Printing character percent in a text file - python

I just wrote a function which prints character percent in a text file. However, I got a problem. My program is counting uppercase characters as a different character and also counting spaces. That's why the result is wrong. How can i fix this?
def count_char(text, char):
count = 0
for character in text:
if character == char:
count += 1
return count
filename = input("Enter the file name: ")
with open(filename) as file:
text = file.read()
for char in "abcdefghijklmnopqrstuvwxyz":
perc = 100 * count_char(text, char) / len(text)
print("{0} - {1}%".format(char, round(perc, 2)))

You should try making the text lower case using text.lower() and then to avoid spaces being counted you should split the string into a list using: text.lower().split(). This should do:
def count_char(text, char):
count = 0
for word in text.lower().split(): # this iterates returning every word in the text
for character in word: # this iterates returning every character in each word
if character == char:
count += 1
return count
filename = input("Enter the file name: ")
with open(filename) as file:
text = file.read()
totalChars = sum([len(i) for i in text.lower().split()]
for char in "abcdefghijklmnopqrstuvwxyz":
perc = 100 * count_char(text, char) / totalChars
print("{0} - {1}%".format(char, round(perc, 2)))
Notice the change in perc definition, sum([len(i) for i in text.lower().split()] returns the number of characters in a list of words, len(text) also counts spaces.

You can use a counter and a generator expression to count all letters like so:
from collections import Counter
with open(fn) as f:
c=Counter(c.lower() for line in f for c in line if c.isalpha())
Explanation of generator expression:
c=Counter(c.lower() for line in f # continued below
^ create a counter
^ ^ each character, make lower case
^ read one line from the file
# continued
for c in line if c.isalpha())
^ one character from each line of the file
^ iterate over line one character at a time
^ only add if a a-zA-Z letter
Then get the total letter counts:
total_letters=float(sum(c.values()))
Then the total percent of any letter is c[letter] / total_letters * 100
Note that the Counter c only has letters -- not spaces. So the calculated percent of each letter is the percent of that letter of all letters.
The advantage here:
You are reading the entire file anyway to get the total count of the character in question and the total of all characters. You might as well just count the frequency of all character as you read them;
You do not need to read the entire file into memory. That is fine for smaller files but not for larger ones;
A Counter will correctly return 0 for letters not in the file;
Idiomatic Python.
So your entire program becomes:
from collections import Counter
with open(fn) as f:
c=Counter(c.lower() for line in f for c in line if c.isalpha())
total_letters=float(sum(c.values()))
for char in "abcdefghijklmnopqrstuvwxyz":
print("{} - {:.2%}".format(char, c[char] / total_letters))

You want to make the text lower case before counting the char:
def count_char(text, char):
count = 0
for character in text.lower():
if character == char:
count += 1
return count

You can use the built in .count function to count the characters after converting everything to lowercase via .lower. Additionally, your current program doesn't work properly as it doesn't exclude spaces and punctuation when calling the len function.
import string
filename = input("Enter the file name: ")
with open(filename) as file:
text = file.read().lower()
chars = {char:text.count(char) for char in string.ascii_lowercase}
allLetters = float(sum(chars.values()))
for char in chars:
print("{} - {}%".format(char, round(chars[char]/allLetters*100, 2)))

Related

Scrabble cheater: scoring wildcard characters to zero in Python

I'm new to python world, and I made a code of scrabble finder with two wildcards (* and ?) in it. When scoring the word, I would like to score wildcard letters to zero, but it looks like it doesn't work. I'm wondering what is missing here.
When you look into the line after "# Add score and valid word to the empty list", I tried to code if a letter in the word is not in the rack, I removed the letter so that I can only score other characters that are not coming from wildcards and matches with the letter in the rack. For example, if I have B* in my rack and the word is BO, I would like to remove O and only score B so that I can score wildcard to zero.
But the result is not what I expected.
import sys
if len(sys.argv) < 2:
print("no rack error.")
exit(1)
rack = sys.argv[1]
rack_low = rack.lower()
# Turn the words in the sowpods.txt file into a Python list.
with open("sowpods.txt","r") as infile:
raw_input = infile.readlines()
data = [datum.strip('\n') for datum in raw_input]
# Find all of the valid sowpods words that can be made
# up of the letters in the rack.
valid_words = []
# Call each word in the sowpods.txt
for word in data:
# Change word to lowercase not to fail due to case.
word_low = word.lower()
candidate = True
rack_letters = list(rack_low)
# Iterate each letter in the word and check if the letter is in the
# Scrabble rack. If used once in the rack, remove the letter from the rack.
# If there's no letter in the rack, skip the letter.
for letter in word_low:
if letter in rack_letters:
rack_letters.remove(letter)
elif '*' in rack_letters:
rack_letters.remove('*')
elif '?' in rack_letters:
rack_letters.remove('?')
else:
candidate = False
if candidate == True:
# Add score and valid word to the empty list
total = 0
for letter in word_low:
if letter not in rack_letters:
word_strip = word_low.strip(letter)
for letter in word_strip:
total += scores[letter]
valid_words.append([total, word_low])
I'm going to go a slightly different route with my answer and hopefully speed the overall process up. We're going to import another function from the standard library -- permutations -- and then find possible results by trimming the total possible word list by the length of the rack (or, whatever argument is passed).
I've commented accordingly.
import sys
from itertools import permutations # So we can get our permutations from all the letters.
if len(sys.argv) < 2:
print("no rack error.")
exit(1)
rack = sys.argv[1]
rack_low = rack.lower()
# Turn the words in the sowpods.txt file into a Python list.
txt_path = r'C:\\\\\sowpods.txt'
with open(txt_path,'r') as infile:
raw_input = infile.readlines()
# Added .lower() here.
data = [i.strip('\n').lower() for i in raw_input]
## Sample rack of 7 letters with wildcard character.
sample_rack = 'jrnyoj?'
# Remove any non-alphabetic characters (i.e. - wildcards)
# We're using the isalpha() method.
clean_rack = ''.join([i for i in sample_rack if i.isalpha()])
# Trim word list to the letter count in the rack.
# (You can skip this part, but it might make producing results a little quicker.)
trimmed_data = [i for i in data if len(i) <= len(clean_rack)]
# Create all permutations from the letters in the rack
# We'll iterate over a count from 2 to the length of the rack
# so that we get all relevant permutations.
all_permutations = list()
for i in range(2, len(clean_rack) + 1):
all_permutations.extend(list(map(''.join, permutations(clean_rack, i))))
# We'll use set().intersection() to help speed the discovery process.
valid_words = list(set(all_permutations).intersection(set(trimmed_data)))
# Print sorted list of results to check.
print(f'Valid words for a rack containing letters \'{sample_rack}\' are:\n\t* ' + '\n\t* '.join(sorted(valid_words)))
Our output would be the following:
Valid words for a rack containing letters 'jrnyoj?' are:
* jo
* jor
* joy
* no
* nor
* noy
* ny
* on
* ony
* or
* oy
* yo
* yon
If you want to verify that the results are actually in the sowpods.txt file, you can just index the sowpods.txt list by where the word you want to look up is indexed:
trimmed_data[trimmed_data.index('jor')]
When you are totalling the scores you are using the words from the wordlist and not the inputted words:
total=0
for letter in word_low:
...
Rather, this should be:
total=0
for letter in rack_low:
...
Also, You do not need to loop and remove the letters with strip at the end.
you can just have:
total = 0
for letter in rack_low:
if letter not in rack_letters:
try:
total += scores[letter]
except KeyError: # If letter is * or ? then a KeyError occurs
pass
valid_words.append([total, word_low])

Python: Select particular characters from a file and add them to a new variable

I'm fairly new to python and I'm not sure why what I'm doing is wrong.
numberOfOrders = 0
numberOfProducts = 0
allOrders = open("file.txt", "r") #A .txt file in the same directory as the .py file.
#file.txt:
#(A->[a:20,a:20,b:10,c:25,c:25])
#(B->[d:100,e:70])
#(C->[f:10000,g:200000])
while True:
theline = allOrders.readline()
for theline in allOrders:
for char in theline: #Iterate over each character of a line.
listProducts = "" #Empty string, will be the concatenation of the wanted characters.
if char == "[": #Wanted character.
listProducts = listProducts + "["
elif char == ":": #To keep count of no. of products in a list.
numberOfProducts += 1
elif is_number(char) == True: #Function that checks whether char is a number.
listProducts = listProducts + str(char) #Add to the string "listProducts".
elif char == ",": #Wanted character.
listProducts = listProducts + str(char)
elif char == "]":#Wanted character, to end the string.
listProducts = listProducts +str(char)
break
numberOfOrders += 1 #To keep track of no. of orders. Each line of file is an order.
if len(theline) == 0:
break
allOrders.close()
print(numberOfProducts)
print(numberOfOrders)
print(listProducts)
I basically only want the numbers and commas within brackets. That's my biggest issue here.
The output I get for
print(listProducts)
is
]
Thank you.
Regarding your code, the solution is to:
Removing the for theline in allOrders which is not coherent
Moving the initialization of listProducts before the while loop
Of course this can be widely optimized using regex for example, as suggested by #tdelaney.
You can keep the digits and commas by stripping everything else out using a regular expression. Then you have a string of decimals and commas that you can split to give a list of products in each order line.
import re
with open('file.txt') as all_orders:
# substitute '' for all non-digit non-comma then split
orders = [re.sub(r'[^\d,]', '', line).split(',')
for line in all_orders]
number_of_orders = len(orders)
number_of_products = sum(map(len, orders))
print('orders', number_of_orders, 'products', number_of_products)

Need assistance with cleaning words that were counted from a text file

I have an input text file from which I have to count sum of characters, sum of lines, and sum of each word.
So far I have been able to get the count of characters, lines and words. I also converted the text to all lower case so I don't get 2 different counts for same word where one is in lower case and the other is in upper case.
Now looking at the output I realized that, the count of words is not as clean. I have been struggling to output clean data where it does not count any special characters, and also when counting words not to include a period or a comma at the end of it.
Ex. if the text file contains the line: "Hello, I am Bob. Hello to Bob *"
it should output:
2 Hello
2 Bob
1 I
1 am
1 to
Instead my code outputs
1 Hello,
1 Hello
1 Bob.
1 Bob
1 I
1 am
1 to
1 *
Below is the code I have as of now.
# Open the input file
fname = open('2013_honda_accord.txt', 'r').read()
# COUNT CHARACTERS
num_chars = len(fname)
# COUNT LINES
num_lines = fname.count('\n')
#COUNT WORDS
fname = fname.lower() # convert the text to lower first
words = fname.split()
d = {}
for w in words:
# if the word is repeated - start count
if w in d:
d[w] += 1
# if the word is only used once then give it a count of 1
else:
d[w] = 1
# Add the sum of all the repeated words
num_words = sum(d[w] for w in d)
lst = [(d[w], w) for w in d]
# sort the list of words in alpha for the same count
lst.sort()
# list word count from greatest to lowest (will also show the sort in reserve order Z-A)
lst.reverse()
# output the total number of characters
print('Your input file has characters = ' + str(num_chars))
# output the total number of lines
print('Your input file has num_lines = ' + str(num_lines))
# output the total number of words
print('Your input file has num_words = ' + str(num_words))
print('\n The 30 most frequent words are \n')
# print the number of words as a count from the text file with the sum of each word used within the text
i = 1
for count, word in lst[:10000]:
print('%2s. %4s %s' % (i, count, word))
i += 1
Thanks
Try replacing
words = fname.split()
With
get_alphabetical_characters = lambda word: "".join([char if char in 'abcdefghijklmnopqrstuvwxyz' else '' for char in word])
words = list(map(get_alphabetical_characters, fname.split()))
Let me explain the various parts of the code.
Starting with the first line, whenever you have a declaration of the form
function_name = lambda argument1, argument2, ..., argumentN: some_python_expression
What you're looking at is the definition of a function that doesn't have any side effects, meaning it can't change the value of variables, it can only return a value.
So get_alphabetical_characters is a function that we know due to the suggestive name, that it takes a word and returns only the alphabetical characters contained within it.
This is accomplished using the "".join(some_list) idiom which takes a list of strings and concatenates them (in other words, it producing a single string by joining them together in the given order).
And the some_list here is provided by the generator expression [char if char in 'abcdefghijklmnopqrstuvwxyz' else '' for char in word]
What this does is it steps through every character in the given word, and puts it into the list if it's alphebetical, or if it isn't it puts a blank string in it's place.
For example
[char if char in 'abcdefghijklmnopqrstuvwyz' else '' for char in "hello."]
Evaluates to the following list:
['h','e','l','l','o','']
Which is then evaluates by
"".join(['h','e','l','l','o',''])
Which is equivalent to
'h'+'e'+'l'+'l'+'o'+''
Notice that the blank string added at the end will not have any effect. Adding a blank string to any string returns that same string again.
And this in turn ultimately yields
"hello"
Hope that's clear!
Edit #2: If you want to include periods used to mark decimal we can write a function like this:
include_char = lambda pos, a_string: a_string[pos].isalnum() or a_string[pos] == '.' and a_string[pos-1:pos].isdigit()
words = "".join(map(include_char, fname)).split()
What we're doing here is that the include_char function checks if a character is "alphanumeric" (i.e. is a letter or a digit) or that it's a period and that the character preceding it is numeric, and using this function to strip out all the characters in the string we want, and joining them into a single string, which we then separate into a list of strings using the str.split method.
This program may help you:
#I created a list of characters that I don't want \
# them to be considered as words!
char2remove = (".",",",";","!","?","*",":")
#Received an string of the user.
string = raw_input("Enter your string: ")
#Make all the letters lower-case
string = string.lower()
#replace the special characters with white-space.
for char in char2remove:
string = string.replace(char," ")
#Extract all the words in the new string (have repeats)
words = string.split(" ")
#creating a dictionary to remove repeats
to_count = dict()
for word in words:
to_count[word]=0
#counting the word repeats.
for word in to_count:
#if there is space in a word, it is white-space!
if word.isalpha():
print word, string.count(word)
Works as below:
>>> ================================ RESTART ================================
>>>
Enter your string: Hello, I am Bob. Hello to Bob *
i 1
am 1
to 1
bob 2
hello 2
>>>
Another way is using Regex to remove all non-letter chars (to get rid off char2remove list):
import re
regex = re.compile('[^a-zA-Z]')
your_str = raw_input("Enter String: ")
your_str = your_str.lower()
regex.sub(' ', your_str)
words = your_str.split(" ")
to_count = dict()
for word in words:
to_count[word]=0
for word in to_count:
if word.isalpha():
print word, your_str.count(word)

counting letters in a text file in python

So I'm trying to do this problem
Write a program that reads a file named text.txt and prints the following to the
screen:
 The number of characters in that file
 The number of letters in that file
 The number of uppercase letters in that file
 The number of vowels in that file
I have gotten this so far but I am stuck on step 2 this is what I got so far.
file = open('text.txt', 'r')
lineC = 0
chC = 0
lowC = 0
vowC = 0
capsC = 0
for line in file:
for ch in line:
words = line.split()
lineC += 1
chC += len(ch)
for letters in file:
for ch in line:
print("Charcter Count = " + str(chC))
print("Letter Count = " + str(num))
You can do this using regular expressions. Find all occurrences of your pattern as your list and then finding the length of that list.
import re
with open('text.txt') as f:
text = f.read()
characters = len(re.findall('\S', text))
letters = len(re.findall('[A-Za-z]', text))
uppercase = len(re.findall('[A-Z]', text))
vowels = len(re.findall('[AEIOUYaeiouy]', text))
The answer above uses regular expressions, which are very useful and worth learning about if you haven't used them before. Bunji's code is also more efficient, as looping through characters in a string in Python is relatively slow.
However, if you want to try doing this using just Python, take a look at the code below. A couple of points: First, wrap your open() inside a using statement, which will automatically call close() on the file when you are finished. Next, notice that Python lets you use the in keyword in all kinds of interesting ways. Anything that is a sequence can be "in-ed", including strings. You could replace all of the string.xxx lines with your own string if you would like.
import string
chars = []
with open("notes.txt", "r") as f:
for c in f.read():
chars.append(c)
num_chars = len(chars)
num_upper = 0;
num_vowels = 0;
num_letters = 0
vowels = "aeiouAEIOU"
for c in chars:
if c in vowels:
num_vowels += 1
if c in string.ascii_uppercase:
num_upper += 1
if c in string.ascii_letters:
num_letters += 1
print(num_chars)
print(num_letters)
print(num_upper)
print(num_vowels)

Counting words starting with a character

Write a function that accepts a string and a character as input and
returns the count of all the words in the string which start with the
given character. Assume that capitalization does not matter here. You
can assume that the input string is a sentence i.e. words are
separated by spaces and consists of alphabetic characters.
This is my code:
def count_input_character (input_str, character):
input_str = input_str.lower()
character = character.lower()
count = 0
for i in range (0, len(input_str)):
if (input_str[i] == character and input_str[i - 1] == " "):
count += 1
return (count)
#Main Program
input_str = input("Enter a string: ")
character = input("Enter character whose occurances are to be found in the given input string: ")
result = count_input_character(input_str, character)
#print(result)
The only part missing here is that how to check if the first word of the sentence is stating with the user given character. consider this output:
Your answer is NOT CORRECT Your code was tested with different inputs. > For example when your function is called as shown below:
count_input_character ('the brahman the master of the universe', 't')
####### Your function returns ############# 2 The returned variable type is: type 'int'
### Correct return value should be ######## 3 The returned variable type is: type 'int'
You function misses the first t because in this line
if (input_str[i] == character and input_str[i - 1] == " "):
when i is 0, then input_str[i - 1] is input_str[-1] which Python will resolve as the last character of the string!
To fix this, you could change your condition to
if input_str[i] == character and (i == 0 or input_str[i - 1] == " "):
Or use str.split with a list comprehension. Or a regular expression like r'(?i)\b%s', with (?i) meaning "ignore case", \b is word boundary and %s a placeholder for the character..
Instead of looking for spaces, you could split input_str on whitespace, this would produce a list of words that you could then test against character. (Pseudocode below)
function F sentence, character {
l = <sentence split by whitespace>
count = 0
for word in l {
if firstchar(word) == character {
count = count + 1
}
}
return count
}
Although it doesn't fix your specific bug, for educational purposes, please note you could rewrite your function like this using list comprehension:
def count_input_character (input_str, character):
return len([x for x in input_str.lower().split() if x.startswith(character.lower())])
or even more efficiently(thanks to tobias_k)
def count_input_character (input_str, character):
sum(w.startswith(character.lower()) for w in input_str.lower().split())
def c_upper(text, char):
text = text.title() #set leading char of words to uppercase
char = char.upper() #set given char to uppercase
k = 0 #counter
for i in text:
if i.istitle() and i == char: #checking conditions for problem, where i is a char in a given string
k = k + 1
return k

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