给定以下无序制表符分隔文件：
Asia Srilanka
Srilanka Colombo
Continents Europe
India Mumbai
India Pune
Continents Asia
Earth Continents
Asia India
目标是生成以下输出（制表符分隔）：
Earth Continents Asia India Mumbai
Earth Continents Asia India Pune
Earth Continents Asia Srilanka Colombo
Earth Continents Europe
我创建了以下脚本来实现目标：
root={} # this hash will finally contain the ROOT member from which all the nodes emanate
link={} # this is to hold the grouping of immediate children
for line in f:
line=line.rstrip('\r\n')
line=line.strip()
cols=list(line.split('\t'))
parent=cols[0]
child=cols[1]
if not parent in link:
root[parent]=1
if …在Unix shell脚本中,可以从内部脚本本身重定向stderr和stdout,如下所示:
#!/bin/ksh
# script_name: test.sh
export AUTO_LOGFILE=`basename $0 .sh`.log
# stdout and stderr Redirection. This will save the old stdout on FD 3, and the old stderr on FD 4.
exec 3>&0 4>&1 >>$AUTO_LOGFILE 2>&1
echo "Hello World"
# The above echo will be printed to test.log
实际上,test.sh可以简单地执行:
test.sh
代替:
test.sh >> test.log 2>&1
我试图在批处理脚本中做类似的事情.我的批处理代码如下:
@echo off & setlocal enableextensions enabledelayedexpansion
REM script_name=test.bat
set AUTO_LOGFILE=%~n0.log
REM How to do the stdout and stderr redirection from within the script itself …我遇到了以下示例.我试着谷歌但找不到太多,所以我在这里发布这个问题.
这是代码:
#!/bin/ksh
#! -*- perl -*-
eval 'exec $PERLLOCATION/bin/perl -x $0 ${1+"$@"} ;'
if 0;
print "hello world\n";
# how can I make it behave like a "normal" shell script from this point onwards? What needs to be done?
# echo "hello world" ### this results in error
我想用下面的代码打印python 3.x中的列表列表,但是它给出了一个错误.
lol=[[1,2],[3,4],[5,6],['five','six']]
for elem in lol:
print (":".join(elem))
# this is the error I am getting-> TypeError: sequence item 0: expected str instance, int found
我期待这个输出:
1:2
3:4
5:6
five:six
我可以使用下面的perl代码实现相同的输出(这仅供参考):
for (my $i=0;$i<scalar(@{$lol});$i++)
{
print join(":",@{$lol->[$i]})."\n";
}
我怎么在python 3.x中做到这一点？
我遇到了以下片段(并且可以追溯到https://docs.python.org/3/library/itertools.html#itertools.product):
def cartesian_product(pools):
result = [[]]
for pool in pools:
result = [x+[y] for x in result for y in pool]
return result
a_list=[1, 2, 3]
b_list=[4, 5]
all_list=[a_list, b_list]
print (cartesian_product(all_list)) # [[1, 4], [1, 5], [2, 4], [2, 5], [3, 4], [3, 5]]
如果我们更改以下行:
result = [[]]
对此:
result = []
然后代码不起作用.
现在考虑下面的一段代码,其中变量my_list初始化为my_list=[]和不是,my_list=[[]]但我们仍然得到预期的结果:
my_list=[]
my_list.append([1,2])
my_list.append([3,4])
print (my_list) # [[1, 2], [3, 4]]
所以在cartesian_product我上面提到的函数中,有result=[[]]和没有意义是result=[]什么？
我执行下面提到的代码时遇到以下错误:
TestGenerics1.java:40: error: cannot find symbol
arr.get(i).eat();
^
symbol: method eat()
location: class Object
1 error
我将面临的问题在以下示例代码的帮助下重现:
import java.util.*;
abstract class Animal
{
void eat() { System.out.println("animal eating"); }
}
class Dog extends Animal
{
void bark() { }
}
class Cat extends Animal
{
void meow() { }
}
class RedCat extends Cat { }
public class TestGenerics1
{
public static void main(String[] args)
{
new TestGenerics1().go();
}
public void go()
{
List<Cat> arrAnimals = new ArrayList<Cat>(Arrays.asList(new RedCat(), new …