unique_words = list(set(words)) print(unique_words) 问题3:列表为空 原因:列表中没有元素。 解决方法:在使用列表之前,检查列表是否为空。 代码语言:txt 复制 if not words: print("列表为空") else: print(words) 参考链接 Python官方文档 - 列表 Python官方文档 - 字符串 希望这些信息对你有所帮助!如...
When it comes to sorting, there’s no shortage of solutions. In this section, we’ll cover three of my favorite ways to sort a list of strings in Python.Sort a List of Strings in Python by Brute ForceAs always, we can try to implement our own sorting method. For simplicity, we’ll...
unique_lengths = map(len,strings) #{} unique_lengths = set(map(len,strings)) #{1, 2, 3, 4, 6} unique_lengths_dict = dict(map(len,strings)) #error unique_lengths_list = list(map(len,strings)) #error 作为一个字典推导式的例子,我们可以创建一个字符串的查找映射表以确定它在列表中的...
Python Code: # Define a function called 'unique_values_in_list_of_lists' that extracts unique values from a list of lists.defunique_values_in_list_of_lists(lst):result=set(xforlinlstforxinl)# Flatten the list of lists and create a set to remove duplicates.returnlist(result)# Convert th...
This time we're using a period as a delimiter and using addr to store our list of octets. But there's one problem. We have a bunch of strings. We're going to have to do math with this to calculate things such as the broadcast parameters, the netmask, and, ultimately, our IPs. So...
类型变量classStack(Generic[T]):def__init__(self):self._container:List[T]=[]defpush(self,item:T)->None:self._container.append(item)defpop(self)->T:returnself._container.pop()# 使用stack_of_ints=Stack[int]()stack_of_ints.push(1)stack_of_strings=Stack[str]()stack_of_strings.push...
交集: np.intersect1d(s, t, assume_unique=True) # 返回排序的、去重的两个list的交集,尽可能保证传入的两个list是去重的,这可以加快运算速度。 差集: np.setdiff1d(s, t, assume_unique=True) # 返回排序的,去重的差集,assume_unique参数同上。
everything = [] for chunk in list_of_lists: everything.extend(chunk) 要比串联方法快: everything = [] for chunk in list_of_lists: everything = everything + chunk 排序 你可以用 sort 函数将一个列表原地排序(不创建新的对象): In [61]: a = [7, 2, 5, 1, 3] In [62]: a.so...
If sep is not specified or is None, any whitespace string is a separator and empty strings are removed from the result. """ return [] def splitlines(self, keepends=False): """ 根据换行分割 """ """ S.splitlines(keepends=False) -> list of strings Return a list of the lines in S,...
words (list): A list of strings containing the words of the input text. Returns: vocabulary (list): A list of every unique character in the list of input words. '''vocabulary =list(set(''.join(words)))returnvocabularydeffind_pair_frequencies(self, corpus):''' Find the frequency of ea...