[ python面试考试[1] ]

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# Scoring criteria:
# 5: pythonic     meeting grade 4, and coding in pythonic way
# 4: complete     correct logic and result. no bug
# 3: minor bug        very close to correct answer. missing corner case or a bit format/cosmetic issue
# 2: medium bug   algorithm can run but answer not right, missing conditions, logical flaw etc
# 1: major bug        algorithm totally wrong or very bad coding practice
# 0: nothing      unfinished major body of the code

Quiz 1:

Goal: Convert all text longer than 8 chars in all cells of the given csv file to end with '…’ with same length alignment (8 chars) and print out.

Hints: - Text length in cell could be of any length - For text shorter or equal to 8 characters, you should not make any change

目标：

将给定csv文件的所有单元格中超过8个字符的所有文本转换为以“…”结尾，并使用相同的长度对齐方式（8个字符）打印出来。 提示：

-单元格中的文本长度可以是任意长度 -对于短于或等于8个字符的文本，不应进行任何更改

import csv

def csv_filter(data_list):
    # print(len(data_list))
    if len(data_list) > 8:
        data_list = data_list[:8] + '...'
        # pass
    return data_list.ljust(11)

def print_truncated_csv(csv_file):
    with open(csv_file, newline='') as csvfile:
        spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
        with open('sample_filter.csv', 'w', newline='') as csvfile_write:
            pamwriter = csv.writer(csvfile_write, delimiter=',',
                                quotechar='|', quoting=csv.QUOTE_MINIMAL)
            for row in spamreader:
                if row:
                    row_iter = map(csv_filter, list(row))
                    pamwriter.writerow(list(row_iter))

source_csv = 'H:\面试\sample.csv'
print_truncated_csv(source_csv)

Quiz 2:

Goal: Calculate N factorial and print in the format of three-digits-segmentation, e.g. 123,456 计算N阶乘，并以三位数分割的格式打印，例如123,456

Hint: 1. n! = n * (n-1)! 2. 0! = 1 3. 1! = 1

from functools import reduce
def factorial(n):
    # Add your code here
    if n:
        return reduce(lambda x,y: x*y, range(1, n+1))
    return 1
def reformat(s):
    # Add your code here
    return ",".join([s[i:i+3] for i in range(0, len(s), 3)])

print(reformat(str(factorial(10))))

Quiz 3: 两个列表a, b, 设计一个函数分别返回对方不存在的元素

解题思路: 继承了字典的Counter类, 返回每个
from collections import Counter

list_1 = ['a', 'b', 'c']

list_2 = ['a', 'b', 'd']

def func(a, b):
    a_count = Counter(a)
    b_count = Counter(b)
    return [k for k in b_count if not a_count.get(k, None)], [k for k in a_count if not b_count.get(k, None)]

print(func(list_1, list_2))

Quiz4: 字符串a="()()", b="())(()",设计一个函数输入一个值, 返回字符串是否符号规则 例: a真, b假

a = "()()"
b = "())(()"

class Stack:
    """
    栈:后进先出
    """
    def __init__(self) -> None:
        self.list_stack = []

    def append(self, data):
        """
        进栈
        """
        self.list_stack.append(data)

    def pop(self):
        """
        出栈
        """
        return self.list_stack.pop()

    def judgment_empty(self):
        """
        空判断
        """
        return self.list_stack


stack = Stack()

def process_parentheses(a):
    for i in a:
        if i == "(":
            stack.append(i)
        if i == ")":
            if stack.judgment_empty():
                stack.pop()
            else:
                return False
    return stack.list_stack == []

print(process_parentheses(a))

解题思路:
设计一个栈(后进, 先出):
值为(时进入栈, 值为)时出栈
最后为栈为空表示符合规则, 有值或者提前返回False为不符合规则