难题:
键入一个数组,輸出该数组的第二大的数据 ,而且撰写有关的功能测试
留意:
1.假如list带有非int, float元素必须remove
2.假如list有反复的较大 元素,必须自身解决,内嵌的
list.sort(reverse=True)和
heapq.nlargest排序 ,元素数量不会改变。
另附编码
removeInvalidItems 去掉并不是int或float种类的值。
留意:不可以像下面那样用一次循环系统,由于remove某一元素,字符发生了更改 ,一些值并不可以清除
for item in l: # remove non_value items if not isinstance(item, (int, float)): l.remove(item)
下面是能用编码,文件夹名称为 findSecondUtil.py
def removeInvalidItems(l): tmpl = list() for item in l: if not isinstance(item, (int, float)): tmpl.append(item) for item in tmpl: l.remove(item) return l
findSecondItem.py 完成寻找第二数字的第一种方式 。这类方式不用去掉反复元素。
import sys from hyang.python3.interview.boyan.findSecondUtil import removeInvalidItems #from findSecondUtil import removeInvalidItems def findSecond(l): l = removeInvalidItems(l) length = len(l) if length == 0: print("there is no number item in the list") return None elif length == 1: print("there is only one number item, it's ", l[0]) return None elif length > sys.maxsize: print("out of scope") return None largest, second = max(l), min(l) if largest == second: return None for item in l: if item > second and item < largest: second = item return second
findSecondTest.py 测试程序
import unittest from hyang.python3.interview.boyan.findSecondItem import findSecond #from findSecondItem import findSecond class Test_findSecondItem(unittest.TestCase): # 假如跑全部测试用例,只运作一次必要条件和完毕标准。则用setupclass()和teardownclass() @classmethod def setUpClass(cls): print("在全部的功能测试实行以前 ,只实行一次============") @classmethod def tearDownClass(cls): print("在全部的功能测试实行以后,只实行一次============") # empty list def test_findSecondItem_01(self): l1 = list() assert (findSecond(l1) == None) # one item in list def test_findSecondItem_02(self): l1 = [2] assert (findSecond(l1) == None) # No item in list after remove non-number items def test_findSecondItem_03(self): l1 = [None, "abc", "xyz"] assert (findSecond(l1) == None) # one item in list after remove non-number items def test_findSecondItem_04(self): l1 = [None, 3, "abc"] assert (findSecond(l1) == None) # duplated largest number def test_findSecondItem_05(self): l1 = [32, None, 12, "abc", 8, 6, 36, 3, 32, 4, 36, 9, 25, '35', 36] assert (findSecond(l1) == 32) # python3中写不写都是会实行的 if __name__ == '__main__': unittest.main()
方式二:findSecondNum.py 从目录去掉全部的较大 元素,再在目录中找一个较大 便是第二大元素 。
import sys from hyang.python3.interview.boyan.findSecondUtil import removeInvalidItems def findSecond(l): l = removeInvalidItems(l) length = len(l) if length == 0: print("there is no number item in the list") return None elif length == 1: print("there is only one number item, it's ", l[0]) return None elif length > sys.maxsize: print("out of scope") return None largest=max(l) largest_count=l.count(largest) while largest_count>0: #remove all the largest item l.remove(largest) largest_count-=1 if len(l)==0: return None else: return max(l)
方式三:运用内嵌的list.sort ,可是要去掉反复元素
import sys from hyang.python3.interview.boyan.findSecondUtil import removeInvalidItems def findSecond(l): removeInvalidItems(l) l2 = list(set(l)) # remove duplicated items l2.sort(reverse=True) length=len(l2) if length>=2: return l2[1] else: return None
方式四:与方式三相近 ,运用内嵌的heapq.nlargest,也必须去掉反复元素
import sys, heapq from hyang.python3.interview.boyan.findSecondUtil import removeInvalidItems def findSecond(l): removeInvalidItems(l) l2 = list(set(l)) # remove duplicated items length = len(l) length = len(l2) if length >= 2: return heapq.nlargest(2, l2)[1] else: return None
最终必须留意程序执行所属途径,见下面的图 ,能够融合自身的配备来调节。