kate | Because Clair needs a friend | Continuous Deployment library
kandi X-RAY | kate Summary
kandi X-RAY | kate Summary
CoreOS Clair is an open source project for the static analysis of vulnerabilities in application containers (currently including appc and docker). Turns out if you throw CoreOS Clair into your Kubernetes cluster, with the help of a friend, Kate will automatically scan all newly launched containers. Kate will also rescan all the images every couple of hours just to let you know if the CVE situation has changed. This allows you to identify vulnerabilities that exist in production, as apposed to fixes that may exit on your upstream platforms.
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Top functions reviewed by kandi - BETA
- Main entry point
- scanContainer scans an image
- scanWorker is a long running routine to scan containers
- GetDefaultIP returns the default gateway address .
- getDefaultGatewayIfaceName returns the default interface name
- initScanWorker initializes scan worker .
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Trending Discussions on kate
QUESTION
Here is my code
...ANSWER
Answered 2021-Jun-14 at 21:50Create a CTE that returns for each Block_id the step of the first John.
Then join the table to the CTE:
QUESTION
I have the following dataframe
...ANSWER
Answered 2021-Jun-14 at 14:44One method uses an updatable CTE:
QUESTION
I am trying to generate an RMarkdown document. I have a list freqsByYear and I would like the user to select from a drop down menu (or some similar method) and this will get stored as Q from here I can pass it to a ggplot function and make the plot as follows.
ANSWER
Answered 2021-Jun-13 at 14:27You could use shiny runtime which allows to create a selectInput and to react to changes to this input with renderPlot:
QUESTION
Trying to figure out how to subtract a constant from a column based on the presence of a value in another DataFrame. For example, if I have the below DataFrame a that contains a column called person name and count:
ANSWER
Answered 2021-Jun-10 at 15:33a["count"] -= a.person.isin(b.person)
QUESTION
I created spreadsheet with Svelte, but it seems like it won't update the cell properly.
I appended the "2" postfix to every cell, and when I edit the content (say to fix first letter in jim to be uppercased) it should end up with two "2" as "Jim22". But it end up with just one "2" as "Jim2".
I also have JSON output to inspect the value of rows, and it changed, so the component was updated correctly, but the editable cell was not updated to "Jim22" it still "Jim2".
Why, and how to fix that?
Code
...ANSWER
Answered 2021-Jun-09 at 10:23Just add the key for each statement. according to document
QUESTION
I have an Access query (qr1) that returns the following data:
dateField stringField1 stringField2 booleanField 11/09/20 17:15 John Nick 0 12/09/20 17:00 John Mary -1 13/09/20 17:30 Ann John 0 13/09/20 19:30 Kate Alan 0 19/09/20 19:30 Ann Missy 0 20/09/20 17:15 Jim George 0 20/09/20 19:30 John Nick 0 27/09/20 15:00 John Mary -1 27/09/20 17:00 Ann John -1 27/09/20 19:30 Kate Alan 0 28/09/20 18:30 Ann Missy -1 03/10/20 18:30 Jim George -1 04/10/20 15:00 John Nick 0 04/10/20 17:15 John Mary 0 04/10/20 20:45 Ann John 0 05/10/20 18:30 Kate Alan 0 17/10/20 15:00 Jim George 0 17/10/20 17:15 John Nick 0 18/10/20 15:00 John Mary -1 18/10/20 17:15 Ann John 0Notes:
- The string data may by repetitive or not.
- The date data are stored as string. I use a function to convert it as date.
ANSWER
Answered 2021-Jun-02 at 22:13One obvious problem is:
QUESTION
I have the following data. It is all in one excel file.
...ANSWER
Answered 2021-Jun-06 at 10:14First, read both Excel sheets.
QUESTION
I'm trying to match job candidates to mentors based on different several variables that would hopefully create a good match. There are two Pandas DataFrames (one for candidates and one for mentors) that I'm trying to connect based on experience, location, desired job, etc.
For example I have a mentor DataFrame that might look something like the below:
...ANSWER
Answered 2021-Jun-04 at 00:08@Henry is on the right path. You'll need to modify your candidate dataframe to a) make sure all arrays are the same length (or add NaNs if you don't have them, and b) tweak a bit to make sure you actually have some matches.
I used your mentor_df, and the following candidate_df:
QUESTION
I have a person table:
...ANSWER
Answered 2021-Jun-01 at 17:28simple group by and conditional aggregation :
QUESTION
users = [
{'id': 0, "name": "Hero"},
{'id': 1, "name": "Dunn"},
{'id': 2, "name": "Sue"},
{'id': 3, "name": "Chi"},
{'id': 4, "name": "Thor"},
{'id': 5, "name": "Clive"},
{'id': 6, "name": "Hicks"},
{'id': 7, "name": "Devin"},
{'id': 8, "name": "Kate"},
{'id': 9, "name": "Klein"},
]
# list of users
friendship_pairs = [(0, 1), (0, 2), (1, 2), (1, 3), (2, 3), (3, 4), (4, 5),
(5, 6), (5, 7), (6, 8), (7, 8),
(8, 9)] # list of friendship pairs
# it is easy to make a dictionary list of friendships for the IDs as the friendship_pairs and written in terms of IDs
# what if I wanted this dict to show for example "Hero": ["dunn","Sue"] instead of 0:[1,2]
friendships = {user['id']: [] for user in users}
for i, j in friendship_pairs:
friendships[i].append(j)
friendships[j].append(i)
display(friendships)
# I made a dictonary of intergers:names
user_Ids = list(user['id'] for user in users)
user_Names = list(user['name'] for user in users)
converter = dict(zip(user_IDS, user_Names))
# How can I use this to turn 'friendship_pairs' into something like 'friendship_pairs_names'
# which has the same information has friendship_pairs but uses their names
ANSWER
Answered 2021-May-30 at 12:11You can try this
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