listLen =len(lst)iflistLen <=1:returnTrue#由首个元素和末尾元素猜测可能的排序规则iflst[0] < lst[-1]:#列表元素升序key =lambdadif: dif >=0else:#列表元素降序或相等key =lambdadif: dif <=0threshold, sortedFlag =10000,TrueimportnumpyiflistLen <= thresholdorlistLen <= randLen*2ornotrandN...
element_to_check= 3ifbisect_left(sorted_list, element_to_check):print(f"{element_to_check} 存在于列表中。")else:print(f"{element_to_check} 不存在于列表中。")7. 使用 numpy 库 对于数值型列表,numpy 提供了强大的数组操作,包括成员判定。importnumpy as np#使用 numpy 库element_to_check = 3...
可以使用sorted()方法。 def check_if_anagram(first_word, second_word): first_word = first_word.lower() second_word = second_word.lower() return sorted(first_word) == sorted(second_word) print(check_if_anagram("testinG", "Testing")) # True print(check_if_anagram("Here", "Rehe")) ...
if "a" in Tuple: print("Yes, 'a' is present") # Yes, 'a' is present if "p" not in Tuple: print("No, 'p' is not present") # No, 'p' is not present 5. Sorting a Tuple 使用语言内置sorted()方法对元组内的元素进行排序。 排序元组 Tuple = ("a", "c", "b", "d", "f...
, text_vector_b in new_vectors: sim_score =similarity(text_vector_a, text_vector_b)[0][1] student_pair= sorted((student_a, student_b)) score = (student_pair[0], student_pair[1],sim_score) plagiarism_results.add(score) return plagiarism_results for data in check_plagiarism(): print...
d = {'c':3, 'd':4, 'b':2, 'a':1}sorted(d.items())#=> [('a', 1), ('b', ...
If you want to sort a sentence by words, then you can use Python’s .split() method: Python >>> string_value = "I like to sort" >>> sorted_string = sorted(string_value.split()) >>> sorted_string ['I', 'like', 'sort', 'to'] In this example, you use .split() to co...
Return False if the counts do not match for any element, True otherwise. Sample Solution: Python Code: # Define a function to check if two lists contain the same elements regardless of order.defcheck_same_contents(nums1,nums2):# Loop through the set of elements in the combined lists.for...
排序:字典可使用函数sorted()并且指定键或值,进行升序或降序排序;集合排序直接调用 sorted(set) 即可。 合并字典 代码语言:javascript 代码运行次数:0 运行 AI代码解释 # 两个字典 dict1={'a':10,'b':8}dict2={'d':6,'c':4}# 方法一: dict1.update(dict2)print(dict1)# 方法二: ...
a=[1,2,3,4,5]b=[]foriina:ifi%2==0:b.append(i) 那么如果使用filter的话,使用filter函数使得代码变得更简洁: 代码语言:python 代码运行次数:1 运行 AI代码解释 a=[1,2,3,4,5]defcheck(i):returni%2==0b=list(filter(check,a))