Haskell-如何迭代列表并保留原始列表，而不仅仅是尾部

如何解决Haskell-如何迭代列表并保留原始列表，而不仅仅是尾部

我对Haskell非常陌生。 下面是我的功能，用于打印列表xs的所有元素，但仍希望保留原始xs，而不是每转一圈都有尾巴。我不知道该怎么做。有没有一种方法可以在迭代之前保存/复制xs？谢谢。

from django.conf.urls.static import static
from django.urls import path

from . import views

urlpatterns = [
                  path('',views.index,name='index.html'),] + static(settings.STATIC_URL,document_root=settings.STATIC_ROOT)


views.py


from django.shortcuts import render


def index(request):
    return render(request,'Mainapp/index.html')

settings.py

import os

BASE_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))



DEBUG = True

ALLOWED_HOSTS = []

# Application deFinition

INSTALLED_APPS = [
    'django.contrib.admin','django.contrib.auth','django.contrib.contenttypes','django.contrib.sessions','django.contrib.messages','django.contrib.staticfiles','Mainapp',]

MIDDLEWARE = [
    'django.middleware.security.SecurityMiddleware','django.contrib.sessions.middleware.SessionMiddleware','django.middleware.common.CommonMiddleware','django.middleware.csrf.CsrfViewMiddleware','django.contrib.auth.middleware.AuthenticationMiddleware','django.contrib.messages.middleware.MessageMiddleware','django.middleware.clickjacking.XFrameOptionsMiddleware',]



TEMPLATES = [
    {
        'BACKEND': 'django.template.backends.django.DjangoTemplates','Dirs': [],'APP_Dirs': True,'OPTIONS': {
            'context_processors': [
                'django.template.context_processors.debug','django.template.context_processors.request','django.contrib.auth.context_processors.auth','django.contrib.messages.context_processors.messages',],},]



# Static files (CSS,JavaScript,Images)
# https://docs.djangoproject.com/en/3.0/howto/static-files/

STATIC_URL = '/static/'

STATIC_ROOT = os.path.join(BASE_DIR,'static')

输出：

listElement :: [Int] -> [(Int,Int)]
listElement [] = []
listElement (x:xs) = [(x,length xs)] ++ listElement xs

预期：

listElement [1,2,3]
[(1,2),(2,1),(3,0)]

解决方法

在通常情况下，您可以创建一个临时函数（在这种情况下为t），然后将原始列表和“当前尾部”作为参数传递：

t :: [Int] -> [Int] -> [(Int,Int)]
t [] _ = []
t (x:xs) original = [(x,length original)] ++ t xs original

listElement :: [Int] -> [(Int,Int)]
listElement xs = t xs xs

main = print $ listElement [1,2,3]

Try it online!

请注意，通常最好预先计算列表的长度，然后将它们中的每个附加到每个元素上并加上一个map。

在这种情况下，看起来编译器可以优化代码，并且只计算一次长度。

listElement xs = map (\x -> (x,length xs)) xs

listElement xs = map (\x -> (x,len)) xs where len = length xs