We Keep Coding

Python - program for searching for relevant cells in excel does not work correctly

I've written a code to search for relevant cells in an excel file. However, it does not work as well as I had hoped.
In pseudocode, this is it what it should do:
Ask for input excel file
Ask for input textfile containing keywords to search for
Convert input textfile to list containing keywords
For each keyword in list, scan the excelfile
If the keyword is found within a cell, write it into a new excelfile
Repeat with next word
The code works, but some keywords are not found while they are present within the input excelfile. I think it might have something to do with the way I iterate over the list, since when I provide a single keyword to search for, it works correctly. This is my whole code: https://pastebin.com/euZzN3T3
This is the part I suspect is not working correctly. Splitting the textfile into a list works fine (I think).
#IF TEXTFILE
elif btext == True:
#Split each line of textfile into a list
file = open(txtfile, 'r')
#Keywords in list
for line in file:
keywordlist = file.read().splitlines()
nkeywords = len(keywordlist)
print(keywordlist)
print(nkeywords)
#Iterate over each string in list, look for match in .xlsx file
for i in range(1, nkeywords):
nfound = 0
ws_matches.cell(row = 1, column = i).value = str.lower(keywordlist[i-1])
for j in range(1, worksheet.max_row + 1):
cursor = worksheet.cell(row = j, column = c)
cellcontent = str.lower(cursor.value)
if match(keywordlist[i-1], cellcontent) == True:
ws_matches.cell(row = 2 + nfound, column = i).value = cellcontent
nfound = nfound + 1
and my match() function:
def match(keyword, content):
"""Check if the keyword is present within the cell content, return True if found, else False"""
if content.find(keyword) == -1:
return False
else:
return True
I'm new to Python so my apologies if the way I code looks like a warzone. Can someone help me see what I'm doing wrong (or could be doing better?)? Thank you for taking the time!

Splitting the textfile into a list works fine (I think).
This is something you should actually test (hint: it does but is inelegant). The best way to make easily testable code is to isolate functional units into separate functions, i.e. you could make a function that takes the name of a text file and returns a list of keywords. Then you can easily check if that bit of code works on its own. A more pythonic way to read lines from a file (which is what you do, assuming one word per line) is as follows:
with open(filename) as f:
keywords = f.readlines()
The rest of your code may actually work better than you expect. I'm not able to test it right now (and don't have your spreadsheet to try it on anyway), but if you're relying on nfound to give you an accurate count for all keywords, you've made a small but significant mistake: it's set to zero inside the loop, and thus you only get a count for the last keyword. Move nfound = 0 outside the loop.
In Python, the way to iterate over lists - or just about anything - is not to increment an integer and then use that integer to index the value in the list. Rather loop over the list (or other iterable) itself:
for keyword in keywordlist:
...
As a hint, you shouldn't need nkeywords at all.
I hope this gets you on the right track. When asking questions in future, it'd be a great help to provide more information about what goes wrong, and preferably enough to be able to reproduce the error.

Reading data from a text file in Python according to the parameters provided

I have a text file something like this
Mqtt_allowed=true
Mqtt_host=192.168.0.1
Mqtt_port=2223
<=============>
cloud_allowed=true
cloud_host=m12.abc.com
cloud_port=1232
<=============>
local_storage=true
local_path=abcd
I needed to get each of the value w.r.t parameter provided by the user.
What i am doing right now is:
def search(param):
try:
with open('config.txt') as configuration:
for line in configuration:
if not line:
continue
function, f_input=line.split("=")
if function == param:
result=f_input.split()
break
else:
result="0"
except FileNotFoundError:
print("File not found: ")
return result
mqttIsAllowed=search("Mqtt_allowed")
print mqttIsAllowed
Now when i call only mqt stuff it is working fine but when i call cloud or anything after the "<==========>" separation it throws an error. Thanks

Just skip all the lines starting with <:
if not line or line.lstrip().startswith("<"):
continue
Or, if you really, really want to match the separator exactly:
if line.strip() == "<=============>":
continue
I think the first variant is better because if someone slightly modified the separator by accident, the second piece of code won't work at all.

Because you are trying to split on the = character in a style that seems to be standard INI format, it is safe to assume that your pairs will be at max size 2. I'm not a fan of using methods that rely on character checking (unless specifically called for), so give this a whirl:
def search(param):
result = '0' # declare here
try:
with open('config.txt') as configuration:
for line in configuration:
if not line:
continue
f_pair = line.strip().split("=") # remove \r\n, \n
if len(f_pair) > 2: # your separator will be much longer
continue
else if f_pair[0] == param:
result = f_pair[1]
# result = f_input.split() # why the 'split()' here?
break
except FileNotFoundError:
print("File not found: ")
return result
mqttIsAllowed=search("Mqtt_allowed")
I'm pretty sure the error you were getting was a ValueError: too many values to unpack.
Here is how I know that:
When you call this function for any of the Mqtt_* values, the loop never encounters the separator string <=============>. As soo as you try to call anything below that first separator (for example a cloud_* key), the loop eventually reaches the first separator and tries to execute:
function, f_input = line.split('=')
But that wont work, in fact it will tell you:
ValueError: too many values to unpack (expected 2)
And that is because you are forcing the split() call to push into only 2 variables, but a split('=') on your separator string will return a list of 15 elements (a '<', a '>' and 13 ''). Thus, doing what I have posted above ensures that your split('=') still goes off, but checks to see if you hit a separator or not.

Python .find() to return the end of found string

So I want to pick some data out of a text file, which looks like this:
##After some other stuff which could change
EASY:[5,500]
MEDIUM:[10,100]
HARD:[20,1000]
EXPERT:[30,2000]
EXTREME:[50,5000]
I'm writing a function which uses the difficulty ('EASY' 'HARD' e.t.c) to return the following list. My current code looks like this:
def setAI(difficulty): #difficulty='EASY' or 'HARD' or...e.t.c)
configFile=open('AISettings.txt')
config=configFile.read()
print(config[(config.find(difficulty)):(config.find(']',(config.find(difficulty))))]) #So it will return the chunk between the difficulty, and the next closed-square-bracket after that
This produces the following output:
>>> HARD:[20,1000
I tried fixing it like this:
print(config[(config.find(difficulty)+2):(config.find(']',(config.find(difficulty)+2))+1)])
which returns:
>>>RD:[20,1000]
The issue I'm trying to adress is that I want it to start after the colon, I am aware that I could use the length of the difficulty string to solve this, but is there a simpler way of returning the end of the string when using the .find() command?
P.S: I couldn't find any duplicates for this, but it is a slightly odd question, so sorry if it's already on here somewhere; Thanks in advance
EDIT: Thanks for the replies, I think you basically all solved the problem, but the chosen answer was becasue I like the iteration line-by-line idea, Cheers guys :)

Well if the file look like this, why not just iterate line by line and do something like:
def setAI(difficulty): #difficulty='EASY' or 'HARD' or...e.t.c)
configFile=open('AISettings.txt')
config=configFile.readlines()
for line in config:
if line.startswith(difficulty.upper()):
print(line[len(difficulty) + 1:])

Find returns the location. But ranges assume that their end number should not be included. Just add one to the end.
config = """
##After some other stuff which could change
EASY:[5,500]
MEDIUM:[10,100]
HARD:[20,1000]
EXPERT:[30,2000]
EXTREME:[50,5000]
"""
difficulty = 'HARD'
begin = config.find(difficulty)
end = config.find(']', begin)
print(config[begin:end+1])

The function find will always give you the position of the first letter of the string. Also consider that the notation string[start:end] will give you the substring including the character at start but excluding the character at end. Therefore you could use something like the following:
def setAI(difficulty):
configFile = open('AISettings.txt')
config = configFile.read()
start = config.find(difficulty) + len(difficulty) + 1
end = config.find(']', start) + 1
print(config[start:end])

Python issue with replace statement?

I've been write this practice program for while now, the whole purpose of the code is to get user input and generate passwords, everything almost works, but the replace statements are driving me nuts. Maybe one of you smart programmers can help me, because I'm kinda new to this whole field of programming. The issue is that replace statement only seems to work with the first char in Strng, but not the others one. The other funcs blower the last run first and then the middle one runs.
def Manip(Strng):
#Strng = 'jayjay'
print (Strng.replace('j','h',1))
#Displays: 'hayjay'
print (Strng.replace('j','h',4))
#Displays: 'hayhay'
return
def Add_nums(Strng):
Size=len(str(Strng))
Total_per = str(Strng).count('%')
# Get The % Spots Position, So they only get replaced with numbers during permutation
currnt_Pos = 0
per = [] # % position per for percent
rGen = ''
for i in str(Strng):
if i == str('%'):
per.append(currnt_Pos)
currnt_Pos+=1
for num,pos in zip(str(self.ints),per):
rGen = Strng.replace(str(Strng[pos]),str(num),4);
return rGen
for pos in AlphaB: # DataBase Of The Positions Of Alphabets
for letter in self.alphas: #letters in The User Inputs
GenPass=(self.forms.replace(self.forms[pos],letter,int(pos)))
# Not Fully Formatted yet; you got something like Cat%%%, so you can use another function to change % to nums
# And use the permutations function to generate other passwrds and then
# continue to the rest of this for loop which will generate something like cat222 or cat333
Add_nums(GenPass) # The Function That will add numbers to the Cat%%%
print (rGen);exit()

get the list and input from one function and run them in different function

i have a programm that generate the list and then i ask them to tell me what they want to do from the menu and this is where my problem start i was able to get the input form the user to different function but when i try to use the if else condition it doesn't check, below are my code
def menu(x,l):
print (x)
if x == 1:
return make_table(l)
if x == 2:
y= input("enter a row (as a number) or a column (as an uppercase letter")
if y in [ "1",'2','3']:
print("Minmum is:",minimum(y,l))
if x== 3:
print ('bye')
def main():
bad_filename = True
l =[]
while bad_filename == True:
try:
filename = input("Enter the filename: ")
fp = open(filename, "r")
for f_line in fp:
f_str=f_line.strip()
f_str=f_str.split(',')
for unit_str in f_str:
unit=float(unit_str)
l.append(unit)
bad_filename = False
except IOError:
print("Error: The file was not found: ", filename)
#print(l)
condition=True
while condition==True:
print('1- open\n','2- maximum')
x=input("Enter the choice")
menu(x,l)
main()
from the bottom function i can get list and i can get the user input and i can get the data and move it in second function but it wont work after that.thank you

I think your problem is simple, and has nothing to do with how you're passing values between functions.
In main, you're reading a value from the user like this:
x=input("Enter the choice")
The input function:
… reads a line from input, converts it to a string (stripping a trailing newline), and returns that.
So, if the user types 1 at the prompt, you get back the string "1".
Now, you pass that value—perfectly correctly—to menu.
In menu, you then try to compare it to various numbers, like this:
if x == 1:
But this will never be true. A string, like "1", is never equal to a number, like 1. They're not even the same kind of value, much less the same value.
So, you need to do one of two things:
Convert the input to an number. For example, change menu(x,l) to menu(int(x), l). OR…
Write menu to expect strings. For example, change if x == 1: to if x == "1":.
You may be wondering why that print (x) didn't help you debug the problem.
print(x) prints out the end-user-friendly string representation of whatever you give it. That is, it automatically calls the str function for you. For debugging purposes, you often want to use repr instead of str, to get the programmer-friendly string representation instead of the end-user-friendly string representation.
For example, print(str("10")) will print out 10—just like print(str(10)), so you can't tell them apart. But print(repr("10")) will print out '10', unlike print(repr(10)), while prints 10, so you can tell them apart. repr can also help you spot things like strings with invisible characters in them, having special "node" objects from a parser instead of just strings, etc.