我有一个看起来像这样的DataFrame：
import numpy as np
import pandas as pd
df=pd.DataFrame([['vt 40462',5,6],[5,6,6],[5,5,8],[4,3,1],['vl 6450',5,6],[5,6,7],
[1,2,3],['vt 40462',5,6],[5,5,8],['vl 658',6,7],[5,5,8],[4,3,1],['vt 40461',5,6],[5,5,8],
[7,8,5]],columns=['A','B','C'])
df
A B C
0 vt 40462 5 6
1 5 6 6
2 5 5 8
3 4 3 1
4 vl 6450 5 6
5 5 6 7
6 1 2 3
7 vt 40462 5 6
8 5 5 8
9 vl 658 6 7
10 5 5 8
11 4 3 1
12 vt 40461 5 6
13 5 5 8
14 7 8 5
我想给索引vt中vl和A之间的值，并创建一个新列：
A B C D
0 vt 40462 5 6 vt 40462
1 5 6 6 vt 40462
2 5 5 8 vt 40462
3 4 3 1 vt 40462
4 vl 6450 5 6 vl 6450
5 5 6 7 vl 6450
6 1 2 3 vl 6450
7 vt 40462 5 6 vt 40462
8 5 5 8 vt 40462
9 vl 658 6 7 vl 658
10 5 5 8 vl 658
11 4 3 1 vl 658
12 vt 40461 5 6 vt 40461
13 5 5 8 vt 40461
14 7 8 5 vt 40461
答案 0 :(得分：3)
使用var items = document.querySelectorAll('#clock');
for (var i = 0, len = items.length; i < len; i++) {
(function () {
var e = new Date("2018-12-31").getTime(),
t = this.querySelector("[data-js=countdown]"),
n = this.querySelector("[data-js=countdown-endtext]"),
day = this.querySelector("[data-js=countdown-day]"),
hour = this.querySelector("[data-js=countdown-hour]"),
min = this.querySelector("[data-js=countdown-minute]"),
sec = this.querySelector("[data-js=countdown-second]"),
s = this.gjs_countdown_interval;
s && s && clearInterval(s);
var l = function (e, t, n, s) {
day.innerHTML = e < 10 ? "0" + e : e,
hour.innerHTML = t < 10 ? "0" + t : t,
min.innerHTML = n < 10 ? "0" + n : n,
sec.innerHTML = s < 10 ? "0" + s : s
},
u = function () {
var day = (new Date).getTime(),
hour = e - day,
min = Math.floor(hour / 864e5),
sec = Math.floor(hour % 864e5 / 36e5),
s = Math.floor(hour % 36e4 / 6e4),
u = Math.floor(hour % 6e4 / 1e3);
l(min, sec, s, u), hour < 0 && (clearInterval(c),
n.innerHTML = "EXPIRED",
t.style.display = "none",
n.style.display = "")
};
if (e) {
var c = setInterval(u, 1e3);
this.gjs_countdown_interval = c,
n.style.display = "none",
t.style.display = "", u()
} else l(0, 0, 0, 0)
}.bind(items[i]))();
}
，如果未找到''，则返回NaN使用<section class="flex-sect">
<div id="clock" class="countdown">
<span data-js="countdown" class="countdown-cont">
<div class="countdown-block">
<div data-js="countdown-day" class="countdown-digit"></div>
<div class="countdown-label">days</div>
</div>
<div class="countdown-block">
<div data-js="countdown-hour" class="countdown-digit"></div>
<div class="countdown-label">hours</div>
</div>
<div class="countdown-block">
<div data-js="countdown-minute" class="countdown-digit"></div>
<div class="countdown-label">minutes</div>
</div>
<div class="countdown-block">
<div data-js="countdown-second" class="countdown-digit"></div>
<div class="countdown-label">seconds</div>
</div>
</span>
<span data-js="countdown-endtext" class="countdown-endtext"></span>
</div>
</section>
来填充NaN并将字段连接在一起并分配给'D'：
str.split输出：
ffill答案 1 :(得分：3)
另一种方法是将assign列D设为以字母开头的A的所有值，然后使用df.ffill()摆脱{{1 }}：
NaN或多或少等效，但分为2个步骤：
df.assign(D=df.loc[df.A.str.contains('^[A-Za-z]', na=False), 'A']).ffill()
A B C D
0 vt 40462 5 6 vt 40462
1 5 6 6 vt 40462
2 5 5 8 vt 40462
3 4 3 1 vt 40462
4 vl 6450 5 6 vl 6450
5 5 6 7 vl 6450
6 1 2 3 vl 6450
7 vt 40462 5 6 vt 40462
8 5 5 8 vt 40462
9 vl 658 6 7 vl 658
10 5 5 8 vl 658
11 4 3 1 vl 658
12 vt 40461 5 6 vt 40461
13 5 5 8 vt 40461
14 7 8 5 vt 40461