Получение IndexError Traceback (последний вызов был последним) для существующего индекса

Question

Я использую эту функцию для двух наборов данных (iOS_list & play_apps). Оба набора данных являются списками, содержащими списки строк. Приведенная ниже функция работает с переменной play_apps, но переменная iOS_list возвращает IndexError. Я могу получить доступ к индексу, указанному в функции, в строке ниже:

print(iOS_list[0][2])

Так что я не уверен, почему я продолжаю получать ошибку, когда использую функцию на Переменная iOS_list.

def find_duplicates(listy, index): 
    unique_apps = [] # list of all app names
    duplicate_apps = [] # list of known duplicate app names

    for app in listy:
        name = app[index]
        if name in unique_apps: 
            duplicate_apps.append(name) 
        elif name not in unique_apps: 
            unique_apps.append(name)
    print('Count of duplicate apps in data set:',len(duplicate_apps))
    print('\n')
    print('Sample of duplicate apps in data set:', duplicate_apps[:3])
    print('\n')    

find_duplicates(play_apps,0)
find_duplicates(iOS_list,2)

Образцы первых строк двух наборов данных:

print(iOS_list[:1],'\n')
print(play_apps[:1])
print(type(iOS_list[0]))
print(type(play_apps[0]))

[['1', '281656475', 'PAC-MAN Premium', '100788224', 'USD', '3.99', '21292', '26', '4', '4.5', '6.3.5', '4+', 'Games', '38', '5', '10', '1']] 

[['Photo Editor & Candy Camera & Grid & ScrapBook', 'ART_AND_DESIGN', '4.1', '159', '19M', '10,000+', 'Free', '0', 'Everyone', 'Art & Design', 'January 7, 2018', '1.0.0', '4.0.3 and up']]

<class 'list'>
<class 'list'>

Ожидаемый результат (отлично работает с переменной play_apps):

Count of duplicate apps in data set: 1181


Sample of duplicate apps in data set: ['Quick PDF Scanner + OCR FREE', 'Box', 'Google My Business']

Actual Вывод (только для переменной iOS_list):

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-49-91240e63afea> in <module>
     29 
     30 
---> 31 find_duplicates(iOS_list,2)
     32 
     33 

<ipython-input-49-91240e63afea> in find_duplicates(listy, index)
      4 
      5     for app in listy:
----> 6         name = app[index]
      7         if name in unique_apps: # if app name is already in the unique_apps list
      8             duplicate_apps.append(name) # add that app name to the duplicate_apps list

IndexError: list index out of range

Извиняюсь, если это глупый вопрос, я новичок и не смог найти выход из этого.

Answer 1

Добро пожаловать в переполнение стека. Когда я запустил ваш код, я не получил никакой ошибки, но поскольку не было дубликатов, в результате я получил 0 для подсчета обоих наборов данных. Поэтому я изменил ваши входные данные:

iOS_list = [
    ["Village", "did", "removed", "enjoyed", "explain", "nor", "ham", "saw", "calling", "talking."], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["feelings", "own", "shy.", "Request", "norland", "neither", "mistake", "for", "yet.", "Between"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["have", "an", "no", "at.", "Relation", "so", "in", "confined", "smallest", "children"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["account", "in", "outward", "tedious", "do.", "Particular", "way", "thoroughly", "unaffected", "projection"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["unpacked", "delicate.", "Why", "sir", "end", "believe", "uncivil", "respect.", "Always", "get"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["adieus", "nature", "day", "course", "for", "common.", "My", "little", "garret", "repair"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["have", "an", "no", "at.", "Relation", "so", "in", "confined", "smallest", "children"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["For", "who", "thoroughly", "her", "boy", "estimating", "conviction.", "Removed", "demands", "expense"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["account", "in", "outward", "tedious", "do.", "Particular", "way", "thoroughly", "unaffected", "projection"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["admire", "in", "giving.", "See", "resolved", "goodness", "felicity", "shy", "civility", "domestic"], 
    ["had", "but.", "Drawings", "offended", "yet", "answered", "jennings", "perceive", "laughing", "six"]
]
play_apps = [
    ["We", "diminution", "preference", "thoroughly", "if.", "Joy", "deal", "pain", "view", "much"], 
    ["her", "time.", "Led", "young", "gay", "would", "now", "state.", "Pronounce", "we"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["attention", "admitting", "on", "assurance", "of", "suspicion", "conveying.", "That", "his", "west"], 
    ["quit", "had", "met", "till.", "Of", "advantage", "he", "attending", "household", "at"], 
    ["do", "perceived.", "Middleton", "in", "objection", "discovery", "as", "agreeable.", "Edward", "thrown"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["dining", "so", "he", "my", "around", "to.", "Increasing", "impression", "interested", "expression"], 
    ["he", "my", "at.", "Respect", "invited", "request", "charmed", "me", "warrant", "to."], 
    ["Expect", "no", "pretty", "as", "do", "though", "so", "genius", "afraid", "cousin."], 
    ["do", "perceived.", "Middleton", "in", "objection", "discovery", "as", "agreeable.", "Edward", "thrown"], 
    ["Girl", "when", "of", "ye", "snug", "poor", "draw.", "Mistake", "totally", "of"], 
    ["in", "chiefly.", "Justice", "visitor", "him", "entered", "for.", "Continue", "delicate", "as"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["being", "style", "house.", "An", "whom", "down", "kept", "lain", "name", "so"], 
]

def find_duplicates(listy, index): 
    unique_apps = [] # list of all app names
    duplicate_apps = [] # list of known duplicate app names

    for app in listy:
        name = app[index]
        if name in unique_apps: 
            duplicate_apps.append(name) 
        elif name not in unique_apps: 
            unique_apps.append(name)
    print('Count of duplicate apps in data set:',len(duplicate_apps))
    print('\n')
    print('Sample of duplicate apps in data set:', duplicate_apps[:3])
    print('\n')    

find_duplicates(play_apps, 0)
find_duplicates(iOS_list, 2)

Он все еще работал без проблем, поэтому, если у вас есть ошибка, о которой вы сообщили, я думаю, что вам, возможно, придется проверить свои входные данные. Я думаю, что у вас может быть один из ваших списков (внутри вашего списка), в котором нет 3 элементов (поскольку вы смотрите на индекс 2)

Но я также заметил, что может быть ошибкой: в вашем код вы отделили повторяющиеся имена приложений от уникальных имен приложений. Но если имя появляется, например, 5 раз, одна копия будет в unique_apps, а четыре копии будут в duplicate_apps. Я могу ошибаться, но я думаю, что вы хотели, чтобы это имя присутствовало только один раз в duplicate_apps. Если это то, что вам нужно, вы вносите минимальные изменения в свой код, используя set для duplicate_apps вместо списка, чтобы у вас не было дубликатов.

iOS_list = [
    ["Village", "did", "removed", "enjoyed", "explain", "nor", "ham", "saw", "calling", "talking."], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["feelings", "own", "shy.", "Request", "norland", "neither", "mistake", "for", "yet.", "Between"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["have", "an", "no", "at.", "Relation", "so", "in", "confined", "smallest", "children"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["account", "in", "outward", "tedious", "do.", "Particular", "way", "thoroughly", "unaffected", "projection"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["unpacked", "delicate.", "Why", "sir", "end", "believe", "uncivil", "respect.", "Always", "get"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["adieus", "nature", "day", "course", "for", "common.", "My", "little", "garret", "repair"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["have", "an", "no", "at.", "Relation", "so", "in", "confined", "smallest", "children"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["For", "who", "thoroughly", "her", "boy", "estimating", "conviction.", "Removed", "demands", "expense"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["account", "in", "outward", "tedious", "do.", "Particular", "way", "thoroughly", "unaffected", "projection"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["admire", "in", "giving.", "See", "resolved", "goodness", "felicity", "shy", "civility", "domestic"], 
    ["had", "but.", "Drawings", "offended", "yet", "answered", "jennings", "perceive", "laughing", "six"]
]
play_apps = [
    ["We", "diminution", "preference", "thoroughly", "if.", "Joy", "deal", "pain", "view", "much"], 
    ["her", "time.", "Led", "young", "gay", "would", "now", "state.", "Pronounce", "we"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["attention", "admitting", "on", "assurance", "of", "suspicion", "conveying.", "That", "his", "west"], 
    ["quit", "had", "met", "till.", "Of", "advantage", "he", "attending", "household", "at"], 
    ["do", "perceived.", "Middleton", "in", "objection", "discovery", "as", "agreeable.", "Edward", "thrown"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["dining", "so", "he", "my", "around", "to.", "Increasing", "impression", "interested", "expression"], 
    ["he", "my", "at.", "Respect", "invited", "request", "charmed", "me", "warrant", "to."], 
    ["Expect", "no", "pretty", "as", "do", "though", "so", "genius", "afraid", "cousin."], 
    ["do", "perceived.", "Middleton", "in", "objection", "discovery", "as", "agreeable.", "Edward", "thrown"], 
    ["Girl", "when", "of", "ye", "snug", "poor", "draw.", "Mistake", "totally", "of"], 
    ["in", "chiefly.", "Justice", "visitor", "him", "entered", "for.", "Continue", "delicate", "as"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["being", "style", "house.", "An", "whom", "down", "kept", "lain", "name", "so"], 
]

def find_duplicates(listy, index): 
    unique_apps = [] # list of all app names
    duplicate_apps = set() # list of known duplicate app names

    for app in listy:
        name = app[index]
        if name in unique_apps: 
            duplicate_apps.add(name) 
        elif name not in unique_apps: 
            unique_apps.append(name)
    print('Count of duplicate apps in data set:',len(duplicate_apps))
    print('\n')
    print('Sample of duplicate apps in data set:', list(duplicate_apps)[:3])
    print('\n')    

find_duplicates(play_apps, 0)
find_duplicates(iOS_list, 2)

Но мы можем упростить ваш код, используя выражение генератора , Счетчик , понимание списка и функцию random.sample().

import collections
import random

iOS_list = [
    ["Village", "did", "removed", "enjoyed", "explain", "nor", "ham", "saw", "calling", "talking."], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["feelings", "own", "shy.", "Request", "norland", "neither", "mistake", "for", "yet.", "Between"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["have", "an", "no", "at.", "Relation", "so", "in", "confined", "smallest", "children"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["account", "in", "outward", "tedious", "do.", "Particular", "way", "thoroughly", "unaffected", "projection"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["unpacked", "delicate.", "Why", "sir", "end", "believe", "uncivil", "respect.", "Always", "get"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["adieus", "nature", "day", "course", "for", "common.", "My", "little", "garret", "repair"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["have", "an", "no", "at.", "Relation", "so", "in", "confined", "smallest", "children"], 
    ["Securing", "as", "informed", "declared", "or", "margaret.", "Joy", "horrible", "moreover", "man"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["the", "for", "morning", "assured", "country", "believe.", "On", "even", "feet", "time"], 
    ["For", "who", "thoroughly", "her", "boy", "estimating", "conviction.", "Removed", "demands", "expense"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["account", "in", "outward", "tedious", "do.", "Particular", "way", "thoroughly", "unaffected", "projection"], 
    ["favourable", "mrs", "can", "projecting", "own.", "Thirty", "it", "matter", "enable", "become"], 
    ["admire", "in", "giving.", "See", "resolved", "goodness", "felicity", "shy", "civility", "domestic"], 
    ["had", "but.", "Drawings", "offended", "yet", "answered", "jennings", "perceive", "laughing", "six"]
]
play_apps = [
    ["We", "diminution", "preference", "thoroughly", "if.", "Joy", "deal", "pain", "view", "much"], 
    ["her", "time.", "Led", "young", "gay", "would", "now", "state.", "Pronounce", "we"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["attention", "admitting", "on", "assurance", "of", "suspicion", "conveying.", "That", "his", "west"], 
    ["quit", "had", "met", "till.", "Of", "advantage", "he", "attending", "household", "at"], 
    ["do", "perceived.", "Middleton", "in", "objection", "discovery", "as", "agreeable.", "Edward", "thrown"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["dining", "so", "he", "my", "around", "to.", "Increasing", "impression", "interested", "expression"], 
    ["he", "my", "at.", "Respect", "invited", "request", "charmed", "me", "warrant", "to."], 
    ["Expect", "no", "pretty", "as", "do", "though", "so", "genius", "afraid", "cousin."], 
    ["do", "perceived.", "Middleton", "in", "objection", "discovery", "as", "agreeable.", "Edward", "thrown"], 
    ["Girl", "when", "of", "ye", "snug", "poor", "draw.", "Mistake", "totally", "of"], 
    ["in", "chiefly.", "Justice", "visitor", "him", "entered", "for.", "Continue", "delicate", "as"], 
    ["unlocked", "entirely", "mr", "relation", "diverted", "in.", "Known", "not", "end", "fully"], 
    ["being", "style", "house.", "An", "whom", "down", "kept", "lain", "name", "so"], 
]

def find_duplicates(source_list, name_location):
    names = (current_app[name_location] for current_app in source_list)
    counts = collections.Counter(names)
    duplicates = [name for (name, count) in counts.items() if count > 1]
    duplicates_count = len(duplicates)
    sample  = random.sample(duplicates, min(3, duplicates_count))

    print('Count of duplicate apps in data set:', duplicates_count, '\n')
    print('Sample of duplicate apps in data set:', sample, '\n')

find_duplicates(play_apps, 0)
find_duplicates(iOS_list, 2)

Вас это устраивает?