Python: How to pass in only one variable into a lambda function?

Question

My code is reading in a text file line by line. Each line is then trimmed of all whitespace to one space character and based on whether it matches the pattern it is then written to either the matched_data_file or the unmatched_data_file. I have to make use of lambda in this particular example. I think the error lies with the following line but I'm not 100% sure:

success(line=row) if pattern.match(line) else failure(line=row)

Any help is greatly appreciated, thanks in advance!

I get the following error message:

Traceback (most recent call last): File "model_dev_txt_to_csv.py", line 26, in process(source_filename) File "model_dev_txt_to_csv.py", line 23, in process process_line(line, lambda: write_csv(m, line), lambda: write_csv(u, line)) File "model_dev_txt_to_csv.py", line 12, in process_line return success(line=row) if pattern.match(line) else failure(line=row) TypeError: () got an unexpected keyword argument 'line'

The following is my current code:

import re
import csv

pattern = re.compile("([0-9]+) +([0-9\.-]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+) +([0-9\.\-+Ee]+)")
source_filename = "track_param_hist.txt"
matched_data_file = "good_hist_csv.csv"
unmatched_data_file = "bad_hist_csv.csv"

def process_line(line, success, failure):
    # Make sure all whitespace is reduced to one space character
    row = (' '.join(line.split())).split(' ')
    success(line=row) if pattern.match(line) else failure(line=row)

def write_csv(file, line):
    csv.writer(file).writerow(line)

def process(source):
    print("Script for splitting text file into two separate csvs...")
    with open(matched_data_file, 'w') as m:
        with open(unmatched_data_file, 'w') as u:
            with open(source) as f:
                for line in f:
                    process_line(line, lambda: write_csv(m, line), lambda: write_csv(u, line))

if __name__ == "__main__":
    process(source_filename)

Your lambdas don't define any parameters - try e.g. lambda line: write_csv(...) — jonrsharpe
– jonrsharpe, Commented Nov 6, 2015 at 10:51
Or since they already have access to line, call them with no arguments. — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Nov 6, 2015 at 10:52
Seeing as your process_line turns a line into a row, I think it'd be clearer if you used the name row as a variable from that point. So the variable name in the lambda is row and doesn't shadow the original line — GP89
– GP89, Commented Nov 6, 2015 at 10:56
success(line=row) if pattern.match(line) else failure(line=row) doesn't make any sense. Maybe you mean if pattern.match(line): success(line=row) else: failure(line=row) — Open AI - Opting Out
– Open AI - Opting Out, Commented Nov 6, 2015 at 10:57

Mariusz Jamro · Accepted Answer · 2015-11-06 13:46:00Z

2

Lambda expression's syntax in Python is:

lambda [list of arguments]: <expression>

In your code you do not have any arguments defined for your lambdas. You need to add an argument named line before the : character to make your code work:

lambda line: write_csv(m, line), lambda line: write_csv(u, line)

edited Nov 6, 2015 at 13:46

answered Nov 6, 2015 at 10:53

Mariusz Jamro

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