Truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()

Having issue filtering my result dataframe with an or condition. I want my result df to extract all column _var_ values that are above 0.25 and below -0.25. This logic below gives me an ambiguous truth value however it work when I split this filtering in two separate operations. What is happening here? not sure where to use the suggested a.empty(), a.bool(), a.item(),a.any() or a.all() . re

一个系列的真值不明确。 使用a.empty,a.bool(),a.item(),a.any

有问题过滤我的结果数据框or条件。 我想要我的结果df提取大于0.25且小于-0.25的所有列_var_值。 下面的逻辑给我一个模棱两可的真值,但是当我在两个单独的操作中分割这个过滤时,它是有效的。 这里发生了什么? 不知道在哪里使用建议的a.empty(), a.bool(), a.item(),a.any() or a.all() 。 result = result[(result['var']>0.25) or (result['var']<-0.25)] 的or与and Python的语句需要truth -值。 对于pandas这

What do * and ** before a variable name mean in a function signature?

Possible Duplicate: Understanding kwargs in Python I have read a piece of python code, and I don't know what does * and ** mean in this code : def functionA(self, *a, **kw): // code here I just know about one use of *: extract all attribute it has to parameter of method or constructor. If this true for above function, so what does the rest : ** ? Inside a function header : * co

变量名称在函数签名中的含义是什么?

可能重复: 在Python中理解kwargs 我已经阅读了一段python代码,并且我不知道*和**在这段代码中的含义: def functionA(self, *a, **kw): // code here 我只知道*的一种用法:提取它具有的方法或构造函数参数的所有属性。 如果这对于上述功能是正确的,那么剩下的是什么:**? 在函数头里面: *收集元组中的所有位置参数 **收集字典中的所有关键字参数 >>> def functionA(*a, **kw): print(a)

How can I slice a list till end with negative indexes

I learned that Python lists can also be traversed using negative index, so I tried to slice/sublist a list using negative index, but I cannot slice it till end. My list is: areas = ["hallway", 11.25, "kitchen", 18.0, "living room", 20.0, "bedroom", 10.75, "bathroom", 9.50] Knowing that slicing syntax is [start:end] and end index is not calculated, I did upstairs = areas[-4:0] but this doesn&#

我怎样才能切分列表,直到消极索引结束

我了解到Python列表也可以使用负向索引遍历,因此我试图使用负向索引对列表进行切片/子列表,但是我无法切片到最后。 我的清单是: areas = ["hallway", 11.25, "kitchen", 18.0, "living room", 20.0, "bedroom", 10.75, "bathroom", 9.50] 知道切片语法是[start:end]和结束索引不计算,我做upstairs = areas[-4:0]但这并不给我列表的最后一个元素。 areas[-4:0]转换为areas[len(areas) - 4: 0] ,该区域从较高索引切换到

Python Slice operator: Negative start combined with Positive stop

I began to work with Python and started to concern about the slicing operator with negative values. Let me show my problem with the following code example: a = [1,2,3,4,5] a[-1] # a) returns [5], which is obvious a[-1:1] # b) returns an empty list [], but why? So what I already know from other questions here on StackOverflow is, how slicing works with a negative start index and an omitted st

Python Slice运算符:Negative start与Positive stop结合使用

我开始与Python合作,开始关注具有负值的切片运算符。 让我用下面的代码示例显示我的问题: a = [1,2,3,4,5] a[-1] # a) returns [5], which is obvious a[-1:1] # b) returns an empty list [], but why? 所以我在StackOverflow的其他问题上已经知道,切片如何与负开始索引和省略的停止索引一起工作。 但我的问题是:为什么情况b)(负面启动和正面停止)不会返回类似[5, 1] ? 因为我选择了列表的最后一个元素,并且选

What are the default values for extended slice with negative and positive step

This question already has an answer here: What are the default slice indices in Python *really*? 6 answers Why does list[::-1] not equal list[:len(list):-1]? 2 answers

具有负面和正面步骤的扩展切片的默认值是多少?

这个问题在这里已经有了答案: Python中的默认切片索引*真*是什么? 6个答案 为什么list [:: - 1]不等于list [:len(list): - 1]? 2个答案

List Slicing in Python Without Using Built

I'm very new to programming and am currently stuck on a practice problem involving list slicing without using the built-in function, so the idea is that you create your own slice function to be called upon by the user. Below is the question and following that is my attempt to solve it. Any help is appreciated. Thank you! Python provides slicing functionality for lists, but for this quest

在不使用内建的情况下在Python中列表切片

我对编程非常陌生,目前我坚持一个涉及列表切片而不使用内置函数的实践问题,所以我们的想法是创建自己的切片函数供用户调用。 以下是我试图解决的问题。 任何帮助表示赞赏。 谢谢! Python为列表提供了切片功能,但对于这个问题,您将实现自己的能够生成列表切片的功能(注意:您不能在解决方案中使用切片操作符)。 该函数应该被称为slice,并按照以下特定顺序接受以下三个输入: 将从中创建切片的列表,来源。 这个

Is list[i:j] guaranteed to be an empty list if list[j] precedes list[i]?

The Python tutorial explains slice behavior when indices are negative, but I can't find documentation describing the behavior when the end index precedes the start index. (I've also looked at Explain Python's slice notation, and perhaps I'm not reading carefully enough, but the answers there don't seem to address this point.) The behavior that I observe is that an empty lis

如果list [j]在list [i]之前,list [i:j]保证是一个空列表吗?

Python教程解释了索引为负数时的切片行为,但我无法找到描述当结束索引位于开始索引之前时的行为的文档。 (我也看过解释Python的切片符号,也许我没有仔细阅读,但那里的答案似乎没有解决这一点。) 我观察到的行为是返回一个空列表,这对我来说似乎是合理的。 然而,对我来说这似乎是合理的,因为它会以相反顺序返回i和j之间的项目列表,或者仅仅引发异常。 如果list[j]在list[i]之前, list[i] list[i:j]保证是一个空列

Python list extend functionality using slices

I'm teaching myself Python ahead of starting a new job. Its a Django job, so I have to stick to 2.7. As such, I'm reading Beginning Python by Hetland and don't understand his example of using slices to replicate list.extend() functionality. First, he shows the extend method by a = [1, 2, 3] b = [4, 5, 6] a.extend(b) produces [1, 2, 3, 4, 5, 6] Next, he demonstrates extend by s

Python列表使用切片扩展功能

在开始一项新工作之前,我正在自学Python。 它是Django的工作,所以我必须坚持2.7。 因此,我正在阅读Hetland的Beginning Python,并且不理解他使用slice来复制list.extend()功能的例子。 首先,他展示了extend方法 a = [1, 2, 3] b = [4, 5, 6] a.extend(b) 产生[1, 2, 3, 4, 5, 6] 接下来,他演示了通过切片延伸 a = [1, 2, 3] b = [4, 5, 6] a[len(a):] = b 它产生与第一个例子完全相同的输出。 这个怎么用? A

slice/stride by variable?

background I'm attempting to code a basic letter game in python. In the game, the computer moderator picks a word out of a list of possible words. Each player (computer AI and human) is shown a series of blanks, one for each letter of the word. Each player then guesses a letter and a position, and are told one of the following: That letter belongs in that position (the best outcome)

切片/跨度由变量?

背景 我试图在python中编写一个基本的字母游戏。 在游戏中,计算机主持人从可能的单词列表中选出一个单词。 每个玩家(计算机AI和人)都会显示一系列空白,每个字母都有一个空格。 每个玩家然后猜测一个字母和一个位置,并被告知以下之一: 那封信属于那个位置(最好的结果) 那封信是在这个词里,但不是在那个位置上 这封信不在任何剩余的空白处 当这个词已经完全显示出来时,玩家正确猜出最多的字母会赢得一分。

Python lists/arrays: disable negative indexing wrap

While I find the negative number wraparound (ie A[-2] indexing the second-to-last element) extremely useful in many cases, when it happens inside a slice it is usually more of an annoyance than a helpful feature, and I often wish for a way to disable that particular behaviour. Here is a canned 2D example below, but I have had the same peeve a few times with other data structures and in other nu

Python列表/数组:禁用负向索引换行

虽然我发现在很多情况下,负片数环绕(即A[-2]索引倒数第二个元素)非常有用,但当它发生在片内时,通常比有用的功能更麻烦,而且我经常希望有一种方法来禁用该特定行为。 下面是一个2D的封装示例,但是我已经和其他数据结构以及其他维度数量相同。 import numpy as np A = np.random.randint(0, 2, (5, 10)) def foo(i, j, r=2): '''sum of neighbours within r steps of A[i,j]''' return A[i-r:i+r+1, j-r:j+r+1].sum()