namedict | Generate Chinese names for the newborn baby | Translation library
kandi X-RAY | namedict Summary
kandi X-RAY | namedict Summary
Before you use this utility, please try to answer the three questions below. If all answers are "Yes", just go ahead. Congratulations, Mr. Yes! I think this tool will meet your requirements and helps you give your child an Easy-to-Read Chinese name. That's the initial idea, just to make it pronounceable for foreigners, and provide a consistent pronunciation in Chinese and other languages. It's not trying to catch up with the international trend.
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Community Discussions
Trending Discussions on namedict
QUESTION
I'm a bit confused by the following code. I would have expected that only the 'D' value in the 'Bob' outer dictionary would have been changed, but instead all the 'D' items have been updated. Any ideas as to what is going on here?
...ANSWER
Answered 2021-Mar-20 at 14:58Assignment statements in Python do not copy objects, they create bindings between a target and an object. So when you are doing nameDict[n] = zeroItemDict
it means a shallow copy of zeroItemDict
is assigned to every key of nameList
. Any changes made to zeroItemDict
will reflect in all of the copies made.
To avoid this use copy.deepcopy
when you're assigning zeroItemDict
like this nameDict[n] = copy.deepcopy(zeroItemDict)
.
Note: The difference between shallow and deep copying is only relevant for compound objects
QUESTION
What I am trying to do is create a dictionary within a dictionary. It is supposed to be a movie lover's club in which you can add movies to a member's account, but when I try to add it, it gets overwritten. Below is my code:
...ANSWER
Answered 2020-Apr-10 at 16:18Edit2: With your updated code, here's the solution to your problem. You were actually really close, the only issue was that you were using .update
on the movies
dictionary, rather than the movies[name]
dictionary, so you would replace movies[name]
with your new movie dict each time. The solution here is to update movies[name]
instead. Also, I made a change to your option_five
function so that when you add a new member, they have an empty dictionary by default rather than an empty string, so it can be updated:
QUESTION
I have a dictionary of dictionaries, d
:
ANSWER
Answered 2019-Dec-06 at 21:38Using the code with a couple of minor calculations, I was able to get [9.5, 9.0, 9.75, 8.5]
The two changes I made was that I had to cast the output of dictionary d
to a float()
, and I changed nameDict
to d
, since nameDict
was not defined in the example code provided. Here is the resulting code below:
QUESTION
I'm new to Python and trying to wirte a simple password reset code sample to learn about sets, dictionaries, and exception handlers. To reset a password, the program must first take in both the student_id and user_id and confirms their ID against the list. If the student_id and user_id match, it will prompt the user for their admission term as an extra security measure. If all things match, the program will greet the user by name and prompt their to enter their new password. The new password cannot match any of the user's previous passwords. The program should not quit until the user asks to quit or successfully logs in (note, resetting the password should not be the end and we may give the user a "quit" option if they decide they do not want to change their password.).
The program should run similar to the following:
...ANSWER
Answered 2019-Nov-18 at 08:16Ok so quick disclaimer all the code in this is sudo code and is for example to give you the idea of my train of thought.
So that out of the way I see your problems as the following.
- User Input (Actions)
- DataBase Handling
If I am correct on this I would suggest that you take the following approach.
Take care of the Database, by writing a class to handle the getting and setting of specific user in an out of the database that you are using and into a dictionary or any other data structure allows you to forget the structure of the underlying csv etc and will make you actions clearer and simpler later on. something like below would do the trick.
QUESTION
I'm attempting to convert two lists to json.
For example :
...ANSWER
Answered 2019-Jul-10 at 14:35Disclaimer: Since we don't know where the g
comes from, I will only build the root children
array.
Since your arrays have the same length, you can use a plain for
and use with the index to play with both arrays. Just build an array and check each iteration if the "child" already exists. If not, create it.
QUESTION
I'm following the 'Apache Spark with Scala - Hands on with Big Data' course on Udemy.
In one of the lectures, you have to set up an EMR environment and submit a JAR file to the cluster.
When submitting the code, I get the following error.
Edit: code does actually proceed after the error.
ANSWER
Answered 2018-Nov-06 at 13:11spark-submit --class com.sundogsoftware.spark.MovieSimilarities1M /path/to/jar/MovieSimilarities1M-assembly-1.0.jar
you need to provide the class name as well
QUESTION
I have a broadcast variable which is constructed in the following manner
...ANSWER
Answered 2017-Feb-24 at 20:09One way to create a local map is to use collectAsMap
:
QUESTION
- Print a dataset with Movie Name & Number of times it has been rated.
- That's a simple way to get the most "popular" movie
- One file called "u.data" with movieID, userID, ratings, timestamp
- One file called "u.item" with movieID and movie name and information about - each movie
- Create a dictionnary key = MovieID, values = Name from the u.item files
- Broadcast the dictionary to the executioner nodes on the cluster
- Create a rdd with the MovieID and 1 on each line
- Reduce this rdd by movieID and sum each one
- Flip the key(movieID) and the value(Total) to sort the dataset by this total
- Then I should map the movieID with the broadcasted dictionary but I get a syntax error on this line:
sortedMoviesWithNames = sortedMovies.map(lambda (count, movie) : (nameDict.value[movie], count))
This code example is from cookbook for Apache Spark and Python. All others codings exercises work perfectly on my environment. Windows 10 / Canopy / Python 3.5 / Spark 2.3.2
I've checked the broadcasted dictionary it's ok, and already print the sortedMovies RDD which is ok too. I've checked the online errata of the book, nothing too.
I'm wondering if this is a syntax error due to the Python version or something like that.
...ANSWER
Answered 2018-Dec-26 at 21:53I believe the correct syntax for a lambda with multiple arguments is:
QUESTION
I tried my best to look for other answers, but I don't manage to find out how I can do it.
I have a dictionary like this:
...ANSWER
Answered 2018-Nov-10 at 17:17Use dict.get
with list comprehesion
as:
QUESTION
I have been trying to do Problem 22 on Project Euler, but I cannot seem to get the right answer or see any problem with my code. I copied and pasted the contents of the txt file rather than accessing it directly from my code.
...ANSWER
Answered 2018-May-14 at 13:16You are close. There are a few problems with your code:
namedict
is not a dictionary, it is a list.- The requirement is to sort alphabetically by name, not by value.
- You need to use or store indices of your sorted list. I use
enumerate
for this.
Combining these elements:
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Install namedict
On a UNIX-like operating system, using your system’s package manager is easiest. However, the packaged Ruby version may not be the newest one. There is also an installer for Windows. Managers help you to switch between multiple Ruby versions on your system. Installers can be used to install a specific or multiple Ruby versions. Please refer ruby-lang.org for more information.
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