从保存在 2D 列表中的三元组单词构造文本

Question

我目前有一个文本，其单词保存为二维列表中的三元组。

我的二维列表：

[['Python', 'is', 'an'], ['interpreted,', 'high-level', 'and'], ['general-purpose', 'programming', 'language.'], ["Python's", 'design', 'philosophy'], ['emphasizes', 'code', 'readability'], ['with', 'its', 'notable'], ['use', 'of', 'significant'], ['whitespace.', 'Its', 'language'], ['constructs', 'and', 'object-oriented'], ['approach', 'aim', 'to'], ['help', 'programmers', 'write'], ['clear,', 'logical', 'code'], ['for', 'small', 'and'], ['large-scale', 'projects.']]

我正在创建一个 Python 代码，它随机选择一组这些三元组，然后尝试通过使用最后两个单词并选择一个以这两个单词开头的三元组来创建一个新的随机文本。最后，我的程序在写完 200 个单词或无法选择其他三元组时结束。

到目前为止的代码：

import random

with open(r'c:\\python\4_TRIPLETS\Sample.txt', 'r') as file:
    data = file.read().replace('\n', '').split()
    lines = [data[i:i + 3] for i in range(0, len(data), 3)]
    random.shuffle([random.shuffle(i) for i in lines])

first_triplet = random.choice(lines)
last_two = first_triplet[1:3]


output_text=[]
while True:
    candidates = [t for t in lines if t[0:2] == last_two]
    if not candidates:
        break
    
    next_triplet = random.choice(candidates)
    last_two = next_triplet[1:3]
    output_text.append(next_triplet)

我无法自动执行搜索匹配项并将它们存储在新列表中的重复过程。

有什么想法吗？

Answer 1

import random
import sys
import os
import json

outlist = []
file_in = sys.argv[1]
file_ot = str(file_in) + ".ot"

with open(file_in, 'r') as file:
    data = file.read().replace('\n', '').split()
    lines = [data[i:i + 3] for i in range(0, len(data), 1)]
print("\nΧωρισμένο Κείμενο σε Λίστα Τριπλετών:\n", lines)


triplet = random.choice(lines)
last_two = triplet[1:3]

print("\nΕπιλεγμένη Τριπλέτα: \n", triplet)
print("\nΔύο Τελευταίες Λέξεις Αυτής:\n", last_two)
outlist.extend(triplet)

proc_list = lines
# first selected, remove from list
proc_list.remove(triplet)

n = 0
while True:

    n += 1
    print("\nΕπανάληψη {0}\n".format(n))

    if proc_list == 0:
        print("a")
        break

    random.shuffle(proc_list)
    
    candidates = []
    
    for element in proc_list:
        if element[:2] == last_two:
            candidates.append(element)
    
    print(candidates)

    if not candidates:
        print("b")
        break
    
    
    if len(outlist) >= 200:   
        print("c")
        break
    
    triplet = random.choice(candidates)
    outlist.append(triplet[-1])
    proc_list.remove(triplet)
    
    last_two = triplet[1:3]
    print(outlist)

with open(file_ot, 'w') as f:
    f.write(json.dumps(outlist, indent=10))

print(" ".join(outlist))

Answer 2

可以使用递归函数（改了部分代码，检查cmets）：

import random

#adding ["is", "an", "experiment"] to check if it works (no other triplet were present that satisfy the condition)
lines = [['Python', 'is', 'an'], ['interpreted,', 'high-level', 'and'], ['general-purpose', 'programming', 'language.'], ["Python's", 'design', 'philosophy'], ['emphasizes', 'code', 'readability'], ['with', 'its', 'notable'], ['use', 'of', 'significant'], ['whitespace.', 'Its', 'language'], ['constructs', 'and', 'object-oriented'], ['approach', 'aim', 'to'], ['help', 'programmers', 'write'], ['clear,', 'logical', 'code'], ['for', 'small', 'and'], ['large-scale', 'projects.'], ["is", "an", "experiment"]]

first_triplet = ['Python', 'is', 'an'] # random.choice(lines)


def appendNextTriplet(output_text, lines):
    if len(output_text) >= 200:
        return output_text
    candidates = [t for t in lines if t[:2] == output_text[-2:]]
    if not candidates:
        return output_text
    next_triplet = random.choice(candidates)
    output_text += next_triplet # changed from append to concatenation, it was not correct
    return appendNextTriplet(output_text, lines)

print(appendNextTriplet(first_triplet, lines)) # ['Python', 'is', 'an', 'is', 'an', 'experiment']