gender | Predict Gender from Names Using Historical Data | Development Tools library
kandi X-RAY | gender Summary
kandi X-RAY | gender Summary
Data sets, historical or otherwise, often contain a list of first names but seldom identify those names by gender. Most techniques for finding gender programmatically rely on lists of male and female names. However, the gender associated with names can vary over time. Any data set that covers the normal span of a human life will require a historical method to find gender from names. This R package uses historical datasets from the U.S. Social Security Administration, the U.S. Census Bureau (via IPUMS USA), and the North Atlantic Population Project to provide predictions of gender for first names for particular countries and time periods.
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gender Examples and Code Snippets
def predict_gender():
"""Predict the gender of the faces showing in the image"""
# create a new cam object
cap = cv2.VideoCapture(0)
while True:
_, img = cap.read()
# resize the image, uncomment if you want to resize
def predict_age_and_gender(input_path: str):
"""Predict the gender of the faces showing in the image"""
# Initialize frame size
# frame_width = 1280
# frame_height = 720
# Read Input Image
img = cv2.imread(input_path)
# re
private static Dataset aggregateYearlySalesByGender(Dataset dataset) {
Dataset aggDF = dataset.groupBy(column("year"), column("source"), column("gender"))
.sum("transaction_amount")
.withColumnRenamed("sum(transaction
Community Discussions
Trending Discussions on gender
QUESTION
I want to produce a plot via R plotly with independent legends while respecting the colorscale.
This is what I have:
...ANSWER
Answered 2022-Mar-19 at 15:21This isn't exactly what you're looking for. I was able to create a meaningful color bar, though.
I removed the call for interaction between the groups and created a separate trace. Then I created legend groups and named them to create separate legends for gender
and age
. When I pull color =
out of the call to create a colorbar, this synced the color scales.
However, it assigns colors to the labels for age and gender and that's not meaningful! There are a few things that don't line up with your request, but someone may be able to build on this information.
QUESTION
ANSWER
Answered 2022-Mar-21 at 14:44Does this work for you?
QUESTION
I Am having the following data in my SSAS
cube.
My need is to get the value of the measure based on two conditions with two different dimensions using the
MDX
.
In this example data, I need to get the
Reseller Sales Amount
value where the value ofTitle
dimension is equal to Sales Representative and the value of theGender
dimension is equal to Male condition.
I have tried to achieve the requirement with the Case statement and IIF() function available in the MDX
but it is not working.
Please find the queries I have tried with different functions.
Using Case statement:
...ANSWER
Answered 2022-Feb-25 at 12:03You can use a Tuple either directly in the calculated measure:
QUESTION
I've been working on some survey data using the survey
package. I read the documentation available on post-stratification and calibration, however I got stuck trying to calibrate the sampling weights on a total known for the population that is not the population total.
To make my self clear I prepared an example: Let's say I have income information for a sample stratified by sex, which lets me create the svydesign
object:
ANSWER
Answered 2022-Feb-17 at 04:09Here is a workaround.
all your data is stored at dis$variables
, from there you can export it and make your calculations. I hope this can inspire better solutions
QUESTION
I have the following table in R which lists a person race, gender, age, and cholesterol test. age and cholesterol test are displayed as dummy variables. age can be categorized as low, medium, or high, while cholesterol tests can be categorized as low or high. I want to transform the age and cholesterol columns to be single columns where low is categorized as 1, medium is categorized as 2, and high is categorized as 3. Cholesterol test can be neigh low or high if a person never took one and should be N/A in the expected output. I want the solution to be dynamic so that if I have multiple columns in this format, the code would still work (i.e. there may be some new tests, which can be categorized as high, low, or medium as dummy variables).
How can I do this in R?
input:
...ANSWER
Answered 2022-Jan-03 at 18:33Perhaps this helps
QUESTION
I am a bit stuck, I have a working function that can be utilised using .apply()
, however, I cannot seem to get it to work with .assign()
. I'd like this to work with assign, so I can chain a number of transformations together.
Could anyone point me in the right direction to resolving the issue?
This works
...ANSWER
Answered 2021-Dec-14 at 17:39From the documentation of DataFrame.assign
:
DataFrame.assign(**kwargs)
(...)
Parameters **kwargs : dict of {str: callable or Series}
The column names are keywords. If the values are callable, they are computed on the DataFrame and assigned to the new columns. The callable must not change input DataFrame (though pandas doesn’t check it). If the values are not callable, (e.g. a Series, scalar, or array), they are simply assigned.
This means that in
QUESTION
I have a controller that accepts ObjectNode
as @RequestBody
.
That ObjectNode
represents json
with some user data
ANSWER
Answered 2021-Nov-28 at 12:22Register Jackson ParameterNamesModule, which will automatically map JSON attributes to the corresponding constructor attributes and therefore will allow you to use immutable classes.
QUESTION
I am trying to create an R Shiny app that calculates a score using ridge regression and then uses that in a random forest model. I saved both models as RDS and kept them in the same folder where the app.R is.
Then I read the models and data of predicted probabilities and define some functions:
...ANSWER
Answered 2021-Oct-11 at 01:32The orignial error is occuring because you are not asking for the same input$id
as the id
you assign in textInput
.
QUESTION
I'm working on a chat application using node js that displays a list of online users. Each user has the following attributes:
ipv4, rank, score, login time
ipv4 is used when accessing a user.
Users should be listed according to this sorting strategy:
- Users with higher rank should be at the top of the list.
- If the user rank is equal, the users with the highest score should be at the top of the list after high-rank users.
- If the user score is equal, users with less login time should be at the top of the list after high-score users.
for example:
1. Login userA (ipv4:1.1.1.1, rank:0, score:3000, loginTime:1631966424070)
- list: [userA]
2. Login userB (ipv4:2.2.2.2, rank:1, score:2000, loginTime:1631966424080)
- list: [userB, userA]
3. Login userC (ipv4:3.3.3.3, rank:1, score:1000, loginTime:1631966424090)
- list: [userB, userC, userA]
4. Login userD (ipv4:4.4.4.4, rank:0, score:3000, loginTime:1631966424100)
- list: [userB, userC, userA, userD]
Access example:
Get user(ipv4:3.3.3.3) -> return userC
There are many users and I use the functional-red-black-tree data structure to improve performance. In this data structure, each item has a key and a value. The constructor of this tree takes a comparator function -like array sort comparator- with two parameters, each of which are keys, as a parameter.
I have defined the user as follows:
...ANSWER
Answered 2021-Sep-18 at 16:06In your rand
example, it is normal that the key (10, 0, 0, 0) will not match, even though ipv4=10 is somewhere in the tree. This is because that node will be more like (10, 66, 35, 19)... which will not be encountered, as the ordering of the tree is mainly based on rank, not on ipv4. The decision to where to drill down in the tree will be based on rank (0), moving to the left side of the tree, while the node you intend to find may be at the right (higher rank 66).
What you can do is to accompany the tree with a Map
, which will map an ipv4 value with an existing Key
instance. So every time you insert in the tree, also add the corresponding mapping in that Map. And every time you want to find or want to delete, first retrieve the Key
instance from that Map by the ipv4 value you have, and pass that Key
instance to the tree's get
or remove
method.
Here is a wrapper class that makes this combination. Extend it with any other method you want to include:
QUESTION
I am using Angular, Passport and sessions for authentication. I can create a new user, login but,
Problem
when checking if the user is logged using Passport's isAuthenticated() function always returns false. I followed the instructions from various online resources and I am unsure if:
Question
I can use passport and sessions with my Angular app? or I am just doing something wrong in my code? I would really appreciate it if you guys have any idea/hint why it's not working. Thank you.
Backend code
...ANSWER
Answered 2021-Sep-13 at 15:00I'm just guessing what's wrong because you didn't precise the port of your application in your question, but I assume it's 4200 as I see that you have added these lines to the original project:
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