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Python: While reading the file and counting the words in a line, I want to count words coming between " " or ' ' as a single word

I have a file in which I have to count the number of words in each line, but there is a trick, whatever comes in between ' ' or " ", should be counted as a single word.
Example file:
TopLevel
DISPLAY "In TopLevel. Starting to run program"
PERFORM OneLevelDown
DISPLAY "Back in TopLevel."
STOP RUN.
For the above file the count of words in each line has to be as below:
Line: 1 has: 1 words
Line: 2 has: 2 words
Line: 3 has: 2 words
Line: 4 has: 2 words
Line: 5 has: 2 words
But I am getting as below:
Line: 1 has: 1 words
Line: 2 has: 7 words
Line: 3 has: 2 words
Line: 4 has: 4 words
Line: 5 has: 2 words
from os import listdir
from os.path import isfile, join
srch_dir = r'C:\Users\sagrawal\Desktop\File'
onlyfiles = [srch_dir+'\\'+f for f in listdir(srch_dir) if isfile(join(srch_dir, f))]
for i in onlyfiles:
index = 0
with open(i,mode='r') as file:
lst = file.readlines()
for line in lst:
cnt = 0
index += 1
linewrds=line.split()
for lwrd in linewrds:
if lwrd:
cnt = cnt +1
print('Line:',index,'has:',cnt,' words')

If you only have this simple format (no nested quotes or escaped quotes), you could use a simple regex:
lines = '''TopLevel
DISPLAY "In TopLevel. Starting to run program"
PERFORM OneLevelDown
DISPLAY "Back in TopLevel."
STOP RUN.'''.split('\n')
import re
counts = [len(re.findall(r'\'.*?\'|".*?"|\S+', l))
for l in lines]
# [1, 2, 2, 2, 2]
If not, you have to write a parser

If you are looking for a not regex solution, this is my method for you:
# A simple function that will simply count words in each line
def count_words(line):
# Check the next function
line = manage_quotes(line)
words = line.strip()
# In case of several spaces in a row, We need to filter empty words
words = [word for word in words if len(word) > 0]
return len(words)
# This method will manage the quotes
def manage_quotes(line):
# We do not mind the escaped quotes, They are like a simple char
# Also since the changes will be local we can replace words in line
line = line.replace("\\\"", "Q").replace("\\\'", "q")
# As all words between 2 quotes act as one word we can replace them with 1 simple word and we start with `"`
# This loop will help to find all quotes in one line
while True:
i1 = line.find("\"")
if (i1 == -1): # No `"` anymore
break
i2 = line[i1+1:].find("\"") # Search after the previous one
if (i2 == -1): # What shall we do with not paired quotes???
# raise Exception()
break
line = line[:i1-1] + "QUOTE" + line[i2:]
# Now search for `'`
while True:
i1 = line.find("\'")
if (i1 == -1): # No `'` anymore
break
i2 = line[i1+1:].find("\'") # Search after the previous one
if (i2 == -1): # What shall we do with not paired quotes???
# raise Exception()
break
line = line[:i1-1] + "quote" + line[i2:]
return line
This is how this method works, For example, You have a line like this DISPLAY "Part One \'Test1\'" AND 'Part Two \"Test2\"'
At first, we remove escaped quotes:
DISPLAY "Part One qTest1q" AND 'Part Two QTest2Q'
Then we replace double quotations:
DISPLAY QUOTE AND 'Part Two QTest2Q'
Then the other one:
DISPLAY QUOTE AND quote
And now we count this which is 4

You can solve this without regex if you keep some marker if you are inside a quoted area or not.
str.split() - splitts at spaces, returns a list
str.startswith()
str.endswith() - takes a (tuple of) string(s) and returns True if it starts/ends with (any of) it
Code:
# create input file
name = "file.txt"
with open(name, "w") as f:
f.write("""TopLevel
DISPLAY "In TopLevel. Starting to run program"
PERFORM OneLevelDown
DISPLAY "Back in TopLevel."
STOP RUN.""")
# for testing later
expected = [(1,1),(2,2),(3,2),(4,2),(5,2)] # 1 base line/word count
# program that counts words
counted = []
with open(name) as f:
for line_nr, content in enumerate(f,1): # 1 based line count
splt = content.split()
in_quotation = []
line_count = 0
for word in splt:
if not in_quotation:
line_count += 1 # only increments if list empty
if word.startswith(("'",'"')):
in_quotation.append(word[0])
if word.endswith(("'","'")):
in_quotation.pop()
counted.append((line_nr, line_count))
print(expected)
print(counted)
print("Identical: ", all(a == expected[i] for i,a in enumerate(counted)))
Output:
[(1, 1), (2, 2), (3, 2), (4, 2), (5, 2)]
[(1, 1), (2, 2), (3, 2), (4, 2), (5, 2)]
Identical: True
You can tinker with the code - currently it does not well behave if you space out your " - it does not know if something ends or starts and both tests are True.

It seems that the code attached above doesn't care about ' or ".
And here is the definition of str.split in Python here.
If sep is not specified or is None, a different splitting algorithm is applied: runs of consecutive whitespace are regarded as a single separator, and the result will contain no empty strings at the start or end if the string has leading or trailing whitespace. Consequently, splitting an empty string or a string consisting of just whitespace with a None separator returns [].
Code:
input_str = '''TopLevel
DISPLAY "In TopLevel. Starting to run program"
PERFORM OneLevelDown
DISPLAY "Back in TopLevel."
STOP RUN.'''
input_str_list = input_str.split('\n')
print(input_str_list)
def get_trick_word(s: str):
num = 0
quotation_list = []
previous_char = ' '
for c in s:
need_set_previous = False
if c == ' ' and len(quotation_list) == 0 and previous_char != ' ':
num = num + 1
else:
has_quato = len(quotation_list)
if c == '\'' or c == '"':
if len(quotation_list) != 0 and quotation_list[-1] == c:
quotation_list.pop()
else:
quotation_list.append(c)
if has_quato and len(quotation_list) == 0:
num = num + 1
need_set_previous = True
previous_char = c
if need_set_previous:
previous_char = ' '
if previous_char != ' ' and len(quotation_list) == 0:
num = num + 1
return num
result = [get_trick_word(s) for s in input_str_list]
print(result)
And the result is:
# ['TopLevel ', ' DISPLAY "In TopLevel. Starting to run program" ', ' PERFORM OneLevelDown ', ' DISPLAY "Back in TopLevel." ', ' STOP RUN.']
# [1, 2, 2, 2, 2]

How to group consecutive letters in a string in Python?

For example: string = aaaacccc, then I need the output to be 4a4c. Is there a way to do this without using any advanced methods, such as libraries or functions?
Also, if someone knows how to do the reverse: turning "4a4c: into aaaacccc, that would be great to know.

This will do the work in one iteration
Keep two temp variable one for current character, another for count of that character and one variable for the result.
Just iterate through the string and keep increasing the count if it matches with the previous one.
If it doesn't then update the result with count and value of character and update the character and count.
At last add the last character and the count to the result. Done!
input_str = "aaaacccc"
if input_str.isalpha():
current_str = input_str[0]
count = 0
final_string = ""
for i in input_str:
if i==current_str:
count+=1
else:
final_string+=str(count)+current_str
current_str = i
count = 1
final_string+=str(count)+current_str
print (final_string)

Another solution and I included even a patchwork reverse operation like you mentioned in your post. Both run in O(n) and are fairly simple to understand. The encode is basically identical one posted by Akanasha, he was just a bit faster in posting his answer while i was writing the decode().
def encode(x):
if not x.isalpha():
raise ValueError()
output = ""
current_l = x[0]
counter = 0
for pos in x:
if current_l != pos:
output += str(counter) + current_l
counter = 1
current_l = pos
else:
counter += 1
return output + str(counter) + current_l
def decode(x):
output = ""
i = 0
while i < len(x):
if x[i].isnumeric():
n = i + 1
while x[n].isnumeric():
n += 1
output += int(x[i:n])*x[n]
i = n
i += 1
return output
test = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaasasggggggbbbbdd"
test1 = encode(test)
print(test1)
test2 = decode(test1)
print(test2)
print(test == test2)

yes, you do not need any libraries:
list1 = list("aaaacccc")
letters = []
for i in list1:
if i not in letters:
letters.append(i)
string = ""
for i in letters:
string += str(list1.count(i))
string+=str(i)
print(string)
Basically, it loops through the list, finds the unique letters and then prints the count with the letter itself. Reversing would be the same function, just print the amount.

Python XML search for string

I tried to construct my own string.find() method/function in Python. I did this for a computer science class I'm in.
Basically, this program opens a text file, gets a user input on this the text they want to search for in the file, and outputs the line number on which the string resides, or outputs a 'not found' if the string doesn't exist in the file.
However, this takes about 34 seconds to complete 250,000 lines of XML.
Where is the bottleneck in my code? I made this in C# and C++ as well, and this runs in about 0.3 seconds for 4.5 million lines. I also performed this same search using the built-in string.find() from Python, and this takes around 4 seconds for 250,000 lines of XML. So, I'm trying to understand why my version is so slow.
https://github.com/zach323/Python/blob/master/XML_Finder.py
fhand = open('C:\\Users\\User\\filename')
import time
str = input('Enter string you would like to locate: ') #string to be located in file
start = time.time()
delta_time = 0
def find(str):
time.sleep(0.01)
found_str ='' #initialize placeholder for found string
next_index = 0 #index for comparison checking
line_count = 1
for line in fhand: #each line in file
line_count = line_count +1
for letter in line: #each letter in line
if letter == str[next_index]: #compare current letter index to beginning index of string you want to find
found_str += letter #if a match, concatenate to string placeholder
#print(found_str) #print for visualization of inline search per iteration
next_index = next_index + 1
if found_str == str: #if complete match is found, break out of loop.
print('Result is: ', found_str, ' on line %s '%(line_count))
print (line)
return found_str #return string to function caller
break
else:
#if a match was found but the next_index match was False, reset the indexes and try again.
next_index=0 # reset indext back to zero
found_str = '' #reset string back to empty
if found_str == str:
print(line)
if str != "":
result = find(str)
delta_time = time.time() - start
print(result)
print('Seconds elapsed: ', delta_time)
else:
print('sorry, empty string')

Try this:
with open(filename) as f:
for row in f:
if string in row:
print(row)

The following code runs on a text file of size comparable to the size of your file. Your code doesn't run too slowly on my computer.
fhand = open('test3.txt')
import time
string = input('Enter string you would like to locate: ') #string to be located in file
start = time.time()
delta_time = 0
def find(string):
next_index_to_match = 0
sl = len(string)
ct = 0
for line in fhand: #each line in file
ct += 1
for letter in line: #each letter in line
if letter == string[next_index_to_match]: #compare current letter index to beginning index of string you want to find
# print(line)
next_index_to_match += 1
if sl == next_index_to_match: #if complete match is found, break out of loop.
print('Result is: ', string, ' on line %s '%(ct))
print (line)
return True
else:
#if a match was found but the next_index match was False, reset the indexes and try again.
next_index_to_match=0 # reset indext back to zero
return False
if string != "":
find(string)
delta_time = time.time() - start
print('Seconds elapsed: ', delta_time)
else:
print('sorry, empty string')

How to compare sentence character by character in python?

I want to write a code to count number of words in a given sentence by using character comparison and below is the code I have written as I am not allowed to use some fancy utilities like split(), etc. So, could you please guide me where am I making mistakes' I am a novice in python and currently trying to fiigure out how to do charactery by character comparison so as to find out simple counts of words, lines, strings withous using built in utitilites. So, kindly guide me about it.
Input Sentence : I am XYZ
Input_Sentence = raw_input("Enter your sentence: ")
print Input_Sentence
count = 0
i=0
while(Input_Sentence[i] != "\n"):
if(Input_Sentence[i] == ' '):
count=count+1
i+=1
else:
i+=1
print ('Number of Words in a given sentence is :' +str(count))

At first I wouldn't use a while loop in this context. Why not using a for loop?
for char in Input_sentence:
With this you iterate over every letter.
Then you can use the rest of you code and check:
if char == ' ':

# initialize the counter
word_count = 0
last_space_index = 0
# loop through each character in the sentence (assuming Input_Sentence is a string)
for i, x in enumerate(Input_Sentence): # enumerate to get the index of the character
# if a space is found (or newline character for end of sentence)
if x in (' ', '\n'):
word_count += 1 # increment the counter
last_space_index = i # set the index of the last space found
if len(Input_Sentence) > (last_space_index + 1): # check if we are at the end of the sentence (this is in case the word does not end with a newline character or a space)
word_count += 1
# print the total number of words
print 'Number of words:', word_count

The following will avoid errors if there's an space at the beginning or the end of the sentence.
Input_Sentence = raw_input("Enter your sentence: ")
print Input_Sentence
count = 0
sentence_length = len(Input_Sentence)
for i in range(sentence_length):
if Input_Sentence[i] == " ":
if i not in (0, sentence_length - 1):
count += 1
count += 1
print "There are %s words in the sentence \"%s\"." % (count, Input_Sentence)

You may use try-except syntax.
In your code you used while(Input_Sentence[i] != "\n") to find when the sentence comes to an end. If you just print the output at every step before i+ = 1 like this:
...
while(Input_Sentence[i] != "\n"):
...
print i,Input_Sentence[i]
i+=1
else:
print i,Input_Sentence[i],'*'
i+=1
...
you can see for yourself that the output is something like this:
Enter your sentence: Python is good
Python is good
0 P *
1 y *
2 t *
3 h *
4 o *
5 n *
6
7 i *
8 s *
9
10 g *
11 o *
12 o *
13 d *
Traceback (most recent call last):
File "prog8.py", line 19, in <module>
while(Input_Sentence[i] != "\n"):
IndexError: string index out of range
which means that the code that you have written works fine upto the length of the input sentence. After that when i is increased by 1 and it is demanded of the code to check if Input_Sentence[i] == "\n" it gives IndexError. This problem can be overcome by using exception handling tools of Python. Which leaves the option to neglect the block inside try if it is an exception and execute the block within except instead.
Input_Sentence = raw_input("Enter your sentence: ")
print Input_Sentence
count = 0
i=0
try:
while (Input_Sentence[i] != "\n"):
if (Input_Sentence[i] == ' '):
count=count+1
i+=1
else:
i+=1
except:
count = count+1
print ('Number of Words in a given sentence is :' +str(count))

How to count the number of times a digit appears at the beginning of each number in a file? (python)

I am trying to count the number of times 1,2,3,...,9 appear at the beginning of each number in a file. This is how my code goes:
DECIMAL_NUM='123456789'
def main():
#get the file name from the user
file_name=str(input("Enter a file name: "))
#open the file to read
input_file= open(str(file_name),'r')
#reads the first line of the file
line=input_file.readline().strip()
one=0
two=0
three=0
four=0
five=0
six=0
seven=0
eight=0
nine=0
i=0
while line!="":
if line[0]==DECIMAL_NUM[0]:
one+=1
elif line[0]==DECIMAL_NUM[1]:
two+=1
elif line[0]==DECIMAL_NUM[2]:
three+=1
elif line[0]==DECIMAL_NUM[3]:
four+=1
elif line[0]==DECIMAL_NUM[4]:
five+=1
elif line[0]==DECIMAL_NUM[5]:
six+=1
elif line[0]==DECIMAL_NUM[6]:
seven+=1
elif line[0]==DECIMAL_NUM[7]:
eight+=1
elif line[0]==DECIMAL_NUM[8]:
nine+=1
line=input_file.readline().strip()
i+=1
input_file.close()
print(one)
print(two)
main()
I am also counting how many numbers are there in the file, so that I can calculate percentage of appearance of each digit. I think my codes are a little bit wordy and there might be a better way to do it. The input file has the following numbers:
1292
1076
188040
1579
3510
2597
3783
64690
For some reason, I am getting the number of times 1 is appearing as 1, when it should be 5. Could someone please give me some pointers? Thanks

Here is one way of approaching this task:
# Get non-empty lines from input file:
relevant_lines = [line for line in open(file_name).readlines() if line.strip()]
# Count them:
num_lines = len(relevant_lines)
import defaultdict
# If a key does not exist in a defaultdict when adding a value for it,
# it will be added with a default value for the given data type
# (0 in case of int):
d = defaultdict(int)
# Iterate through lines; get first character of line
# and increment counter for this character by one in defaultdict:
for line in relevant_lines:
d[line[0]] += 1
# Print results:
for key, value in d.items():
print(k + ' appears ' + value + ' times in file.')
If you are not allowed to use dicts, here's how to fix your code:
DECIMAL_NUM='123456789'
def main():
# Get file name from user
file_name = input("Enter a file name: ")
# Open the file to read, and get a list of all lines:
lines = open(file_name, 'r').readlines()
one = 0
two = 0
three = 0
four = 0
five = 0
six = 0
seven = 0
eight = 0
nine = 0
for line in lines:
if line.strip(): # Check if line is not empty
if line[0] == DECIMAL_NUM[0]:
one += 1
elif line[0] == DECIMAL_NUM[1]:
two += 1
elif line[0] == DECIMAL_NUM[2]:
three += 1
elif line[0] == DECIMAL_NUM[3]:
four += 1
elif line[0] == DECIMAL_NUM[4]:
five += 1
elif line[0] == DECIMAL_NUM[5]:
six += 1
elif line[0] == DECIMAL_NUM[6]:
seven += 1
elif line[0] == DECIMAL_NUM[7]:
eight += 1
elif line[0] == DECIMAL_NUM[8]:
nine += 1
print(one)
print(two)
main()

You code is fine. It's your data file that's giving you problem. Remove the blank lines and your program should give you the right results.
1292
1076
188040
1579
3510
2597
3783
64690
After you processed the first line, the next line is read. But that's a blank line and your while loop ends.