Find character at which string can be differentiated from list of strings

Question

For every string in a df column, I need the character at which this string becomes unique, that is, its uniqueness point (UP). For illustration, here is a toy dataframe:

import pandas as pd
import numpy as np
df = pd.DataFrame({
'word':['can', 'cans', 'canse', 'canpe', 'canp', 'camp'],
'code':['k@n', 'k@n}', 'k@(z', np.nan, 'k@()', np.nan]})

  word  code
0 can    k@n
1 cans  k@n}
2 canse k@(z
3 canpe
4 canp  k@()
5 camp

The expected result is given below. I computed the UP for the two columns word and code:

  word  code wordUP codeUP
0 can    k@n      4      4 # 'can' can be discriminated from 'cans' at the imagined fourth letter, which does not exist
1 cans  k@n}      5      4
2 canse k@(z      5      4
3 canpe           5        # empty cells don't have a UP
4 canp  k@()      5      4
5 camp            3

My current implementation works, but is too slow for my 100k row dataframe. You can see it below. Can you come up with something faster?

def slice_prefix(a, b, start=0, length=1):
    while 1:
        while a[start:start + length] == b[start:start + length]:
            start += length
            length += length
        if length > 1:
            length = 1
        else:
            return start 

df = df.fillna('')

# get possible orthographic and phonetic representations
all_words = df['word'].dropna().to_list()
all_codes = df['code'].dropna().to_list()

# prepare new columns
df['wordUP'] = np.nan
df['codeUP'] = np.nan

# compute UP
for idx,row in df.iterrows():
    word = row['word']
    code = row['code']
    wordUP = max([slice_prefix(word, item) for item in all_words if item != word]) + 1
    codeUP = max([slice_prefix(code, item) for item in all_codes if item != code]) + 1
    df.loc[idx, 'wordUP'] = wordUP
    df.loc[idx, 'codeUP'] = codeUP

df.loc[df['code']=='', 'codeUP'] = 0

score 0 · Answer 1 · answered Mar 12 '22 at 17:14

As it is, your code runs in 0.0012 second in average (10,000 iterations) on my computer.

I suggest you refactor your code in a way which is both faster (0.0008 second, -33%) and more idiomatic, thus readable:

import numpy as np
import pandas as pd


def slice_prefix(a, b, start=0, length=1):
    while 1:
        while a[start:start + length] == b[start:start + length]:
            start += length
            length += length
        if length > 1:
            length = 1
        else:
            return start

def find_up(x, strings):
    """New helper function"""
    return int(max([slice_prefix(x, item) for item in strings if item != x])) + 1

df = df.fillna(" ")

df = (
    df
    .assign(wordUP=df["word"].apply(lambda x: find_up(x, df["word"])))
    .assign(codedUP=df["code"].apply(lambda x: find_up(x, df["code"])))
    .replace(1, "")
)

print(df)
# Output
    word  code  wordUP codedUP
0    can   k@n       4       4
1   cans  k@n}       5       4
2  canse  k@(z       5       4
3  canpe             5        
4   canp  k@()       5       4
5   camp             3

Find character at which string can be differentiated from list of strings

1 Answers1