Question

我有一些基本的代码来处理Perl哈希，在这里我可以处理以下元素：$ data {“ WV2”} {789} {PP1}（或在分配中使用该实际文本）...但是我想要使用Python字典做类似的事情。

Perl和Python中的两个简单程序说明了我一直在尝试复制的内容：-

因此，Perl代码：-

whoami

...以及Python代码：-

# hash.pl

use strict;
use warnings;

use Data::Dumper;

my %data = ();

my @reg_list = ( "MC1", "CA2", "WV2" );
my @site_list = ( 123, 456, 391, 287 );

$data{MC1}{4564}{PP}{1} = "-15,-15C";
$data{MC1}{4564}{PP}{2} = "5,5C";
$data{MC1}{4564}{PP}{3} = "-19,-19C";
$data{MC1}{4564}{PP}{4} = "-12,-12C";

printf("---- One:\n");
print Dumper(%data);                 # Ok, shows the full strucure

printf("---- Two:\n");
print Dumper($data{"MC2"});          # Shows as undef (sensible)

printf("---- Three:\n");
print Dumper($data{"MC1"});          # Ok, showing the key:values for each "site" key

printf("---- Four:\n");
print Dumper($data{"MC1"}{"4564"});  # Ok, shows the actual equality value above


# ---- This works Ok

my %xdata = ();
$xdata{"MC1"}{123}{"PP"} = "-15,-15C";
$xdata{"MC1"}{456}{"PP"} = "5,5C";
$xdata{"MC1"}{391}{"PP"} = "-19,-19C";
$xdata{"MC1"}{287}{"PP"} = "-12,-12C";

printf("---- One:\n");
print Dumper(%xdata);                # Ok, shows the full strucure


#pprint.pprint(data["MC2"]) 
#pprint.pprint(data["MC1"}{391]) 


# [eof]

每个程序的输出如下：-

# dict.py

import pprint
import collections

reg_list  = [ "MC1", "CA2", "WV2" ]
site_list = [ 123, 456, 391, 287 ]

#data = {}
data = collections.defaultdict(dict) # {}
data["MC1"][123] = "-15,-15C"
data["MC1"][456] = "5,5C"
data["MC1"][391] = "-19,-19C"
data["MC1"][287] = "-12,-12C"

print("---- One:")
pprint.pprint(data)              # Ok, shows the full strucure

print("---- Two:")
pprint.pprint(data["MC2"])       # Shows: {} [...Ok, undefined...]

print("---- Three:")
pprint.pprint(data["MC1"])       # Ok, showing the key:values for each "site" key

print("---- Four:")
pprint.pprint(data["MC1"][391])  # Ok, shows the actual equality value above

# ---- Cannot get the following to work

xdata = collections.defaultdict(dict) # {}
xdata["MC1"][123]["PP"] = "-15,-15C"  # ERROR: Key error 123
xdata["MC1"][456]["PP"] = "5,5C"
xdata["MC1"][391]["PP"] = "-19,-19C"
xdata["MC1"][287]["PP"] = "-12,-12C"

#pprint.pprint(data["MC2"]) 
#pprint.pprint(data["MC1"][391]) 

# [eof]

...并且来自Python：-

# Perl Output:

---- One:
$VAR1 = 'MC1';
$VAR2 = {
          '4564' => {
                      'PP' => {
                                '4' => '-12,-12C',
                                '1' => '-15,-15C',
                                '3' => '-19,-19C',
                                '2' => '5,5C'
                              }
                    }
        };
---- Two:
$VAR1 = undef;
---- Three:
$VAR1 = {
          '4564' => {
                      'PP' => {
                                '4' => '-12,-12C',
                                '1' => '-15,-15C',
                                '3' => '-19,-19C',
                                '2' => '5,5C'
                              }
                    }
        };
---- Four:
$VAR1 = {
          'PP' => {
                    '4' => '-12,-12C',
                    '1' => '-15,-15C',
                    '3' => '-19,-19C',
                    '2' => '5,5C'
                  }
        };
---- One:
$VAR1 = 'MC1';
$VAR2 = {
          '391' => {
                     'PP' => '-19,-19C'
                   },
          '456' => {
                     'PP' => '5,5C'
                   },
          '123' => {
                     'PP' => '-15,-15C'
                   },
          '287' => {
                     'PP' => '-12,-12C'
                   }
        };

我尝试查找有关Nesting Dictionaries的信息...但是我看过的所有内容并不能清楚地解释该概念的工作原理（无论如何，在我看来）....尤其是在以下情况下词典的使用级别更高。

我已经写了25年的Perl代码，但仅从Python开始。

在Windows 10 x64上运行ActiveState Perl v5.16.3，Build 1603和Anaconda Python 3.6.5。

非常感谢您的任何想法或建议。

Answer 1

Python不能像Perl的散列那样自动保留多级字典。在第二层和更深的级别上，您必须为空的dict分配一个空的dict，然后再向其添加更多密钥：

xdata = collections.defaultdict(dict)
xdata["MC1"] = collections.defaultdict(dict)
xdata["MC1"][123]["PP"] = "-15,-15C"  # ERROR: Key error 123
xdata["MC1"][456]["PP"] = "5,5C"
xdata["MC1"][391]["PP"] = "-19,-19C"
xdata["MC1"][287]["PP"] = "-12,-12C"

Answer 2

解决问题的简单方法似乎是：-

library('rvest')

for (i in 1:40) {

     webpage <- read_html(paste0(("http://search.beaconforfreedom.org/search/censored_publications/result.html?author=&cauthor=&title=&country=7327&language=&censored_year=&censortype=&published_year=&censorreason=&sort=t&page=, i"))

     rank_data_html <- html_nodes(webpage,'tr+ tr td:nth-child(1)')

     rank_data <- html_text(rank_data_html)

     rank_data<-as.numeric(rank_data)

     title_data_html <- html_nodes(webpage,'.censo_list font')

     title_data <- html_text(title_data_html)

     author_data_html <- html_nodes(webpage,'.censo_list+ td font')
     author_data <- html_text(author_data_html)

     country_data_html <- html_nodes(webpage,'.censo_list~ td:nth-child(4) font')

     rcountry_data <- html_text(country_data_html)

     year_data_html <- html_nodes(webpage,'tr+ tr td:nth-child(5) font')

     year_data <- html_text(year_data_html)

     type_data_html <- html_nodes(webpage,'tr+ tr td:nth-child(6) font')

     type_data <- html_text(type_data_html)

}

censorship_df<-data.frame(Rank = rank_data, Title = title_data, Author = author_data, Country = rcountry_data, Type = type_data, Year = year_data)

write.table(censorship_df, file="sample.csv",sep=",",row.names=F)

...但是那仍然意味着我每次“发现”新的“值”时都必须“手动”定义一个字典... 等等

尽管固有的 Gotcha！带有错误的键入和dict内容的（可能）损坏，但What is the best way to implement nested dictionaries?似乎是解决此问题的好方法...尤其是值（无论如何，在我当前的应用程序中）“从来没有看到过”（它们是在进入我的应用程序之前由机器生成和验证的）...因此，一些可能有效的代码可能是：-

xdata = collections.defaultdict(dict) # {}
xdata["MC1"][123] = {}                # Define the dict before using it
xdata["MC1"][123]["PP"] = "-15,-15C"  # Works Ok

感谢大家的建议和指点。

如何在Perl中实现多键哈希作为Python中的嵌套字典？

2 个答案: