有两个版本,返回相反的答案,但总有一个是错误的。我不确定我哪里出了问题。我尝试了一系列其他选项,但这似乎是最接近的。编辑:需要在循环中
目标:识别列表中的元素,识别元素不在列表中时,识别列表为空时,根据情况返回字符串。
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 = search_for_string(['a', 'b', 'c'], 'd')
print(apple)
其他测试:
apple = search_for_string(['a', 'b', 'c'], 'b')
apple = search_for_string([], 'b')
print(search_term)
)。这可能会帮助您进行调试。 - Kevin Wangenumerate()
。这样可以避免使用i
对象。 - boardrider