如果否定则向上舍入如果正则向下舍入？我有
$rounded =1000
39528,65 round should be --> 39000
和
-30965,77 --> -31000
<?php
// initial value
$week = 50;
$year = 2001;
$store = array();
do
{
$week++;
$result = "$week/$year";
array_push($store,$result);
if($week == 53){
$week = 0;
$year++;//increment year by 1
}
continue;
}
// End of Loop
while ($result !== "2/2002");
?>
print_r($store);
结果想要回归
array("51/2001", "52/2001", "01/2002", "02/2002");
使用do..while时继续使用我的问题是什么？
我有两个或更多的列表.有点像这样:
listX = [('A', 1, 10), ('B', 2, 20), ('C', 3, 30), ('D', 4, 30)]
listY = [('a', 5, 50), ('b', 4, 40), ('c', 3, 30), ('d', 1, 20),
('A', 6, 60), ('D', 7, 70])
我想获得移动这样的重复元素的结果:我的结果是从listX + listY获取所有列表,但是在重复的情况下,例如('A', 1, 10), ('D', 4, 30)listX 的元素被呈现或者在listY中退出.结果就这样吧
result = [('A', 7, 70), ('B', 2, 20), ('C', 3, 30), ('D', 11, 100),
('a', 5, 50), ('b', 4, 40), ('c', 3, 30), ('d', 1, 20)]
(A, 7, 70)通过添加('A', 1, 10)和 …
author_A = [['book_x',1,10],['book_y',2,20],['book_z',3,30]]
author_B = [['book_s',5,10],['book_t',2,20],['book_z',3,30]]
author_A AND author_B = ['book_z',3,30]
author_A = [['book_x',1,10],['book_y',2,20]]
author_B = [['book_s',5,10],['book_t',2,20]]
---------------------------------------------
我想要这样的现有数据
author quantity Amount($)
A&B 3 30
A 3 30
B 7 30
total 13 90
我不想要这样的现有数据!在这种情况下,它是ADDED重复['book_z',3,30]
author quantity Amount($)
A 6 60
B 10 60
total 16 120
这是我的问题,任何人请帮助我解决这个问题.谢谢大家
在In print_r或var_dump中
echo "<pre>";
echo var_dump($groupname);
echo "</pre>";
我得到了结果
array(2) {
[0]=>
object(stdClass)#330 (1) {
["name"]=>
string(3) "AAA"
}
[1]=>
object(stdClass)#332 (1) {
["name"]=>
string(3) "BBB"
}
}
现在我想从数组中得到结果.
AAA | BBB
我是一本字典(字典字典)
old_dict = {
'1':{'A':1, 'check': 0, 'AA':2, 'AAA':3 , 'status':0},
'2':{'A':11,'check': 0, 'AA':22, 'AAA':33 ,'status':1},
'3':{'A':111,'check': 0, 'AA':222, 'AAA':333 ,'status':0},
'4':{'A':1111,'check': 1, 'AA':2222, 'AAA':3333 ,'status':0},
}
我想在['check']!= 0之前和['status']之前得到一本新词典!= 0所以将会是
new_dict = {
'1':{'A':1, 'check': 0, 'AA':2, 'AAA':3 , 'status':0},
'3':{'A':111,'check': 0, 'AA':222, 'AAA':333 ,'status':0},
}
如果是列表,我确实喜欢这个
ouputdata = [d for d in data if d[1] == ' 0' and d[6]==' 0']
我试过了
for field in old_dict.values():
if field['check'] !=0 and field['status'] != 0
line = field['A'] + field['AA']+field['AAA']
#write …我有:
$l = array(
array("A"=>0.1,"B"=>1,"C"=>1,"D"=>1),
array("A"=>0.1,"B"=>1,"C"=>0,"D"=>2),
);
$h = array('h1','h2');
1-我怎么能map(l,h)这样？
$result= $array(
'h1'=> array("A"=>0.1,"B"=>1,"C"=>1,"D"=>1),
'h1'=> array("A"=>0.1,"B"=>1,"C"=>0,"D"=>2),
);
2-所以II可以显示(现在的html表)
-------------------
| A | B | C | D
-------------------
h1 |
-------------------
h2 |
--------------------
我试图输出:
<table>
<tr><td>A</td><td>B</td><td>C</td><td>D</td></tr>
foreach($result as $key=>$value){
<tr>
<tr>
}
<table>
有人可以帮帮我吗？
我已经开始编写一些Python代码了.我有的是:
from math import *
def ex(N):
l = []
sum = 0
N = abs(int(N));
for n in range(1,N):
if N % n == 0:
l.append(n)
sum += n
l.append(N)
print ' of '+str(N),l
print 'SUM', (sum+N)
我不知道这是好还是坏,但这是我试过的:)
是否可以使用列表理解来复制代码的行为？如果是这样,怎么样？
我必须实现一个cmpT应该返回以下结果的函数:
>>> cmpT((1, 2), (1, 2))
True
>>> cmpT((1, 2), (2, 1))
True
>>> cmpT((1, 2), (1, 2, 1))
False
>>> cmpT((1, 2), ())
False
我的代码:
def cmpT(t1, t2):
if t1 == t2:
return True
else:
return False
它没有提供所需的输出,cmpT((1, 2), (2, 1))也没有返回True.怎么了？
我正在学习scala,我尝试使用以下代码.
object Demo7 {
def main(args: Array[String]): Unit = {
class Person(val fullName: String) {
println(s"This is the primary constructor. Name is ${fullName}")
val initial = fullName.substring(0, 1) // Computed during initialization
//def this(firstName: String, lastName: String) = this(s"$firstName $lastName")
}
new Person("Tek Tuk")
new Person("Tek Tuk").fullName
}
}
然后我运行我得到与每次调用相同的返回结果.我理解这一行
new Person("Tek Tuk").fullName
不应该编译,任何人都可以解释为什么这行得到编译并返回与第一行相同的结果？
谢谢.