I am trying to remove the character ‘ from my string by doing the following
kickoff = tree.xpath('//*[@id="page"]/div[1]/div/main/div/article/div/div[1]/section[2]/p[1]/b[1]/text()')
kickoff = kickoff.replace("'", "")
This gives me the error AttributeError: ‘list’ object has no attribute ‘replace’
Coming from a php background I am unsure what the correct way to do this is?
falsetru
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asked Apr 15, 2016 at 9:03
2
xpath method returns a list, you need to iterate items.
kickoff = [item.replace("'", "") for item in kickoff]
answered Apr 15, 2016 at 9:05
falsetrufalsetru
349k62 gold badges701 silver badges621 bronze badges
2
kickoff = tree.xpath('//*[@id="page"]/div[1]/div/main/div/article/div/div[1]/section[2]/p[1]/b[1]/text()')
This code is returning list not a string.Replace function will not work on list.
[i.replace("'", "") for i in kickoff ]
answered Apr 15, 2016 at 9:06
Himanshu duaHimanshu dua
2,4461 gold badge17 silver badges27 bronze badges
This worked for me:
kickoff = str(tree.xpath('//*[@id="page"]/div[1]/div/main/div/article/div/div[1]/section[2]/p[1]/b[1]/text()'))
kickoff = kickoff.replace("'", "")
This error is caused because the xpath returns in a list. Lists don’t have the replace attribute. So by putting str before it, you convert it to a string which the code can handle. I hope this helped!
answered Mar 28, 2019 at 3:29
![]()
In Python, the list data structure stores elements in sequential order. We can use the String replace() method to replace a specified string with another specified string. However, we cannot apply the replace() method to a list. If you try to use the replace() method on a list, you will raise the error “AttributeError: ‘list’ object has no attribute ‘replace’”.
This tutorial will go into detail on the error definition. We will go through an example that causes the error and how to solve it.
Table of contents
- AttributeError: ‘list’ object has no attribute ‘replace’
- Python replace() Syntax
- Example #1: Using replace() on a List of Strings
- Solution
- Example #2: Using split() then replace()
- Solution
- Summary
AttributeError: ‘list’ object has no attribute ‘replace’
AttributeError occurs in a Python program when we try to access an attribute (method or property) that does not exist for a particular object. The part “‘list’ object has no attribute ‘replace’” tells us that the list object we are handling does not have the replace attribute. We will raise this error if we try to call the replace() method on a list object. replace() is a string method that replaces a specified string with another specified string.
Python replace() Syntax
The syntax for the String method replace() is as follows:
string.replace(oldvalue, newvalue, count)
Parameters:
- oldvalue: Required. The string value to search for within string
- newvalue: Required. The string value to replace the old value
- count: Optional. A number specifying how many times to replace the old value with the new value. The default is all occurrences
Let’s look at an example of calling the replace() method to remove leading white space from a string:
str_ = "the cat is on the table"
str_ = str.replace("cat", "dog")
print(str_)
the dog is on the table
Now we will see what happens if we try to use the replace() method on a list:
a_list = ["the cat is on the table"]
a_list = a_list.replace("cat", "dog")
print(a_list)
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
1 a_list = ["the cat is on the table"]
2
----≻ 3 a_list = a_list.replace("cat", "dog")
4
5 print(a_list)
AttributeError: 'list' object has no attribute 'replace'
The Python interpreter throws the Attribute error because the list object does not have replace() as an attribute.
Example #1: Using replace() on a List of Strings
Let’s look at an example list of strings containing descriptions of different cars. We want to use the replace() method to replace the phrase “car” with “bike”. Let’s look at the code:
lst = ["car one is red", "car two is blue", "car three is green"]
lst = lst.replace('car', 'bike')
print(lst)
Let’s run the code to get the result:
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
----≻ 1 lst = lst.replace('car', 'bike')
AttributeError: 'list' object has no attribute 'replace'
We can only call the replace() method on string objects. If we try to call replace() on a list, we will raise the AttributeError.
Solution
We can use list comprehension to iterate over each string and call the replace() method. Let’s look at the revised code:
lst = ["car one is red", "car two is blue", "car three is green"]
lst_repl = [i.replace('car', 'bike') for i in lst]
print(lst_repl)
List comprehension provides a concise, Pythonic way of accessing elements in a list and generating a new list based on a specified condition. In the above code, we create a new list of strings and replace every occurrence of “car” in each string with “bike”. Let’s run the code to get the result:
['bike one is red', 'bike two is blue', 'bike three is green']
Example #2: Using split() then replace()
A common source of the error is the use of the split() method on a string prior to using replace(). The split() method returns a list of strings, not a string. Therefore if you want to perform any string operations you will have to iterate over the items in the list. Let’s look at an example:
particles_str = "electron,proton,muon,cheese"
We have a string that stores four names separated by commas. Three of the names are correct particle names and the last one “cheese” is not. We want to split the string using the comma separator and then replace the name “cheese” with “neutron”. Let’s look at the implementation that will raise an AttributeError:
particles = str_.split(",")
particles = particles.replace("cheese", "neutron")
Let’s run the code to see the result:
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
----≻ 1 particles = particles.replace("cheese", "neutron")
AttributeError: 'list' object has no attribute 'replace'
The error occurs because particles is a list object, not a string object:
print(particles)
['electron', 'proton', 'muon', 'cheese']
Solution
We need to iterate over the items in the particles list and call the replace() method on each string to solve this error. Let’s look at the revised code:
particles = [i.replace("cheese","neutron") for i in particles]
print(particles)
In the above code, we create a new list of strings and replace every occurrence of “cheese” in each string with “neutron”. Let’s run the code to get the result:
['electron', 'proton', 'muon', 'neutron']
Summary
Congratulations on reading to the end of this tutorial! The error “AttributeError: ‘list’ object has no attribute ‘replace’” occurs when you try to use the replace() function to replace a string with another string on a list of strings.
The replace() function is suitable for string type objects. If you want to use the replace() method, ensure that you iterate over the items in the list of strings and call the replace method on each item. You can use list comprehension to access the items in the list.
Generally, check the type of object you are using before you call the replace() method.
For further reading on AttributeErrors involving the list object, go to the article:
- How to Solve Python AttributeError: ‘list’ object has no attribute ‘split’.
- How to Solve Python AttributeError: ‘list’ object has no attribute ‘lower’.
- How to Solve Python AttributeError: ‘list’ object has no attribute ‘get’.
To learn more about Python for data science and machine learning, go to the online courses page on Python for the most comprehensive courses available.
Have fun and happy researching!
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saladdd 1 / 1 / 1 Регистрация: 23.04.2014 Сообщений: 637 |
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25.05.2018, 00:45. Показов 21220. Ответов 12 Метки нет (Все метки)
raceback (most recent call last):
__________________
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437 / 429 / 159 Регистрация: 21.05.2016 Сообщений: 1,338 |
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25.05.2018, 01:19 |
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Напишите print(content) да посмотрите что там. Там список строк, конечно
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saladdd 1 / 1 / 1 Регистрация: 23.04.2014 Сообщений: 637 |
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26.05.2018, 16:29 [ТС] |
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oldnewyear, я имел ввиду как переменную content дать на вход функции replace.
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5403 / 3827 / 1214 Регистрация: 28.10.2013 Сообщений: 9,554 Записей в блоге: 1 |
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26.05.2018, 16:39 |
4 |
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Замените f.readlines() на f.read(), либо обходите список строк в цикле.
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1 / 1 / 1 Регистрация: 23.04.2014 Сообщений: 637 |
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26.05.2018, 18:29 [ТС] |
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Garry Galler, вот так и делаю уже
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Semen-Semenich 3964 / 2946 / 1067 Регистрация: 21.03.2016 Сообщений: 7,452 |
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26.05.2018, 20:53 |
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saladdd,
создает список а не строку о чем вам и говорится в ошибки что список не имеет атрибута replace. справка по методу replace str.replace(old, new[, maxcount]) -> str
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saladdd 1 / 1 / 1 Регистрация: 23.04.2014 Сообщений: 637 |
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26.05.2018, 21:46 [ТС] |
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Semen-Semenich, мне говорят что вот так вот тоже можно
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Semen-Semenich 3964 / 2946 / 1067 Регистрация: 21.03.2016 Сообщений: 7,452 |
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26.05.2018, 21:56 |
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saladdd, вы так и не поняли свою ошибку.
или как сказано выше для readlines()
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saladdd 1 / 1 / 1 Регистрация: 23.04.2014 Сообщений: 637 |
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27.05.2018, 16:20 [ТС] |
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Semen-Semenich, я свою ошибку понял , просто не совсем понмял как происходит присвоение. Добавлено через 35 минут
и вот такой
они оба рабочие от чего это завесит, от версии python.
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Garry Galler
5403 / 3827 / 1214 Регистрация: 28.10.2013 Сообщений: 9,554 Записей в блоге: 1 |
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27.05.2018, 17:15 |
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они оба рабочие Этот вариант
не рабочий.
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saladdd 1 / 1 / 1 Регистрация: 23.04.2014 Сообщений: 637 |
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27.05.2018, 18:18 [ТС] |
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Garry Galler, спасибо я с этим вариантом разобрался , тот вариант про который я говорил выше я нашёл на просторах сети скорее всего он и нерабочий, я хотел поинтересоваться.Почему иногда контент передают через аргумент в выражении?
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Garry Galler
5403 / 3827 / 1214 Регистрация: 28.10.2013 Сообщений: 9,554 Записей в блоге: 1 |
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27.05.2018, 18:32 |
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Почему иногда контент передают через аргумент в выражении Потому что это позволяет синтаксис python’а.
В общем случае такие варианты (через обращение к классу) не особенно нужны.
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saladdd 1 / 1 / 1 Регистрация: 23.04.2014 Сообщений: 637 |
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27.05.2018, 18:51 [ТС] |
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Garry Galler, скажите ,а если способ в моём случае записать разбитую строку content.split(‘ ‘) только чтобы конечный результат
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Уведомления
- Начало
- » Python для новичков
- » Ошибка в коде
#1 Ноя. 19, 2015 23:59:15
Ошибка в коде
Есть код:
path = glob.glob("C:...*.txt") with open("sample_text_collection_metadata.txt", "w", encoding = "utf-8") as metadata_out: for txt in test: f = open(txt, "r", encoding = "utf-8").read() autor_name = f.split("Autor:")[1].replace("r", "").split("n")[0] text_file = file.split("Title:")[1].replace("r", "").split("n")[0] file_name = file.split("/")[-1] metadata_out.write(file_name + "t" + autor_name + "t" + text_file + "n") f.close() with open("sample_text_collection/" + file_name, "w", "utf-8") as file_out: file_out.write(txt)
Загвоздка в том, что когда выполнение кода доходит до autor_name = f.split(“Autor:”).replace(“r”, “”).split(“n”) появляется ошибка AttributeError: ‘list’ object has no attribute ‘replace’
Вопрос: откуда берется ‘list’? если при проверке type(f) мы получаем <class ‘str’>.
Офлайн
- Пожаловаться
#2 Ноя. 20, 2015 02:36:45
Ошибка в коде
Pytonist
Вопрос: откуда берется ‘list’?
>>> 'a|b|c'.split('|') ['a', 'b', 'c'] >>>
Офлайн
- Пожаловаться
#3 Ноя. 20, 2015 15:15:12
Ошибка в коде
Уже разобрался. Причина была банальна, было неверно задано условие … Наверное надо перекурить денек второй …
Офлайн
- Пожаловаться
#4 Ноя. 21, 2015 18:17:27
Ошибка в коде
как сложить все числа, данные в документе?
вот что уже есть:
g=open(‘dfg.txt’, ‘rt’)
s=0
while True:
f = g.readline()
if f == »:
break
n = int(f.strip())
Вот документ:
12
23
45
1
2
зарание спасибо!
Офлайн
- Пожаловаться
#5 Ноя. 21, 2015 18:28:18
Ошибка в коде
Собрать строки в документе в список, попутно преобразовав в числа и просуммировать через sum
In [1]: sum([1,2,3,4,5]) Out[1]: 15
_________________________
Python golden rule: Do not PEP 8 unto others; only PEP 8 thy self.
Don’t let PEP 8 make you insanely intolerant of other people’s code.
Офлайн
- Пожаловаться
Во-первых,
for i in range(len(dicts)):
for x in dicts[i].items():
dicts[i].items.replace('-','_')
Вы получаете исключение, потому что вы пытаетесь достичь метода «replace» в функции «items», а не вызов функции. так должно быть:
dicts[i].items().replace('-','_')
Во-вторых, для меня это неправильный подход.
Поскольку вы используете python, почему бы не сделать это питоническим способом и просто не вернуть новый список новых dicts с новыми значениями?
Представленная ниже функция берет список dicts списков кортежей (это предоставленная вами структура) и заменяет каждое «to_change» на «change_to». Я вызываю это, используя ваши требования, чтобы изменить каждый знак «-» на «_».
d=[
{
'title': [('Agente 007, Moonraker: Operazione spazio', 'it')],
'sub-title': [('Missione nel cosmo per...', 'it')],
},{
'title': [('Agente 007, Vivi e lascia morire', 'it')],
'sub-title': [('Il primo James Bond con...', 'it')]
}
]
def sub_string(dict_list, to_change, change_to):
new_list = []
for d_orig in dict_list:
d_new = {}
for key in d_orig.keys():
new_key = key.replace(to_change, change_to)
new_value = [[s.replace(to_change, change_to) for s in tup] for tup in d_orig[key]]
d_new[new_key] = new_value
new_list.append(d_new)
return new_list
sub_string(d, '-', '_')
Вы это имели в виду?
ERROR:docstamp.cli.cli:Error filling document for 17th item
Traceback (most recent call last):
File «/opt/anaconda3/envs/acpyss/lib/python3.5/site-packages/docstamp/cli/cli.py», line 152, in create
template_doc.fill(item)
File «/opt/anaconda3/envs/acpyss/lib/python3.5/site-packages/docstamp/template.py», line 220, in fill
doc_contents[key] = replace_chars_for_svg_code(content)
File «/opt/anaconda3/envs/acpyss/lib/python3.5/site-packages/docstamp/svg_utils.py», line 26, in replace_chars_for_svg_code
result = result.replace(c, entity)
AttributeError: ‘list’ object has no attribute ‘replace’
ERROR:docstamp.cli.cli:Error filling document for 21th item
Traceback (most recent call last):
File «/opt/anaconda3/envs/acpyss/lib/python3.5/site-packages/docstamp/cli/cli.py», line 152, in create
template_doc.fill(item)
File «/opt/anaconda3/envs/acpyss/lib/python3.5/site-packages/docstamp/template.py», line 220, in fill
doc_contents[key] = replace_chars_for_svg_code(content)
File «/opt/anaconda3/envs/acpyss/lib/python3.5/site-packages/docstamp/svg_utils.py», line 26, in replace_chars_for_svg_code
result = result.replace(c, entity)
AttributeError: ‘list’ object has no attribute ‘replace’
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AttributeError: ‘Response’ object has no attribute ‘replace’ |
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Posts: 404 Threads: 94 Joined: Dec 2017 Reputation: #! python3
# using the inauguration speech of William Henry Harrison analyzed in the previous example, we can write the following code that generates arbitrarily # long Markov chains (with the chain length set to 100) based on the #structure of its text
import requests
from random import randint
def wordListSum(wordList):
sum = 0
for word, value in wordList.items():
sum += value
return sum
def retrieveRandomWord(wordList):
randIndex = randint(1, wordListSum(wordList))
for word, value in wordList.items():
randIndex -= value
if randIndex <= 0:
return word
def buildWordDict(text):
# Remove newlines and quotes
text = text.replace("n", " ")
text = text.replace('"', "")
# Make sure punctuation marks are treated as their own "words"
# so that they will be included in the Markov chain
punctuation = [",", ".", ";", ":"]
for symbol in punctuation:
text = text.replace(symbol, " " + symbol + " ")
words = text.split(" ")
# Filter out empty words
words = [word for word in words if word != ""]
wordDict = {}
for i in range(1, len(words)):
if words[i - 1] not in wordDict:
# Create a new dictionary for this word
wordDict[words[i - 1]] = {}
if words[i] not in wordDict[words[i - 1]]:
wordDict[words[i - 1]][words[i]] = 0
wordDict[words[i - 1]][words[i]] += 1
return wordDict
text = requests.get("http://pythonscraping.com/files/inaugurationSpeech.txt")
if text.status_code == 200:
content = text.text
wordDict = buildWordDict(text)
# Generate a Markov chain of length 100
length = 100
chain = ""
currentWord = "I"
for i in range(0, length):
chain += currentWord + " "
currentWord = retrieveRandomWord(wordDict[currentWord])
print(chain)
There is a replace method in the documentation, does this message means that there is not attribute new line? Posts: 4,229 Threads: 97 Joined: Sep 2016 Reputation: requests.get returns a Request object, which has no replace method. If you want to use the string replace method, you need to get the text attribute of the Request object, and use replace on that. Truman
Posts: 404 Threads: 94 Joined: Dec 2017 Reputation: Thank you, this is how it should be done: text = requests.get("http://pythonscraping.com/files/inaugurationSpeech.txt")
if text.status_code == 200:
content = text.text
wordDict = buildWordDict(content)
Truman
Posts: 404 Threads: 94 Joined: Dec 2017 Reputation: It may be off topic but would anyone be so kind to give me further explanations on lines 36-42. It’s about creating two-dimensional dictionary but it looks confusing to me. Posts: 4,229 Threads: 97 Joined: Sep 2016 Reputation: Line 36 is looping through the indexes of the words list. Not very pythonic. You should loop over items, not the indexes. This loop is meant to be over pairs of consecutive words, but there are ways to do that without resorting to indexes. Lines 37 and 39 create a sub-dictionary for the previous word (note the for loop starts with 1, the index of the second word) if it doesn’t already have one. You could use collections.defaultdict(dict) instead. Lines 40 and 41 creates a zero count for the current word if it’s not in the sub-dictionary for the previous word. Again, you could do this with collections.defaultdict(int) instead, but it’s not clear to me how to initialize the nested defaultdicts. Finally, line 42 adds one to the count of the current word in the previous word’s sub dictionary. So it’s creating a dictionary with counts of pairs of words. Frex, wordDict[‘spam’][‘eggs’] would be the number of times the word ‘eggs’ occurred after the word ‘spam’. Truman
Posts: 404 Threads: 94 Joined: Dec 2017 Reputation: I am a shame of my ignorance but will dare to ask. What does this piece of code means: for word, value in wordList.items():
randIndex -= value
if randIndex <= 0:
return word
why randIndex-value? Posts: 4,229 Threads: 97 Joined: Sep 2016 Reputation:
It’s a way to do a weighted random selection. You have a dictionary of words and their weight (the value). You make a random number from 1 to the total of the weights (that’s what is in randIndex). Then you go through the words and subtract each word’s value from randIndex ( In 3.6+, the random module has choices, which can handle this for you more efficiently. However, prior to that you have to use code like the above. Truman
Posts: 404 Threads: 94 Joined: Dec 2017 Reputation: Does that mean that some words will never be returned? Or they will because dictionaries are unordered? In that case, why do we need to do this calculation if we can randomly choose any word? Posts: 4,229 Threads: 97 Joined: Sep 2016 Reputation: Mar-19-2019, 11:37 PM
The only time a word would never be returned is if it’s weight was 0. Then it would have a probability 0 / t of being returned. It means that some words are more likely to be returned than others, and some words are less like to be returned than others. The standard random.choice() selects every item in the provided sequence with equal likelihood. Truman
Posts: 404 Threads: 94 Joined: Dec 2017 Reputation:
That’s exactly why I don’t understand why the author of this code used weighted random selection. Wouldn’t it be more ‘fair’ that each word has an equal chance of being selected? |
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