Python regex findall list index out of range

Question

import re
with open("day2.txt", "r") as file: 
    line=file.read().split("\n")
    forward=0
    pos=0
    for i in range(0,len(line)-1):
        a=line[i]
        print(a)
        if (re.findall('^f',a)[0]) == 'f':
             forward=forward+int(((re.findall('\d',a)[0])))
        if (re.findall('^u',a)[0]) == 'u':
             pos=pos-int(((re.findall('\d',a)[0])))
        if (re.findall('^d',a)[0]) == 'd':
             pos=pos+int(((re.findall('\d',a)[0])))
print(forward*pos)

Here a or line[i] is a string. Test cases in the input.txt file is this but a few thousand lines of these

forward 6
up 4
forward 8
down 6
forward 9

Ideally the output should be the sum of 6+8 I get an error when i run it as a script i get list index out of range, but no errors when i run it via the shell line by line

The exact error message is :

Traceback (most recent call last):
  File "day2.py", line 10, in <module>
    if (re.findall('^u',a)[0]) == 'u':
IndexError: list index out of range

Where am i going wrong?

It's a text file containing a few thousand lines, of text with a pattern similar to the code box under testcases. Ie forward 6\n up 4\n forward 8\n, made the edit to the question mentioning the same. Thanks — keshav bantu
– keshav bantu, Commented Dec 2, 2021 at 20:24
Ok, can you add a few lines to your question, and make sure they include lines that cause your problem. — defladamouse
– defladamouse, Commented Dec 2, 2021 at 20:25
The issue is that your re.findall('^f',a) finds nothing, so the index 0 is out of range. Do you really need to use regex here? Is your file only forward and up followed by a number? — defladamouse
– defladamouse, Commented Dec 2, 2021 at 20:33

defladamouse · Accepted Answer · 2021-12-02 20:43:19Z

1

It would be possible to do with regex, but in my opinion that would be very over-engineered. Since your file is made of a small number of known strings followed by a number, you can simply isolate the numbers based on the strings.

This should work:

forward = 0
pos = 0
with open("input.txt", 'r') as file:
    elements = file.read().split("\n")

for e in elements:
    if "forward" in e:
        forward += int(e[7:])
    elif "up" in e:
        pos -= int(e[2:])
    elif "down" in e:
        pos += int(e[4:])

print(forward*pos)

answered Dec 2, 2021 at 20:43

defladamouse

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3 Comments

CrunchyLentils Over a year ago

Replace the string splicing index with -1 and you will always access the last value of a string. This removes the need to know the length of the string to start the slice at the correct index. The strength of regex lies in in being able to define a pattern of unknown characters and pull whatever matches that pattern. If you know you are looking for a string 'forward' then using the in keyword is more readable and reduces code complexity. You can access all the elements in a list with in as well, accessing elements in a list by the index can cause errors and reduce readability.

keshav bantu Over a year ago

I accept the solution, it worked. Thanks. But is there a way i can move forward with the over-engineered regex method.

defladamouse Over a year ago

@CrunchyLentils I didn't suggest [-1] because that would break for values greater than 9, but of course if the value is in the range [0, 9] that would be easier.

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