniqueWords:
print i

If the order of the words is not important I recommend you to create a set to store the unique words:
uniqueWords = set(allWords)
As you can see running the code below, it can be much faster, but it may depend on the original list of words:
import timeit
setup="""
word_list = [str(x) for x in range(1000, 2000)]
allWords = []
for word in word_list:
allWords.append(word)
allWords.append(word)
"""
smt1 = "unique = set(allWords)"
smt2 = """
uniqueWords = [ ]
for word in allWords:
if word not in uniqueWords:
uniqueWords.append(word)
else:
uniqueWords.remove(word)
"""
print("SET:", timeit.timeit(smt1, setup, number=1000))
print("LOOP:", timeit.timeit(smt2, setup, number=1000))
OUTPUT:
SET: 0.03147706200002176
LOOP: 0.12346845000001849

maybe this fits your idea:
allWords=['hola', 'hello', 'distance', 'hello', 'hola', 'yes']
uniqueWords=dict()
for word in allWords:
if word not in uniqueWords:
uniqueWords.update({word:1})
else:
uniqueWords[word]+=1
for k, v in uniqueWords.items():
if v==1:
print(k)
Prints:
distance
yes

Iterating through a string word by word

I wanted to know how to iterate through a string word by word.
string = "this is a string"
for word in string:
print (word)
The above gives an output:
t
h
i
s
i
s
a
s
t
r
i
n
g
But I am looking for the following output:
this
is
a
string

When you do -
for word in string:
You are not iterating through the words in the string, you are iterating through the characters in the string. To iterate through the words, you would first need to split the string into words , using str.split() , and then iterate through that . Example -
my_string = "this is a string"
for word in my_string.split():
print (word)
Please note, str.split() , without passing any arguments splits by all whitespaces (space, multiple spaces, tab, newlines, etc).

This is one way to do it:
string = "this is a string"
ssplit = string.split()
for word in ssplit:
print (word)
Output:
this
is
a
string

for word in string.split():
print word

Using nltk.
from nltk.tokenize import sent_tokenize, word_tokenize
sentences = sent_tokenize("This is a string.")
words_in_each_sentence = word_tokenize(sentences)
You may use TweetTokenizer for parsing casual text with emoticons and such.

One way to do this is using a dictionary. The problem for the code above is it counts each letter in a string, instead of each word. To solve this problem, you should first turn the string into a list by using the split() method, and then create a variable counts each comma in the list as its own value. The code below returns each time a word appears in a string in the form of a dictionary.
s = input('Enter a string to see if strings are repeated: ')
d = dict()
p = s.split()
word = ','
for word in p:
if word not in d:
d[word] = 1
else:
d[word] += 1
print (d)

s = 'hi how are you'
l = list(map(lambda x: x,s.split()))
print(l)
Output: ['hi', 'how', 'are', 'you']

You can try this method also:
sentence_1 = "This is a string"
list = sentence_1.split()
for i in list:
print (i)