【每日一题1223】欧姆定律

Amoyshmily · 2021 年12 月 23 日 02:03

已知欧姆定律公式：V = IR，其中V表示电压，单位是V；I表示电流，单位是A；R表示电阻，单位是R。

给定一个字符串，其中包含带有V，A或者R描述的任意两个单位信息，并且使用空格相隔。例如"2R 10V"和"1V 1A"。请编写一个函数，计算出欧姆定律公式中缺少的单位和值，最多保留6位小数。

示例：
输入：“2200R 5V”，输出：“0.002273A”。
输入：“25V 1e-2A”，输出：“2500.0R”。

题目难度：简单
题目来源：CodeWars：Ohm’s Law

def solution(info: str) -> str:
    # your code

assert solution("2200R 5V") == "0.002273A"
assert solution("0.005A 30V") == "6000.0R"
assert solution("30V 5000R") == "0.006A"
assert solution("0R 0A") == "0.0V"

nobugs · 2021 年12 月 23 日 02:34

def solution(info: str) -> str:
    R, V, I = None, None, None
    for i in info.split():
        if 'R' in i:
            R = float(i.replace('R', ''))
        elif 'V' in i:
            V = float(i.replace('V', ''))
        elif 'A' in i:
            I = float(i.replace('A', ''))
        else:
            raise ValueError("传值格式错误")
    if R is None:
        return str(round(V / I, 6)) + 'R'
    elif V is None:
        return str(round(I*R, 6)) + 'V'
    elif I is None:
        return str(round(V / R, 6)) + 'A'
    else:
        raise ValueError("解析错误")


assert solution("2200R 5V") == "0.002273A"
assert solution("0.005A 30V") == "6000.0R"
assert solution("30V 5000R") == "0.006A"
assert solution("0R 0A") == "0.0V"

Huis · 2021 年12 月 23 日 02:43

def solution(info: str) -> str:
    str_arr = info.split(" ")
    _dict = {}
    _dict[str_arr[0][-1]] = float(str_arr[0][:-1])
    _dict[str_arr[1][-1]] = float(str_arr[1][:-1])
    result = -1
    if "R" not in _dict:
        result = str(round(_dict["V"] / _dict["A"], 6)) + "R"
    elif "V" not in _dict:
        result = str(round(_dict["A"] * _dict["R"], 6)) + "V"
    else:
        result = str(round(_dict["V"] / _dict["R"], 6)) + "A"
    return result

Kawi · 2021 年12 月 23 日 02:53

def solution(info: str) -> str:
    V, A, R = None, None, None
    for i in info.split(" "):
        l = len(i) - 1
        if 'R' in i:
            R = float(i[0:l])
        elif 'A' in i:
            A = float(i[0:l])
        elif 'V' in i:
            V = float(i[0:l])
        else:
            raise ValueError("传值格式错误")
    if A is None:
        return str(round(V / R, 6)) + "A"
    elif V is None:
        return str(round(A*R, 6)) + "V"
    else:
        return str(round(V/A, 6)) + "R"


assert solution("2200R 5V") == "0.002273A"
assert solution("0.005A 30V") == "6000.0R"
assert solution("30V 5000R") == "0.006A"
assert solution("0R 0A") == "0.0V"

1092939085 · 2021 年12 月 23 日 03:15

def solution(info: str) -> str:
    info_list = info.split(" ")
    a = info_list[0]
    b = info_list[1]
    if a[-1] in ['R', 'V'] and b[-1] in ['R', 'V']:
        if a[-1] == 'R':
            result = str(round(float(b[:-1]) / float(a[:-1]), 6)) + "A"
        else:
            result = str(round(float(a[:-1]) / float(b[:-1]), 6)) + "A"
    elif a[-1] in ['V', 'A'] and b[-1] in ['V', 'A']:
        if a[-1] == 'A':
            result = str(round(float(b[:-1]) / float(a[:-1]), 6)) + "R"
        else:
            result = str(round(float(a[:-1]) / float(b[:-1]), 6)) + "R"
    else:
        result = str(round(float(a[:-1]) * float(b[:-1]), 6)) + "V"
    return result


assert solution("2200R 5V") == "0.002273A"
assert solution("0.005A 30V") == "6000.0R"
assert solution("30V 5000R") == "0.006A"
assert solution("0R 0A") == "0.0V"

zhonghc · 2021 年12 月 23 日 06:10

import re


def solution(info: str) -> str:
    re_v = re.search(r'(\S+)V', info)
    re_i = re.search(r'(\S+)A', info)
    re_r = re.search(r'(\S+)R', info)
    if not re_v:
        return str(round(float(re_i.group(1)) * float(re_r.group(1)), 6)) + 'V'
    if not re_i:
        return str(round(float(re_v.group(1)) / float(re_r.group(1)), 6)) + 'A'
    if not re_r:
        return str(round(float(re_v.group(1)) / float(re_i.group(1)), 6)) + 'R'


assert solution("2200R 5V") == "0.002273A"
assert solution("0.005A 30V") == "6000.0R"
assert solution("30V 5000R") == "0.006A"
assert solution("0R 0A") == "0.0V"

Mooii · 2021 年12 月 23 日 09:48

def solution(info: str) -> str:
    s1, s2 = info.split(" ")
    a, b = float(s1[:-1]), float(s2[:-1])
    if 'V' in info:
        if 'V' in s1:
            r = a / b
        else:
            r = b / a
        return str(round(r, 6)) + ('A' if 'R' in info else 'R')
    else:
        return str(round(a*b, 6)) + 'V'


assert solution("2200R 5V") == "0.002273A"
assert solution("0.005A 30V") == "6000.0R"
assert solution("30V 5000R") == "0.006A"
assert solution("0R 0A") == "0.0V"

1831718564_5640 · 2021 年12 月 23 日 14:38

def solution7(ss:str):
    try:
        R,V,A='','',''
        for i in ss.split(' '):
            if 'R'in i:
                R=float(i.replace('R',''))
            elif 'V' in i:
                V=float(i.replace('V',''))
            elif 'A' in i:
                A=float(i.replace('A',''))
        if R=='':
            return str(round(V/A,6))+'R'
        elif V=='':
            return str(round(A*R,6)) + 'V'
        elif A=='':
            return str(round(V/R,6)) + 'A'
    except:
        raise ValueError
assert solution7("2200R 5V") == "0.002273A"
assert solution7("0.005A 30V") == "6000.0R"
assert solution7("30V 5000R") == "0.006A"
assert solution7("0R 0A") == "0.0V"

lekaixin · 2022 年1 月 29 日 04:29

def solution(info: str) -> str:
    list1=info.split(' ')
    for i in list1:
        if 'A' in i:
            A=float(i[0:-1])
        elif 'R' in i:
            R=float(i[0:-1])
        else:
            V=float(i[0:-1])
    if 'A' not in info:
        return str(round(V/R,6))+'A'
    if 'V' not in info:
        return str(round(A*R,6))+'V'
    if 'R' not in info:
        return str(round(V/A,6))+'R'

assert solution("2200R 5V") == "0.002273A"
assert solution("0.005A 30V") == "6000.0R"
assert solution("30V 5000R") == "0.006A"
assert solution("0R 0A") == "0.0V"

1120395481 · 2022 年3 月 4 日 05:44

换个口味，这个运行时间会慢一些

def solution(info: str) -> str:
    import re
    V,A,R =0,0,0

    r = re.search(r"(.+)([VAR])\s(.+)([VAR])", info)

    def assignNewValueToV(v):
        nonlocal V
        V = v
    def assignNewValueToA(v):
        nonlocal A
        A = v
    def assignNewValueToR(v):
        nonlocal R
        R = v

    eval('assignNewValueTo{}({})'.format(r.group(2), float(r.group(1))))
    eval('assignNewValueTo{}({})'.format(r.group(4), float(r.group(3))))

    if(V==0):
        return "{}V".format(round(A*R, 6))
    elif(A==0):
        return "{}A".format(round(V / R, 6))
    else:
        return "{}R".format(round(V / A, 6))

    # your code

assert solution("2200R 5V") == "0.002273A"
assert solution("0.005A 30V") == "6000.0R"
assert solution("30V 5000R") == "0.006A"
assert solution("0R 0A") == "0.0V"```