小编Meg*_*gan的帖子

两个版本,返回相反的答案,但总是一个出错.我不确定我哪里出错了.我已经尝试了一系列其他选项,但这似乎是最接近的.编辑:需要循环

目标:识别列表中的元素,识别元素何时不在列表中,识别列表何时为[],相应地返回字符串.

def search_for_string(a_list, search_term):
    i=0
    for search_term in a_list:
        i += 1
        if a_list[i] == search_term: 
            return 'string found!' 
        elif a_list[i] != search_term:
            return 'string not found2'
    if len(a_list) == 0:
        return 'string not found'

apple = search_for_string(['a', 'b', 'c'], 'd')
print(apple)


def search_for_string(a_list, search_term):
    i=0
    for search_term in a_list:
        if a_list[i] == search_term: 
            return 'string found!' 
        elif a_list[i] != search_term:
            return 'string not found2'
        i += 1
    if len(a_list) == 0:
        return 'string not found'

apple = …

试图找出函数以返回Python中直线的斜率。问题的方向是找到具有给定斜率坐标的m的斜率。通读其他一些堆栈溢出文章，但似乎没有一个能解决问题。这是我尝试过的各种变化：

def slope(x1, y1, x2, y2):
    m = 0
    b = (x2 - x1)
    d = (y2 - y1)
    if b != 0:
        m = (d)/(b) 

    return m

slope(2, 3, 6, 7)


def slope(x1, y1, x2, y2):
    m = 0
    a = float(x1)
    b = float(x2)
    c = float(y1)
    d = float(y2)
    m = (d-c)/(b-a)
    return m
slope(2, 3, 6, 7)

def slope(x1, y1, x2, y2):
    m = ''
    a = float(x1)
    b = float(x2)
    c = float(y1)
    d = …