How to refer to a group without ambiguity in a regex?

Suppose I want to realize a replacement in a string: "aabbcc11dd ee" -> "aabbcc112dd ee", I can use a regex like this within StringReplace:

RegularExpression["[a-z](\d+)[a-z]"] -> "$  12" 

well, if only the right-hand side "$ 12" was interpreted as “the group 1 with the digit 2” instead of “the group 12”.

So, in Wolfram’s regex, is there any syntax setup to eliminate this ambiguity? I know that, for example, in Python there is something like '\g<i>', which, because of the angled brackets, allows no room for ambiguity.

extract email from facebook group

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MYSQL get the turnover , income, total expenses from my database and group them by year-month

I have 4 tables :

Products - to record products spending - to record expenses directv  - to record subscription payments to TV channels sales    - to record sales 

Products has 4 columns : idProd is the primary key

+--------+-----------+--------------+-------------+ | idProd | nameProd  | sellingPrice | buyingPrice | +--------+-----------+--------------+-------------+ |      1 | Iphone 8  |          500 |         400 | |      2 | Samsung 2 |          600 |         400 | +--------+-----------+--------------+-------------+ 

Spending – 3 columns and idSpd is the primary key

+-------+--------+------------+ | idSpd | amount | dateSpd    | +-------+--------+------------+ |     1 |   1000 | 2018-11-01 | |     2 |   1000 | 2018-11-01 | |     3 |   1000 | 2018-10-01 | |     4 |   4000 | 2018-10-01 | +-------+--------+------------+ 

Sales – 5 columns with idSale as primary key and idProd to link it to the product’s table

+--------+--------+--------------+----------+------------+ | idSale | idProd | sellingPrice | quantity | dateSale   | +--------+--------+--------------+----------+------------+ |      1 |      1 |          700 |        2 | 2018-11-01 | |      2 |      1 |          700 |        5 | 2018-11-15 | |      3 |      2 |          800 |        2 | 2018-11-16 | +--------+--------+--------------+----------+------------+ 

and directv :

+-------+-----------------+-------+------------+ | idDtv | brand           | price | dateDtv    | +-------+-----------------+-------+------------+ |     1 | channel decoder |   150 | 2018-11-09 | +-------+-----------------+-------+------------+ 

and I wish to get for example :

income|gain of product  |  turnover | Spending | DirecTv | month | year 1000  |     400         | 5500      | 3000     | 50      | 10     |2018 500   |     200         |  1000     | 2000     | 0       | 11     |2018 

my queries :

--for directv select sum(dv.price) , month(dv.dateDtv), year(dv.dateDtv) from directv dv GROUP by year(dv.dateDtv) , month(dv.dateDtv)   --for turnover select sum(sl.quantity*sl.sellingPrice) , month(sl.dateSale) , year(sl.dateSale) from sales sl GROUP by year(sl.dateSale) , month(sl.dateSale)  --for spending select sum(spd.amount) , month(spd.dateSpd) , year(spd.dateSpd) from spending spd GROUP by year(spd.dateSpd) , month(spd.dateSpd)   --for gain of product SELECT sum(s.quantity*(s.sellingPrice-p.buyingPrice)) from sales s JOIN products p on s.idProd = p.idProd GROUP by year(s.dateSale) , month(s.dateSale)  

and income = gain of product + directv – spending

but i want joins all these queries and calculate the income , someone could help me ? it was okay but i was unable to combine with my new table product today

Numbers where there is a unique group with integral character table

Call a number $ n$ special in case there is a unique group of order n whose character table consists of only integers. This should be equivalent to the condition that the number of conjugacy classes coincides with the number of conjugacy classes of cyclic subgroups.

Questions: What is the sequence of special numbers? It starts with 1,2, 4, 6, 12, 18, 54,120. Is it an infinite sequence?

For which $ m$ is $ m!$ special ?(note that the symmetric group has integral character table and thus $ n!$ is a good candidate for being special)

[Side question: For which m are the groups of order m! classified?]

What are the corresponding unique groups of order $ n$ for $ n$ being special?

The sequence of such groups starts with $ \mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_2 \times \mathbb{Z}_2, S_3,D_{12}, (\mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2,(\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2, S_5$

How do I change a Facebook Group name that I’ve set it up via a Facebook page

I have a facebook page, and I set up a group, but want to rename it. When I google how to change the name of it, all the answers are for changing a group name set up by a person, not a page (and thus the options they mention aren’t available when I try).

Does anyone know how to do this?, And if not how to delete a group set up by a facebook page?

Many thanks,

How large can a symmetric generating set of a finite group be?

Let $ G$ be a finite group of order $ n$ and let $ \Delta$ be its generating set. I’ll say that $ \Delta$ generates $ G$ symmetrically if for every permutation $ \pi$ of $ \Delta$ there exists $ f:G\rightarrow G$ an automorphism of $ G$ such that $ f\restriction\Delta=\pi$ .

How large can $ \Delta$ be with respect to $ n$ ? Specifically what’s the asymptotic behaviour? Is there a class of groups (along with their symmetric generating sets) of unbounded order such that $ n$ is polynomially bounded by $ |\Delta|$ . Has any other research been done on these generating sets?

[ Celebrities ] Open Question : Has there been multiple breakout soloists who came from one group? why is it that often only one member excels?

i’ve noticed in bands/group with multiple people, usually only one of the members ends up hitting big (beyonce from destiny’s child, justin timberlake from nsync, it’s the same in foreign countries too, for example taeyeon from girl’s generation, etc.) has there ever been more than one huge star coming from the same group? honestly the closest i can think of this happening was michael and janet jackson, but even so michael jackson is far more successful. i’m wondering why this is seemingly so impossible.