CODEWITHSUNDEEP
pythonpython-3.xlistcompare
Yoda
Yoda

Reputation: 127

How to find the name that corresponds to a list of numbers?

I have a data list like this:

[['Alice', 'Bob', 'Charlie'], [2, 8, 3], [4, 1, 5], [3, 2, 5]]

and I have a list of numbers(key):

[4, 1, 5]

I want to take this (key) and compare it against the data list if found, which in the above example is true then it will print the corresponding name to it (which is Bob) if not found then it just prints error message

How i can do that?

my code:

from sys import argv, exit
import csv
import re
import pandas as pd

argc = len(argv)

if argc != 3:
    print("Invalid file(s)")
    exit(1)


file = open(argv[1] , "r")
file1 = open(argv[2] , "r")

text = file1.read()

strl = []


with file:

    csv = csv.reader(file,delimiter=",")

    for row in csv:
        strl = row[1:9]
        break

    df = pd.read_csv(argv[1])
    data = df.loc[:, df.columns != 'name'].to_numpy().tolist()
    data.insert(0, df["name"].tolist())



numberx = []

for g in range(len(strl)):

    pattern = re.compile('(%s)'%strl[g])
    res = pattern.findall(text)
    repeats = len(res)
    intm = repeats


    pattern = re.compile('(%s)+'%strl[g])
    res = pattern.findall(text)
    repeats = len(res)
    intg=repeats - 1

    number = intm - intg
    numberx.append(number)

in my code case the (key) is numberx and data list is data

Upvotes: 3

Views: 683

Answers (1)

Andrej Kesely
Andrej Kesely

Reputation: 195553

You can use zip() to "tie" name to its list and then do simple == compare:

data = [['Alice', 'Bob', 'Charlie'], [2, 8, 3], [4, 1, 5], [3, 2, 5]]    
numberx = [4, 1, 5]

for name, lst in zip(data[0], data[1:]):
    if numberx == lst:
        print('List found! Name = {}'.format(name))
        break
else:
    print('Error!')

Prints:

List found! Name = Bob

Upvotes: 6

Related Questions