bcos | BCOS平台（Be Credible , Open & Secure） | Cryptography library

I have a data frame in which every column consists of number followed by text, e.g. 533 234r/r.

The following code to get rid off text works well:

I am new to Stackoverflow and I am trying to build a Candle stick Chart using Bokeh from python. I understand that there is a sample code in the Bokeh doc. I have tried to use the sample code and rebuild my graph. I have pasted my code below with arbitrary data for simplicity.

I have the following questions:

1. How do we format the graph to have YYYY-MM-DD on the x-scale
2. Can someone explain the W formula. I know we are getting half of day in m second. But Why ?
3. When the code is executed, the date for the hover for y axis does not display for doji chart (ie meaning when open and close are the same). (this could be bcos there is no vbar) But How do we overcome it ?

Appreciate if someone can help :) enter image description here My code below:

I have a simple mysql procedure below to delete data from a table based on an argument passed. I am getting foreign constraint failed exception as the dependent data is not deleted. I am forming a list of IDs to delete as i want to avoid full table scan on the table when using sub queries. Any help is appreciated.

I am able to validate a single JSON object, but I want to validate an array of JSON objects like below, and console the invalid Pincode city name:

I'm failing to run (or compile) a simple script due to a reportedly missing Text.Pandoc.JSON dependency. This question is similar to another one: Pandoc 'Could not find module ...' for installed module

Here's the script:

I was trying to write a function for the differential equation

y'' + y = bcos(omega *t)

Here is my code

How to count a specific word present in both the columns at once?

ab <- data.frame(one = c("abcd","efg","ijk"), two = c("abcd mmmk","abcd qrst","ijk"))

I need to get the count of abcd only if present in both the columns one and two.

eg: Expected result: count of abcd = 1,

bcos only first row has got both abcd but second row cannot be added since only one row got abcd but the code i tried below gives result as "2"

Please help me to achieve this

Code:

length(grep('abcd',ab$one) & grep('abcd', ab$two))

I I want to compute the permutations 5p2, 5p3, 5p4 and 5p5 from the array [1,2,3,4,5] The function below only runs 5p5. To run 5p2, 5p3, 5p4 I will have to manually iterate through the array using a for...loop. Please help me.