DarrylG

您可以将customerList(作为字符串)直接传递给如下计算价格的函数。

[['potato', 'X', '8'], ['orange', 'X', '7'], ['rice', 'X', '13'], ['bread', 'X', '13']]

计算值的函数:

def calc_purchase_value(customerList_contents, price_list):
    " Compute purchase value "

    # convert customerList_contents to a list of list
    purchase_list = [x.split() for x in customerList_contents.split('\n') if x.strip()]

    # Iterate through the purchase_list looking up items in the price list
    # Each sublist will be of the form [item, 'X', cnt]
    return sum([price_list.get(item, 0) * int(cnt) for item, x, cnt in purchase_list])

使用自定义列表内容(将从文件中读取不同的字符串)

customerList_contents = """
potato X 8
orange X 7
rice X 13
bread X 13
"""

price_list = {"rice": 2.10,
"potato":4.21,
"orange":3.31,
"eggs":1.92,
"cheese":8.10,
"chicken":5.22}

print(calc_purchase_value(purchase_list, price_list))
#>> 84.15

完全解决方案

import glob
import os
import json

def readShopping():
    " Reads each customer shopping list, returning result as a geneerator "
    for file in glob.glob('./shoppinglists/*.txt'):
        with open(file,'r') as customerList:
            contents = customerList.read()
            # return file path and contents
            yield file, contents

def calc_purchase_value(customerList_contents, price_list):
    " Compute purchase value "

    # convert customerList_contents to a list of list
    purchase_list = [x.split() for x in customerList_contents.split('\n') if x.strip()]

    # Iterate through the purchase_list looking up items in the price list
    # Each sublist will be of the form [item, 'X', cnt]
    return sum([price_list.get(item, 0) * int(cnt) for item, x, cnt in purchase_list])

def get_prices():
    " Reads prices from JSON file "
    with open('./shoppinglists/pricelist.json') as json_file:
        return json.load(json_file)

# - Get prices from JSON file
price_list = get_prices()

for file, contents in readShopping():
    # file - path to customer file
    # contents - contents of customer file

    # Extract customer name from file path
    basename = os.path.basename(file)
    customer_name = os.path.splitext(basename)[0]

    # Total price: based upon shopping list and price list
    total_price = calc_purchase_value(contents, price_list)

    # Show results
    print('Customer {} total is: {:.2f}'.format(customer_name, total_price))

在上面,我有两个文本文件在文件夹购物列表“Mary.txt”和“James.txt”(你会说50或更多)。

Mary.txt的内容

orange X 3
rice X 2
potato X 3
eggs X 2

potato X 8
orange X 7
rice X 13
bread X 13

Customer james total is: 84.15
Customer mary total is: 30.60

使用函数搜索搜索(+/-)的二叉树。

预先计算从数组中每个索引到结尾的绝对值之和,允许在搜索的中间节点上终止。

这是因为,如果到目前为止的值之和+从当前索引到数组末尾的值之和<0,那么我们知道剩余数组中没有足够的值来克服当前的负累积值。

def findsums(a, n = -1, sumsofar = 0, signs = [], results = [], cumsum = []):

    """
    finds additions and subtraction of array element which are >= 0
        a - input array\n        
        n - highest index element of array we\'re using on the previous iteration
        sumsofar - sum using element up to index
        signs - signs (+/-) applied to these eleemnt
        results - solutions
        cumsum - cumulative sum of elements from index i to the end of array
    """
    if not cumsum:
        # Base case getting started
        #
        # Cumulative so of a in reverse order
        cumsum = cumsum_calc(a)

        # Use the first number as-is (i.e. no +/- sign)
        signs = [''] # first element is a
        sumsofar = a[0]
        n = 0

    # terminal case
    if n >= len(a)-1:
        if sumsofar >= 0:
            # terminal case (valid solution, so add)
                results.append(signs[:])
        else:
            # invalid solution
            pass # nothing to do 
    elif n == -1 or sumsofar + cumsum[n] >= 0:
        # Viable candidate so far
        # Try +/- alternatives\n        
        # Try + sign

        signs.append(' + ')
        findsums(a, n+1, sumsofar+a[n+1], signs, results, cumsum)
        signs.pop()

        # Try '-' sign
        signs.append(' - ')
        findsums(a, n+1, sumsofar-a[n+1], signs, results, cumsum)
        signs.pop()

    else:
        # Backtrack (sumsofar + cumsum[n] < 0):
        # No valid solution going forward\n        
        # So don\'t go any further with this sequence
        pass

    return results

def cumsum_calc(arr):
    " accum sum from next index forward in the array "
    # Adepted from https://www.geeksforgeeks.org/python-program-to-find-cumulative-sum-of-a-list/\n    
    # maximum sum of elements after i
    b = [abs(x) for x in arr]
    return [sum(b[i+1:]) for i in range(len(b)+1)]

def show_solutions(a, signs):
    " intertwines signs and array to show how they are added "
    # convert a to string, with parentheses around negative values

    converta = list(map(str, a))

    # place sign before each value in array a (converta)
    # function to generate list of sign, value pairs
    create_sign_value_pairs = lambda sign: list(zip(sign, converta))

    # Create sign/value pairs (i.e. [[('+', '(-1)'), ('+', '2')],...]
    sol_with_signs = list(map(create_sign_value_pairs, signs))

    # Convert each solution to a string
    solutions = list(map(lambda sol: ' '.join([''.join(s) for s in sol]), sol_with_signs))

    return "\t" + '\n\t'.join(solutions)

tests = [[2, 3], [-1, 2], [1], [-1], [-1, -2], [1, 2, 3, 4, -5]]

测试=[[2,3],[1,2],[1],[1],[1,-2],[1,2,3,4,-5]]

对于测试中的t: print(“对于数组{},解决方案是:”.format(t)) 打印(显示解决方案(t,s))

输出

For array [2, 3], solutions are:
    2  + 3
For array [-1, 2], solutions are:
    -1  + 2
For array [1], solutions are:
    1
For array [-1], solutions are:

For array [-1, -2], solutions are:
    -1  - -2
For array [1, 2, 3, 4, -5], solutions are:
    1  + 2  + 3  + 4  + -5
    1  + 2  + 3  + 4  - -5
    1  + 2  + 3  - 4  - -5
    1  + 2  - 3  + 4  - -5
    1  + 2  - 3  - 4  - -5
    1  - 2  + 3  + 4  + -5
    1  - 2  + 3  + 4  - -5
    1  - 2  + 3  - 4  - -5
    1  - 2  - 3  + 4  - -5

性能

使用Grzegorz Skibinski(组合法)

当前方法(使用回溯)

72.1 ms±2.34 ms/回路(7次运行的平均值±标准偏差,每个10回路)

使用回溯比测试所有组合快10倍

使用zip将列转换为行,然后选择第三个子列表(行)

lst = [
       [111, 111, 4523.123, 111, 111],
       [111, 111, 4526.15354, 111, 111],
       [111, 111, 4580.112, 111, 111],
       ]

第三栏:

list(zip(*lst))[2] 

使用正则表达式

import re

s = """
biography -> biography
biographical -> biography
biopic -> biography
bio-pic -> biography-pic
I watched a biographical movie -> I watched a biography movie
"""
x = re.sub(r'\b(bio\w*)', 'biography', s)
print(x)

输出

biography -> biography
biography -> biography
biography -> biography
biography-pic -> biography-pic
I watched a biography movie -> I watched a biography movie

import re
from collections import defaultdict, Counter

def create_dict(text):
" Dictionary contains strings for each paragraph using paragraph ID as key"
  d = defaultdict(lambda: "")
  lines = text.splitlines()
  for line in lines:
    matchObj = re.match( r'<P ID=(\d+)>', line)
    if matchObj:
      dictName = matchObj.group(0)
      continue  #skip line containing paragraph ID
    elif re.match(r'</P>', line):
      continue  #skip line containing paragraph ending token
    d[dictName] += line.lower()
  return d

def document_frequency(d):
" frequency of words in document "
  c = Counter()
  for paragraph in d.values():
    words = re.findall(r'\w+', paragraph)
    c.update(words)
  return c

def paragraph_frequency(d):
"Frequency of words in paragraph "
  c = Counter()
  for sentences in d.values():
    words = re.findall(r'\w+', sentences)
    set_words = set(words)  # Set causes at most one occurrence 
                            # of word in paragraph
    c.update(set_words)
  return c

text = """<P ID=1>
I have always wanted to try like, multiple? Different rasteraunts. Not quite sure which kind, maybe burgers!
</P>

<P ID=2>
Nice! I love burgers. Cheeseburgers, too. Have you ever gone to a diner type restauraunt? I have always wanted to try every diner in the country.
</P>

<P ID=3>
I am not related to the rest of these paragraphs at all.
</P>"""

d = create_dict(text)
doc_freq = document_frequency(d)    # Number of times in document
para_freq = paragraph_frequency(d)  # Number of times in paragraphs
print("document:", doc_freq)
print("paragraph: ", para_freq)

结果

document: Counter({'i': 4, 'to': 4, 'have': 3, 'always': 2, 'wanted': 2, 'try': 2, 'not': 2,'burgers': 2, 'diner': 2, 'the': 2, 'like': 1, 'multiple': 1, 'different': 1, 'rasteraunts':1, 'quite': 1, 'sure': 1, 'which': 1, 'kind': 1, 'maybe': 1, 'nice': 1, 'love': 1, 'cheeseburgers': 1, 'too': 1, 'you': 1, 'ever': 1, 'gone': 1, 'a': 1, 'type': 1, 'restauraunt': 1, 'every': 1, 'in': 1, 'country': 1, 'am': 1, 'related': 1, 'rest': 1, 'of': 1, 'these': 1, 'paragraphs': 1, 'at': 1, 'all': 1})
paragraph:  Counter({'to': 3, 'i': 3, 'try': 2, 'have': 2, 'burgers': 2, 'wanted': 2, 'always': 2, 'not': 2, 'the': 2, 'which': 1, 'multiple': 1, 'quite': 1, 'rasteraunts': 1, 'kind': 1, 'like': 1, 'maybe': 1, 'sure': 1, 'different': 1, 'love': 1, 'too': 1, 'in': 1, 'restauraunt': 1, 'every': 1, 'nice': 1, 'cheeseburgers': 1, 'diner': 1, 'ever': 1, 'a': 1, 'type': 1, 'you': 1, 'country': 1, 'gone': 1, 'at': 1, 'related': 1, 'paragraphs': 1, 'rest': 1, 'of': 1,'am': 1, 'these': 1, 'all': 1})