Power Automate - Changing a column's properties removes it from the schema

T

thanby

Before I go tear apart an entire workflow and app to fix this, I wanted to see if there was a simpler solution.

I have a flow in SPO that needs to set a People column with an "Update Item" action. Originally that column was set to allow multiple people, and that was fine. Then I ran into an issue where I had to switch it to single selections instead. That broke the workflow, naturally. At first it was giving me an error saying the field didn't exist anymore and should be removed from the schema of the update. I just deleted and re-added the "Update Item" action, but now the People column isn't even showing up.

My only idea is to completely rip out and recreate the column, but that would bust various parts of the workflow and its associated app. Any thoughts?
 

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My volatility estimate is off by 30bps. My estimate of VaR is off by 5pts

  • Matheus Popst de Campos
  • Finance
  • Replies: 0
I am using parametric volatility to estimate the 90% confidence 3-month VaR for a 2-stock portfolio. (irl, I'll use much more than 2 stocks). I use here both the trailing volatility to estimate volatility, as also the volatilities of the option market.

The problem: my VaR estimate over an 8-year horizon is worse than the actual VaR just 3.5-4.5% of the time, when we expected it'd be ~10% of the time.

My estimate of the volatility is off by only 30bps on average through the period (but it's wrong with a 370bps standard deviation).

What could be happening here?

XDebug desde servidor Hostinger y con VSCode en local

Tengo un servidor compartido en Hostinger (no VPS) y he instalado la extensión de XDebug. Estoy utilizando VS Code como IDE con la extensión php debug.

Si entro en phpinfo(), puedo ver que XDebug está activo, pero solo en el modo desarrollador y no step debugging (debug). Como no puedo editar el archivo php.ini directamente, estoy realizando la configuración en el archivo .htaccess.

PHP Version 8.1.27 y XDebug Version 3.1.4

Enabled Features (through 'xdebug.mode' setting) Feature Enabled/Disabled
Development Helpers ✔ enabled Coverage ✘ disabled GC Stats ✘ disabled Profiler ✘ disabled Step Debugger ✘ disabled Tracing ✘ disabled

El problema es que aunque ponga los breakpoints para que vaya parando e inicio el debug, no va parando. Además, en la configuración de XDebug -> Step Debugger ✘ disabled

¿Alguna idea?

La configuración en .htaccess que tengo es:

Code: 
# Configuración de XDebug para debugging remoto
php_value xdebug.mode develop,debug
php_value xdebug.start_with_request yes
php_value xdebug.client_host MI_IP_LOCAL
php_value xdebug.client_port 9003
php_value xdebug.idekey VSCODE
php_value xdebug.log MIPATHRAIZ/xdebug.log
php_value xdebug.log_level 1

En Visual studio code:

Code: 
{
    "version": "0.2.0",
    "configurations": [
        {
            "name": "Listen for Xdebug",
            "type": "php",
            "request": "launch",
            "port": 9003,
            "log": true,
            "runtimeArgs": [
                "-dxdebug.mode=debug",
                "-dxdebug.start_with_request=yes"
            ],
            "pathMappings": {
                MIPath: "${workspaceFolder}"
            },
            "ignore": [
                "**/vendor/**/*.php"
            ]
        }
    ]
}

JSON formatting column based upon added column of lookup column

  • Jeroen Voortman
  • Technology
  • Replies: 0
I have two SharePoint lists.

List one has the following columns:

  • Productgroup (text)
  • Min (number, percentage)
  • Max (number, percentage)

List two has columns:

  • a lookup column for Productgroup
  • Added columns for Min and Max connected to the chosen productgroup
  • A column for the actual value (number, percentage)

I would like to have a conditional formatting for the column of the actual value so that

  • If below [Productgroup].[Min] Background turns red.
  • If above [Productgroup].[Max] Background turns orange.
  • If between [Productgroup].[Min] and [Productgroup].[Max] background turns green.

I have tried the following as a start, but that didn't do the job.

Code: 
{
  "$schema": "https://developer.microsoft.com/json-schemas/sp/v2/column-formatting.schema.json",
  "elmType": "div",
  "txtContent": "[$Productgroep.Min]"
  "attributes": {
    "class": "=if(@currentField < [Productgroep.Min],'sp-field-severity--warning', '')"
  },
  "children": [
    {
      "elmType": "span",
      "style": {
        "display": "inline-block",
        "padding": "0 4px"
      },
      "attributes": {
        "iconName": "=if(@currentField < [Productgroep.Min],'Error', '')"
      }
    },
    {
      "elmType": "span",
      "txtContent": "@currentField"
    }
  ]
}

Fourier transform relation for spherical convolution

Let $f$ and $g$ be two functions defined over the 2d sphere $\mathbb{S}^2$.

The convolution between $f$ and $g$ is defined as a function $f * g$ over the space $SO(3)$ of 3d rotations as $$(f*g)(R) = \int_{\mathbb{S}^2} f(R^{-1}x) g(x) \mu(dx)$$ where $\mu$ is the Haar measure on $\mathbb{S}^2$.

One can expand $f$ and $g$ in the orthonormal basis of spherical harmonics $Y_n^m$ for $n\geq0$ and $|m|\leq n$ such that $f = \sum \hat{f}_n^m Y_n^m$ where $\hat{f}_n^m = \langle f, Y_n^m \rangle$ (similarly for $g$).

Is there a Fourier relation for the Fourier transform of $f*g$ as is the case for instance for functions over $\mathbb{R}$, for which the Fourier transform of the convolution is the product of the Fourier transforms.

Note that this supposes to have a notion of Fourier transform for functions on $SO(3)$, which is provided by the Wigner D-matrix instead of the spherical harmonics as the Fourier basis.

2-topos “tasks of theory”

Is it possible to obtain any interesting results by associating with each theory T its 2-topos task(T). Consisting of : 1) Objects: Formal task in the theory of T. Condition-data, and a task from the data used. Tasks are considered as a mapping from solutions to answers. (a set of solutions can be considered as a category because solutions are conclusions from data a1→a2→... →answer and morphism between solutions: a1→a2→...→answer ↓ ↓ ↓ ↓ a1'→a2'→... →answer 2) Morphisms: corresponding commutative diagrams between task if and only if the solution to task A follows from task B terminal object: axiom. Product- as a product of categories. Exponential A and B - a task whose answers are morphisms from A to B.Classifier is the sum of two axioms.

Approximation on $H^1_0(B) and cut-off functions

Let $u \in H^1_0(B)$, where $B$ is the unit ball in $\mathbb{R}^N$. Given $\epsilon > 0$, I need to show there exists a function $\chi_\epsilon \in C^\infty_0(\mathbb{R}^N)$ such that $$ \| u - \chi_\epsilon u\|_{H^1_0(B)} < \epsilon $$ and $\text{supp} \chi_\epsilon \subset \subset B$.

My attemps I considered $f \in C^\infty_0(\mathbb{R}^N)$ such that $\text{supp} f \subset B_{1/4}(0)$ and defined $\chi_\epsilon = 1-\epsilon + \epsilon f.$ The bad part: The support of $\chi_\epsilon$ is all $\mathbb{R}^N$.

Extension of scalars for bounded chain complexes of $kG$-modules

I'm wondering if a generalization regarding a statement from Curtis-Reiner holds. The original statement is as follows:

(30.33) Theorem: Let $R$ and $S$ be complete discrete valuation rings, with $S$ an extension of $R$, and suppose that $R/J(R)$ is a finite field. Then for each indecomposable finitely generated $A$-module $M$, the $SA = S\otimes_R A$-module $SM$ is a direct sum of non-isomorphic indecomposable submodules.

I'd like to know if an analogous statement holds for bounded chain complexes rather than (f.g.) modules. The case I'm particularly interested in is when $R = kG$ and $S = k'G$, where $k'/k$ is an extension of finite fields and $G$ is a finite group. Going through the proof of this theorem, I don't know if the following isomorphism holds anymore: $\operatorname{Hom}_{k'G}(k'\otimes_k C,k' \otimes_k D) \cong k' \otimes_k \operatorname{Hom}_{kG}(C,D)$ where $C$ and $D$ are bounded chain complexes of $kG$-modules (see CR Theorem (2.38) for the module-theoretic statement).

Also, if anyone knows what "MO 2e" refers to in Theorem (2.38) of CR, please let me know. Thank you!
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