I have a DataFrame using pandas and column labels that I need to edit to replace the original column labels. I'd like to change the column names in a DataFrame A where the original column names are: ['$a', '$b', '$c', '$d', '$e'] to ['a', 'b', 'c', 'd', 'e']. I have the edited column names stored it in a list, but I don't know how to replace the column names. 只需将其分配给.column
我有一个使用熊猫和列标签的DataFrame，我需要编辑以替换原始列标签。 我想更改DataFrame A中的列名称，其中原始列名称是： ['$a', '$b', '$c', '$d', '$e'] 至 ['a', 'b', 'c', 'd', 'e']. 我将编辑过的列名存储在列表中，但我不知道如何替换列名。 只需将其分配给.columns属性即可： >>> df = pd.DataFrame({'$a':[1,2], '$b': [10,20]}) >>> df.columns = ['a', 'b'] >>> df a b 0 1
I have the following problem I have a dataframe master that contains sentences, such as master Out[8]: original 0 this is a nice sentence 1 this is another one 2 stackoverflow is nice For every row in Master, I lookup into another Dataframe slave for the best match using fuzzywuzzy . I use fuzzywuzzy because the matched sentences between the two dataframes could d
我有以下问题 我有一个包含句子的数据框大师 ，比如 master Out[8]: original 0 this is a nice sentence 1 this is another one 2 stackoverflow is nice 对于Master中的每一行，我使用fuzzywuzzy查找另一个Dataframe 从站以获得最佳匹配。 我使用fuzzywuzzy因为两个数据框之间的匹配句子可能会有点不同（额外的字符等）。 例如， 奴隶可以 slave Out[10]: my_value
According to my interpretation of Python 2.7.2 documentation for Built-In Types 5.7 Set Types, it should be possible to remove the elements of set A from set B by passing A to set.remove(elem) or set.discard(elem) From the documentation for 2.7.2: Note, the elem argument to the __contains__() , remove() , and discard() methods may be a set. I interpret this to mean that I can pass a set to
根据我对内置类型5.7设置类型的Python 2.7.2文档的解释，应该可以通过传递A给set.remove(elem)或set.discard(elem)来从集合B中移除集合A的元素。 从2.7.2的文档： 请注意， __contains__() ， remove()和discard()方法的elem参数可能是一个集合。 我解释这意味着我可以通过一个set来remove(elem)或discard(elem)并且所有这些元素将从目标集中移除。 我会用这个来做一些奇怪的事情，比如从字符串中删除所有的元音，或者从
I came across a question when I solve this LeetCode problem. Although my solution got accepted by the system, I still do not have any idea after searching online for the following question: What is the time complexity of dict.keys() operation? Does it return a view of the keys or a real list (stores in memory) of the keys? In Python 2, it's O(n), and it builds a new list. In Python 3, it
当我解决这个LeetCode问题时，我遇到了一个问题。 尽管我的解决方案已被系统接受，但在线搜索下列问题后，我仍然没有任何想法： What is the time complexity of dict.keys() operation? 它是否会返回密钥的视图或密钥的真实列表（存储在内存中）？ 在Python 2中，它是O（n），它构建了一个新列表。 在Python 3中，它是O（1），但它不返回一个列表。 要从字典的keys绘制一个随机元素，您需要将其转换为列表。 这听起来像
I have a set in Python from which I am removing elements one by one based on a condition. When the set is left with just 1 element, I need to return that element. How do I access this element from the set? A simplified example: S = set(range(5)) for i in range(4): S = S - {i} # now S has only 1 element: 4 return ? # how should I access this element # a lame way is the following # for e i
我有一set在Python从我删除基于条件的一个因素之一。 当这个集合只剩下1个元素时，我需要返回该元素。 我如何从集合中访问这个元素？ 一个简单的例子： S = set(range(5)) for i in range(4): S = S - {i} # now S has only 1 element: 4 return ? # how should I access this element # a lame way is the following # for e in S: # return S 使用set.pop ： >>> {1}.pop() 1 >>> 就你而言，
I am trying to understand algorithms by writing them myself. While trying to replicate merge sort I have run into some trouble: left & right return none-type and an error is raised for the len(left) in the first while loop. I have been fighting with the code and could not figure out what I was missing? Shouldn't it just loop until the size of left and right lists reduce to 1 which woul
我试图通过自己编写算法来理解算法。 在尝试复制合并排序时，我遇到了一些麻烦：left＆right return none-type，并在第一个while循环中为len（左）引发错误。 我一直在与代码战斗，无法弄清楚我错过了什么？ 它不应该只是循环，直到左右列表的大小减少到1，这将允许他们离开if循环并继续下一部分的功能？ def merge_sort(A): if len(A) < 2: return A else: mid= len(A)//2 left= merge_
I understand the basic principle of a recursion, and it is not confusing at all when dealing with simple things such as calculating a number's factorial, which is the most basic usage of a recursion. However, when you start using Recursion in more complicated environments, it starts getting really really confusing for me. In my case, I wanted to create a function which used the "merge
我理解递归的基本原理，并且在处理简单的事情时不会感到困惑，例如计算数字的阶乘，这是递归的最基本用法。 但是，当您在更复杂的环境中开始使用递归时，它开始变得对我非常困惑。 在我的情况下，我想创建一个函数，它使用“合并排序”方式使用递归对列表中的项目进行排序。 所以我开始工作了，这就是我第一次尝试时想到的（不是我必须修正一些错别字）： def merge_sort(ls): size = len(ls) if size <= 1: return
Follow up from this question. How can I modify this code to allow for rewinding of multiple lines? class rewindable_iterator(object): not_started = object() def __init__(self, iterator): self._iter = iter(iterator) self._use_save = False self._save = self.not_started def __iter__(self): return self def next(self): if self._use_save:
从这个问题跟进。 我如何修改此代码以允许重绕多行？ class rewindable_iterator(object): not_started = object() def __init__(self, iterator): self._iter = iter(iterator) self._use_save = False self._save = self.not_started def __iter__(self): return self def next(self): if self._use_save: self._use_save = False else:
I have a generator function that takes iterator object and performs some logic on every item. This runs on a larger list of iterations. It then yields the result back to the calling code, so it can break the for and while loop def func(it): item = next(it) item = item.execute() yield item it = iter(range(1, 10)) condition = True while condition: for item in func(it):
我有一个生成器函数，它接受迭代器对象并对每个项目执行一些逻辑。 这在更大的迭代列表上运行。 然后，它产生的结果返回给调用代码，因此它可以打破for ，并while循环 def func(it): item = next(it) item = item.execute() yield item it = iter(range(1, 10)) condition = True while condition: for item in func(it): condition = item print condition 在Python IDLE中执行此代码，
I am practicing printing out items yielded by a generator function. This works perfectly fine: def fibonacci(): a, b = 0, 1 while True: yield a a, b = b, a + b sumfib = 0 for i in fibonacci(): if i >= 4000000: break elif i % 2 == 0: sumfib += i print(sumfib) But the following code is yielding: list1 = ['f', 'o', 'o', 'b', 'a', 'r'] list2
我正在练习打印由发电机功能产生的物品。 这工作得很好： def fibonacci(): a, b = 0, 1 while True: yield a a, b = b, a + b sumfib = 0 for i in fibonacci(): if i >= 4000000: break elif i % 2 == 0: sumfib += i print(sumfib) 但是下面的代码产生了： list1 = ['f', 'o', 'o', 'b', 'a', 'r'] list2 = ['hello', 'world'] list3 = ['1', '2', '3', '4'] de