how to read lines starting with arithmatic characters in python 3 - python-3.x

I'm using readline to read a file, but some of its lines starts with arithmetic characters, readline skips this lines,
any suggestion?
with open(path) as program_file:
for line in program_file:
line = program_file.readline()
print(line)
some of lines are like this:
2 4 5
+ 3 #skip this line
_ 5 9

It's not the lines starting with characters that's the problems, it's that the for line in x and x.readline statements both read a line from the file, so you'll probably get alternate lines printed out. Remove the readline!
with open(path) as program_file:
for line in program_file:
print(line)

Related

How to skip N central lines when reading file?

I have an input file.txt like this:
3
2
A
4
7
B
1
9
5
2
0
I'm trying to read the file and
when A is found, print the line that is 2 lines below
when B is found, print the line that is 4 lines below
My current code and current output are like below:
with open('file.txt') as f:
for line in f:
if 'A' in line: ### Skip 2 lines!
f.readline() ### Skipping one line
line = f.readline() ### Locate on the line I want
print(line)
if 'B' in line: ## Skip 4 lines
f.readline() ### Skipping one line
f.readline() ### Skipping two lines
f.readline() ### Skipping three lines
line = f.readline() ### Locate on the line I want
print(line)
'4\n'
7
'1\n'
'9\n'
'5\n'
2
>>>
Is printing the values I want, but is printing also 4\n,1\n... and besides that, I need to write several f.realines()which is not practical.
Is there a better way to do this?
My expected output is like this:
7
2
Here is a much simpler code for you:
lines=open("file.txt","r").read().splitlines()
#print(str(lines))
for i in range(len(lines)):
if 'A' in lines[i]:
print(lines[I+2]) # show 2 lines down
elif 'B' in lines[i]:
print(lines[I+4]) # show 4 lines down
This reads the entire file as an array in which each element is one line of the file. Then it just goes through the array and directly changes the index by 2 (for A) and 4 (for B) whenever it finds the line it is looking for.
if you don't like repeated readline then wrap it in a function so the rest of the code is very clean:
def skip_ahead(it, elems):
assert elems >= 1, "can only skip positive integer number of elements"
for i in range(elems):
value = next(it)
return value
with open('file.txt') as f:
for line in f:
if 'A' in line:
line = skip_ahead(f, 2)
print(line)
if 'B' in line:
line = skip_ahead(f, 4)
print(line)
As for the extra output, when the code you have provided is run in a standard python interpreter only the print statements cause output, so there is no extra lines like '1\n', this is a feature of some contexts like the IPython shell when an expression is found in a statement context, in this case f.readline() is alone on it's own line so it is detected as possibly having a value that might be interesting. to suppress this you can frequently just do _ = <expr> to suppress output.

Reading in a file of one-word lines in python

Just curious if there's a cleaner way to do this. I have a list of words in a file, one word per line.
I want to read them in and pass each word to a function.
I've currently got this:
f = open(fileName,"r");
lines = f.readlines();
count = 0
for i in lines:
count += 1
print("--{}--".format(i.rstrip()))
if count > 100:
return
I there a way to read them in faster without using rstrip on each line?
with open(fileName) as f:
lines = (line for _, line in zip(range(100), f.readlines()))
for line in lines:
print('--{}--'.format(line.rstrip()))
This is how I would do it. Note the context manager (the with/as statement), and the generator comprehension giving us only the first 100 lines.
Similar to Patrick's answer:
with open(filename, "r") as f:
for i, line in enumerate(f):
if i >= 100:
break
print("--{}--".format(line[:-1]))
If you don't an .strip() and know the length line terminator, you can use [:-1].

start counting how many lines of show command after line 20

Hello I am trying to write a code for counting how many lines after line 20
def line_count(file_name,start=20):
num_lines = 0
with open(file_name,'r') as f:
for line in f:
num_lines += 1
print(num_lines)
An easy way to do this is simply counting the lines in the program and subtracting 20 from it. This is possible in a one-liner:
>>> len(open("file.txt").read().splitlines()) - 20
6
In the above example, this means that the file itself has 26 lines, which is 6 more than 20.

Python 3.x - don't count carriage returns with len

I'm writing the following code as part of my practice:
input_file = open('/home/me/01vshort.txt', 'r')
file_content = input_file.read()
input_file.close()
file_length_question = input("Count all characters (y/n)? ")
if file_length_question in ('y', 'Y', 'yes', 'Yes', 'YES'):
print("\n")
print(file_content, ("\n"), len(file_content) - file_content.count(" "))
It's counting carriage returns in the output, so for the following file (01vshort.txt), I get the following terminal output:
Count all characters (y/n)? y
0
0 0
1 1 1
9
...or...
Count all characters (y/n)? y
0
00
111
9
In both cases, the answer should be 6, as there are 6 characters, but I'm getting 9 as the result.
I've made sure the code is omitting whitespace, and have tested this with my input file by deliberately adding whitespace and running the code with and without the line:
- file_content.count(" ")
Can anyone assist here as to why the result is 9 and not 6?
Perhaps it isn't carriage returns at all?
I'm also curious as to why the result of 9 is indented by 1 whitespace? The input file simply contains the following (with a blank line at the end of the file, line numbers indicated in the example):
1. 0
2. 0 0
3. 1 1 1
4.
...or...
1. 0
2. 00
3. 111
4.
Thanks.
If you want to ignore all whitespace characters including tabs and newlines and other control characters:
print(sum(not c.isspace() for c in file_content))
will give you the 6 you expect.
Alternatively you can take advantage of the fact the .split() method with no argument will split a string on any whitespace character. So split it into non-space chunks and then join them all back together again without the whitespace characters:
print(len(''.join(file_content.split())))
You're getting 9 because the content of the file could be interpreted like:
file_content = "0\n0 0\n1 1 1\n"
and you're only matching the white spaces (file_content.count(" ")).
In order to count only the characters you'd either:
read line by line the file, or
use a regexp to match white space.
For the indenting of 9: print processes the commas as outlined here

Pulling a list of lines out of a string

Beginning
Line 2
Line 3
Line 4
Line 5
Line 6
End
Trying to pull off line 2 through line 6. Can't do it to save my soul.
a is the saved string I'm searching through.
b = re.findall(r'Beginning(.*?)End', a)
Doesn't give me a thing, just a blank b. I know it's because of the newlines but how do I go about detecting and moving on forward with the newlines. I've tried, not knowing exactly for sure how I'm suppose to use MULTILINE or DOTALL. Nothing changed.
How do I go about getting it to put lines 2 through 6 in b?
To add in this will occur multiple times through the same file that I need to perform this search and pull technique. I have no other easy way of doing this since the information in Lines 2-6 need to be looked through further to pull off data that will be put into a csv file. Since some of the data contains hours and some of the data doesn't contain hours, aka Unavailable, I need to be able to pull off and differentiate between the two occurrences.
string = """Beginning
Line 2
Line 3
Line 4
Line 5
Line 6
End
"""
lines = string.splitlines()
answer = []
flag = False
for line in lines:
line = line.strip()
if not line: continue
if line == "Beginning":
flag = True
continue
if line == "End": flag = False
if not flag: continue
answer.append(line)
Output:
In [209]: answer
Out[209]: ['Line 2', 'Line 3', 'Line 4', 'Line 5', 'Line 6']
You could make a function that takes a multi-line string, then a starting line, and an ending line.
def Function(string, starting_line, ending_line):
if "\n" in string: #Checks for whether or not string is mult-line
pass
else:
return "The string given isn't a multiline string!" #If string isn't multiline, then Function returns a string explaining that string isn't a multi-line string
if ending_line < starting_line: #Checks if ending_line < starting_line
return "ending_line is greater than starting_line!" #If ending_line < starting_line, then Function returns a string explaining that ending_line > starting_line
array = [] #Defines an array
for i in range(len(string)): #Loops through len(string)
if list(string)[i] = "\n": #Checks whether list(string)[i] = a new line
array.append(i) #Appends i to array
return string[array[starting_line - 1]::array[ending_line - 1]]
print(Function(a, 3, 7))
This code should return:
Line 1
Line 2
Line 3
Line 4
Line 5
Line 6

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