Calculating the probability that the Signature column has the status “Y” in the last 30 days
0
$begingroup$
I am doing this formula in R:
for (i in 1:ifinal)
{
df$FPRateSignature30d[i] <-
length(
which(
df$status == "Y" &
` df$Signature==df$Signature[i]& df$date>= (as_datetime(df$date[i])duration(30, days')) &` df$date <
as_datetime(df$date[i])
)
) / length(which(
df$Signature == df$Signature[i] &
df$date >= (as_datetime(df$date[i]) - duration(30, 'days')) &
``df$date < as_datetime(df$date[i])
))
}
but I have ~350K of rows in the real data, and takes me a lot doing this with for loops.
Here it is the data sample:
df <- data.frame(c("1","2","3","4","5","6"),c("Attribute1", "Attribute1", "Attribute2", "Attribute2", "Attribute2", "Attribute1"),
c("2018-11-01 00:00:19", "2018-10-25 00:00:54", "2018-11-01 10:03:27",
"2018-11-01 00:01:23", "2018-11-01 00:01:25","2018-10-21 00:00:55"), c("Y","Y","Y","N","Y","N"))
names(df) <- c("ID","Signature", "date", "status")
df$date <- as.POSIXct(df$date)
ifinal<-nrow(df)
Do you know any other way to do it maybe applying a function?
r
edited 7 hours ago
Jamal♦
30.3k11116226
asked 7 hours ago
user190888user190888
1
New contributor
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
$begingroup$
I am doing this formula in R:
for (i in 1:ifinal)
{
df$FPRateSignature30d[i] <-
length(
which(
df$status == "Y" &
` df$Signature==df$Signature[i]& df$date>= (as_datetime(df$date[i])duration(30, days')) &` df$date <
as_datetime(df$date[i])
)
) / length(which(
df$Signature == df$Signature[i] &
df$date >= (as_datetime(df$date[i]) - duration(30, 'days')) &
``df$date < as_datetime(df$date[i])
))
}
but I have ~350K of rows in the real data, and takes me a lot doing this with for loops.
Here it is the data sample:
df <- data.frame(c("1","2","3","4","5","6"),c("Attribute1", "Attribute1", "Attribute2", "Attribute2", "Attribute2", "Attribute1"),
c("2018-11-01 00:00:19", "2018-10-25 00:00:54", "2018-11-01 10:03:27",
"2018-11-01 00:01:23", "2018-11-01 00:01:25","2018-10-21 00:00:55"), c("Y","Y","Y","N","Y","N"))
names(df) <- c("ID","Signature", "date", "status")
df$date <- as.POSIXct(df$date)
ifinal<-nrow(df)
Do you know any other way to do it maybe applying a function?
r
edited 7 hours ago
Jamal♦
30.3k11116226
asked 7 hours ago
user190888user190888
1
New contributor
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
0
0
$begingroup$
I am doing this formula in R:
for (i in 1:ifinal)
{
df$FPRateSignature30d[i] <-
length(
which(
df$status == "Y" &
` df$Signature==df$Signature[i]& df$date>= (as_datetime(df$date[i])duration(30, days')) &` df$date <
as_datetime(df$date[i])
)
) / length(which(
df$Signature == df$Signature[i] &
df$date >= (as_datetime(df$date[i]) - duration(30, 'days')) &
``df$date < as_datetime(df$date[i])
))
}
but I have ~350K of rows in the real data, and takes me a lot doing this with for loops.
Here it is the data sample:
df <- data.frame(c("1","2","3","4","5","6"),c("Attribute1", "Attribute1", "Attribute2", "Attribute2", "Attribute2", "Attribute1"),
c("2018-11-01 00:00:19", "2018-10-25 00:00:54", "2018-11-01 10:03:27",
"2018-11-01 00:01:23", "2018-11-01 00:01:25","2018-10-21 00:00:55"), c("Y","Y","Y","N","Y","N"))
names(df) <- c("ID","Signature", "date", "status")
df$date <- as.POSIXct(df$date)
ifinal<-nrow(df)
Do you know any other way to do it maybe applying a function?
r
edited 7 hours ago
Jamal♦
30.3k11116226
asked 7 hours ago
user190888user190888
1
New contributor
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am doing this formula in R:
for (i in 1:ifinal)
{
df$FPRateSignature30d[i] <-
length(
which(
df$status == "Y" &
` df$Signature==df$Signature[i]& df$date>= (as_datetime(df$date[i])duration(30, days')) &` df$date <
as_datetime(df$date[i])
)
) / length(which(
df$Signature == df$Signature[i] &
df$date >= (as_datetime(df$date[i]) - duration(30, 'days')) &
``df$date < as_datetime(df$date[i])
))
}
but I have ~350K of rows in the real data, and takes me a lot doing this with for loops.
Here it is the data sample:
df <- data.frame(c("1","2","3","4","5","6"),c("Attribute1", "Attribute1", "Attribute2", "Attribute2", "Attribute2", "Attribute1"),
c("2018-11-01 00:00:19", "2018-10-25 00:00:54", "2018-11-01 10:03:27",
"2018-11-01 00:01:23", "2018-11-01 00:01:25","2018-10-21 00:00:55"), c("Y","Y","Y","N","Y","N"))
names(df) <- c("ID","Signature", "date", "status")
df$date <- as.POSIXct(df$date)
ifinal<-nrow(df)
Do you know any other way to do it maybe applying a function?
r
r
edited 7 hours ago
Jamal♦
30.3k11116226
asked 7 hours ago
user190888user190888
1
New contributor
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
Jamal♦
30.3k11116226
asked 7 hours ago
user190888user190888
1
New contributor
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
Jamal♦
30.3k11116226
edited 7 hours ago
Jamal♦
30.3k11116226
edited 7 hours ago
Jamal♦
30.3k11116226
30.3k11116226
asked 7 hours ago
user190888user190888
1
New contributor
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
user190888user190888
1
asked 7 hours ago
user190888user190888
1
1
New contributor
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user190888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "196"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
user190888 is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f211926%2fcalculating-the-probability-that-the-signature-column-has-the-status-y-in-the%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
user190888 is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
user190888 is a new contributor. Be nice, and check out our Code of Conduct.
user190888 is a new contributor. Be nice, and check out our Code of Conduct.
user190888 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Code Review Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f211926%2fcalculating-the-probability-that-the-signature-column-has-the-status-y-in-the%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown