Question

我有"dinosaur", "dosimetry", and "moist"一词。我正在考虑有数十万个单词的情况。我想返回字符串中任何地方包含"s", "i", "o", "m"的所有单词。该函数应返回"dosimetry", "moist"。

是否有一种有效的方法来执行此操作，或者我必须进行迭代并检查？

Answer 1

A = ['o', 'i', 's', 'm']
words = ["dinosaur", "dosimetry", "moist", "personal", "since",
  "including", "guide", "shop", "directory", "board", "location",
  "change", "white", "text", "small", "emotions", "rating",
  "rate", "movies", "government"]

这里有两种方法返回words中的单词，这些单词包含A中包含的所有字母。

＃1

def select_some(words)
  words.select { |word| A & word.chars == A }
end

select_some(words)
  #=> ["dosimetry", "moist", "emotions", "movies"]

手术线可以更改为

words.select { |word| (A-str.chars).empty? }

＃2

n = 1
H = A.each_with_object({}) do |c,h|
  h[c] = n
  n <<= 1
end
  #=> {"s"=>1, "i"=>2, "o"=>4, "m"=>8} 
N = n - 1
  #=> 15

def select_some(words)
  words.select do |word|
    n = 0
    word.each_char do |c|
      x = H[c]
      n |= x if x
    end
    n == N
  end
end

select_some(words)
  #=> ["dosimetry", "moist", "emotions", "movies"]

Answer 2

只是为了经验

使用正则表达式正向查找

words = %w(dinosaur dosimetry moist)

words.select { |word| word.match?(/(?=.*m)(?=.*s)(?=.*i)(?=.*o).*/) }

#=> ["dosimetry", "moist"]

为了提高搜索速度，我按照English Letter Frequency在正则表达式中排列了字母。

Answer 3

根据要求，以更具可读性/永久性的形式发布我的比较基准。

side = convert(float(format(side, '.3f')))

结果

from tkinter import *
from math import *


# Custom number format code from wwii:
def convert(thing):
    if isinstance(thing, str):
        a,b = thing.split('+')
        b,*d = b.split('.')
        d = '00' if not d else d[0]
        thing = round(float(f'{a}{b}.{d}'), 2)
    elif isinstance(thing, (int,float)):
        thing = str(round(thing, 2))
        thing,*d = thing.split('.')
        d = '00' if not d else d[0]
        thing = thing if len(thing) > 2 else '0'+thing
        thing = f'{thing[:-2]}+{thing[-2:]}.{d}'
    return thing


def show_entry_fields():
    try:
        a, c, d, e = float(e1.get()), float(e3.get()), convert(str(e4.get())), convert(str(e5.get()))
        b = e - d
        s = (a + b + c) / 2
        height = (sqrt (s * (s - a) * (s - b) * (s - c)) * 2) / b
        height = float(format(height, '.3f'))
        height_label['text'] = str(height)
        side =((sqrt ((a ** 2) - (height ** 2))) + b)
        side = convert(float(format(side, '.3f')))
        side_label['text'] = str(side)
    except ValueError:
        pass
    master.after(100, show_entry_fields)


master = Tk()
master.attributes("-topmost", True)
master.title("Triangulation Plotting")

Label(master, text="Measurement #1 Station Line Location").grid(row=1, column=0, sticky=W, pady=4)
e4 = Entry(master)
e4.grid(row=1, column=1, sticky=E)

Label(master, text="Triangulation Measurement #1").grid(row=2, column=0, sticky=W, pady=4)
e1 = Entry(master)
e1.grid(row=2, column=1, sticky=E)

Label(master, text="Measurement #2 Station Line Location").grid(row=3, column=0, sticky=W, pady=4)
e5 = Entry(master)
e5.grid(row=3, column=1, sticky=E)

Label(master, text="Triangulation Measurement #2").grid(row=7, column=0, sticky=W, pady=4)
e3 = Entry(master)
e3.grid(row=7, column=1, sticky=E, pady=4)

Label(master, text="Offset from station line").grid(row=8, column=0, sticky=W, pady=4)
height_label = Label(master, text="")
height_label.grid(row=8, column=1)

Label(master, text="Measurement on Station Line").grid(row=9, column=0, sticky=W, pady=4)
side_label = Label(master, text="")
side_label.grid(row=9, column=1)

master.after(100, show_entry_fields)
master.mainloop()

标识包含列表中每个字符的字符串

3 个答案: