Pandas允许use std::cmp::{min, max};
#[derive(Debug, PartialEq, Clone, Copy)]
struct Interval {
start: usize,
stop: usize,
}
impl Interval {
fn new(start: usize, stop: usize) -> Interval {
Interval {
start: start,
stop: stop,
}
}
pub fn starts_before_disjoint(&self, other: &Interval) -> bool {
self.start < other.start && self.stop < other.start
}
pub fn starts_before_non_disjoint(&self, other: &Interval) -> bool {
self.start <= other.start && self.stop >= other.start
}
pub fn starts_after(&self, other: &Interval) -> bool {
self.start > other.start
}
pub fn starts_after_disjoint(&self, other: &Interval) -> bool {
self.start > other.stop
}
pub fn starts_after_nondisjoint(&self, other: &Interval) -> bool {
self.start > other.start && self.start <= other.stop
}
pub fn disjoint(&self, other: &Interval) -> bool {
self.starts_before_disjoint(other)
}
pub fn adjacent(&self, other: &Interval) -> bool {
self.start == other.stop + 1 || self.stop == other.start - 1
}
pub fn union(&self, other: &Interval) -> Interval {
Interval::new(min(self.start, other.start), max(self.stop, other.stop))
}
pub fn intersection(&self, other: &Interval) -> Interval {
Interval::new(max(self.start, other.start), min(self.stop, other.stop))
}
}
fn main() {
//making vectors
let mut vec = vec![
Interval::new(1, 1),
Interval::new(2, 3),
Interval::new(6, 7),
];
let addition = Interval::new(2, 5); // <- this will take over interval @ 2 and will be adjacent to 3, so we have to merge
let (mut i, len) = (0, vec.len());
while i < len {
let r = &mut vec[i];
if *r == addition {
return; //nothing to do, just a duplicate
}
if addition.adjacent(r) || !addition.disjoint(r) {
//if they are next to each other or overlapping
//lets merge
let mut bigger = addition.union(r);
*r = bigger;
//now lets check what else we can merge
while i < len - 1 {
i += 1;
let next = &vec[i + 1];
if !bigger.adjacent(next) && bigger.disjoint(next) {
//nothing to merge
break;
}
vec.remove(i); //<- FAIL another mutable borrow
i -= 1; //lets go back
vec[i] = bigger.union(next); //<- FAIL and yet another borrow
}
return;
}
if addition.starts_before_disjoint(r) {
vec.insert(i - 1, addition); // <- FAIL since another refence already borrowed @ let r = &mut vec[i]
}
i += 1;
}
}
选择startswith，例如：
DataFrame......可能会产生类似
的内容query = "Ali"
people[people.Name.str.startswith(query)]
但是，我想反转选择中的输入，以查找Name
Ali
Alice
Alicia
Alistair
...
以input中的值开头的行。
类似的东西：
DataFrame例如，这可能会选择名称query = "Ali"
people[query.startswith(people.Name)]
和Al。
这段代码显然不起作用，这对于这个例子来说似乎并不合理，但这是我想要用我的数据实现的。
任何人都知道如何实现这一目标？
答案 0 :(得分：4)
制备
import pandas as pd
people = pd.DataFrame()
people["FullName"] = ["Alice Cooper", "Ali","Aloy"]
input_ = "Alice"
备选方案1：Boolean-indexing
people[[input_.startswith(i) for i in people.FullName]] # passes [False, True, False]
1000个循环，最佳3：937μs/循环
备选方案2 （感谢@Abdou）：使用.apply() and lambda
调用系列值的函数people[people.FullName.apply(lambda s: input_.startswith(s))]
1000次循环，最佳3：每循环1.38 ms
输出：
FullName
1 Ali