金声玉亮2.0模型评测报告

问题

You are an expert Python programmer, and here is your task: Write a function to find the similar elements from the given two tuple lists. Your code should pass these tests:

assert similar_elements((3, 4, 5, 6),(5, 7, 4, 10)) == (4, 5)
assert similar_elements((1, 2, 3, 4),(5, 4, 3, 7)) == (3, 4)
assert similar_elements((11, 12, 14, 13),(17, 15, 14, 13)) == (13, 14)

[BEGIN]
'def similar_elements(test_tup1, test_tup2):
res = tuple(set(test_tup1) & set(test_tup2))
return (res)'
[DONE]

You are an expert Python programmer, and here is your task: Write a python function to identify non-prime numbers. Your code should pass these tests:

assert is_not_prime(2) == False
assert is_not_prime(10) == True
assert is_not_prime(35) == True

[BEGIN]
'import math
def is_not_prime(n):
result = False
for i in range(2,int(math.sqrt(n)) + 1):
if n % i == 0:
result = True
return result'
[DONE]

You are an expert Python programmer, and here is your task: Write a function to find the largest integers from a given list of numbers using heap queue algorithm. Your code should pass these tests:

assert heap_queue_largest( [25, 35, 22, 85, 14, 65, 75, 22, 58],3)==[85, 75, 65]
assert heap_queue_largest( [25, 35, 22, 85, 14, 65, 75, 22, 58],2)==[85, 75]
assert heap_queue_largest( [25, 35, 22, 85, 14, 65, 75, 22, 58],5)==[85, 75, 65, 58, 35]

[BEGIN]
'import heapq as hq
def heap_queue_largest(nums,n):
largest_nums = hq.nlargest(n, nums)
return largest_nums'
[DONE]

You are an expert Python programmer, and here is your task: Write a function to remove characters from the first string which are present in the second string. Your code should pass these tests:

assert remove_dirty_chars("probasscurve", "pros") == 'bacuve'
assert remove_dirty_chars("digitalindia", "talent") == 'digiidi'
assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'

[BEGIN]

答案

模型	模型答案	测试结果
金声玉亮2.0	```python def remove_dirty_chars(string1, string2): # Create a set of characters to remove chars_to_remove = set(string2) # Use a list comprehension to filter out the unwanted characters filtered_string = ''.join([char for char in string1 if char not in chars_to_remove]) return filtered_string ``` [DONE]	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
deepseekr1	To solve this problem, we need to write a function that removes characters from the first string if they are present in the second string. The solution should efficiently filter out the unwanted characters while maintaining the order of the remaining characters in the first string. ### Approach 1. Set Creation: Convert the second string into a set of characters. This allows for O(1) average-time complexity for membership checks, which is efficient even for longer strings. 2. Filtering Characters: Iterate through each character in the first string and check if it is present in the set created from the second string. If a character is not present in the set, it is included in the result. 3. Result Construction: Collect the filtered characters into a list and then join them into a string to form the final result. ### Solution Code ```python def remove_dirty_chars(str1, str2): # Create a set of characters to remove for O(1) lookups remove_chars = set(str2) # Filter characters from str1 not in the remove set filtered = [char for char in str1 if char not in remove_chars] # Join the list into a string and return return ''.join(filtered) ``` ### Explanation - Set Creation: By converting the second string into a set, we efficiently store the characters that need to be removed. This conversion allows us to check if a character from the first string should be removed in constant time. - Filtering: Using a list comprehension, we iterate through each character in the first string. For each character, we check if it is not in the set of characters to remove. This ensures that only characters not present in the second string are included in the result. - Result Construction: The filtered characters are collected into a list, which is then joined into a string to form the final output. This approach maintains the original order of characters from the first string while efficiently removing the unwanted characters. This solution efficiently handles the problem with a time complexity of O(n + m), where n is the length of the first string and m is the length of the second string, due to the set creation and list iteration steps. The space complexity is O(m) for storing the set of characters to remove, plus O(n) for the resulting list, leading to an overall O(n + m) space complexity.	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
gpto1	[BEGIN] 'def remove_dirty_chars(s1, s2): return "".join(c for c in s1 if c not in set(s2))' [DONE]	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
GPT-4o	```python def remove_dirty_chars(str1, str2): # Create a set of characters present in the second string for quick lookup dirty_chars = set(str2) # Use a list comprehension to filter out the characters present in the dirty set result = ''.join([char for char in str1 if char not in dirty_chars]) return result # Test cases assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles' ``` Here, I've written a function `remove_dirty_chars` that takes in two strings. It filters out characters from the first string that are present in the second string by using a set for fast membership checking. The function constructs the result string by including only those characters from the first string that are not in the set of characters from the second string. The function is also tested with the provided test cases to ensure correctness.	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
llama3.1-70B-Instruct	def remove_dirty_chars(s1, s2): for char in s2: s1 = s1.replace(char, '') return s1	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
llama3.1-405B-Instruct	def remove_dirty_chars(s1, s2): for char in s2: s1 = s1.replace(char, '') return s1	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
DeepSeek-Coder-V2	```python def remove_dirty_chars(string1, string2): # Create a set of characters to remove chars_to_remove = set(string2) # Use a list comprehension to filter out the unwanted characters filtered_string = ''.join([char for char in string1 if char not in chars_to_remove]) return filtered_string ``` [DONE]	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
Qwen2-72B-Instruct	'def remove_dirty_chars(str1, str2): return "".join(c for c in str1 if c not in str2)' [DONE]	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
baichuan4	To complete the task of removing characters from the first string that are present in the second string, you can use the following Python function: ```python def remove_dirty_chars(str1, str2): # Convert the second string into a set of characters for efficient lookup chars_to_remove = set(str2) # Use a list comprehension to filter out the characters that should be removed clean_chars = [char for char in str1 if char not in chars_to_remove] # Join the remaining characters to form the cleaned string return ''.join(clean_chars) ``` This function first creates a set of characters from the second string, which allows for O(1) lookup times when checking if a character from the first string should be removed. It then uses a list comprehension to filter out the unwanted characters and finally joins the remaining characters back into a string. Let's verify that it passes the tests you provided: ```python assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles' ``` When executed, no assertion errors should occur, indicating that the function behaves as expected for the given test cases.	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
doubao	'def remove_dirty_chars(str1, str2): result = "" for char in str1: if char not in str2: result += char return result' [DONE]	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
minimax	```python def remove_dirty_chars(str1, str2): return ''.join([char for char in str1 if char not in str2]) # Test cases assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles' ``` This function `remove_dirty_chars` iterates through each character in the first string `str1` and includes it in the result only if it is not present in the second string `str2`. The `join` method is used to concatenate the characters back into a string. The provided test cases should pass with this implementation.	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
kimi	Sure, let's write the function `remove_dirty_chars` that removes characters from the first string which are present in the second string. Here's the code: ```python def remove_dirty_chars(str1, str2): # Use a set comprehension to find characters in str2 and remove them from str1 return ''.join([char for char in str1 if char not in str2]) # Test cases assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles' ``` This function uses a list comprehension to iterate over each character in `str1` and includes it in the result if it is not found in `str2`. The `''.join(...)` part is used to concatenate the filtered characters back into a single string.	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
wenxinyiyan	Here are the solutions to the tasks you provided: 1. Finding similar elements from two tuple lists: ```python def similar_elements(test_tup1, test_tup2): res = tuple(set(test_tup1) & set(test_tup2)) # Sorting the result for consistency with the assertions return tuple(sorted(res)) ``` Note: I added a sorting step to ensure the order of the returned tuple is consistent with the assertions you provided. Sets do not maintain order, so sorting is necessary to guarantee a predictable output. 2. Identifying non-prime numbers: ```python import math def is_not_prime(n): if n < 2: # Numbers less than 2 are not prime return True for i in range(2, int(math.sqrt(n)) + 1): if n % i == 0: return True return False ``` 3. Finding the largest integers from a given list using the heap queue algorithm: ```python import heapq as hq def heap_queue_largest(nums, n): return hq.nlargest(n, nums) ``` 4. Removing characters from the first string which are present in the second string: ```python def remove_dirty_chars(s1, s2): # Create a set of characters from the second string for efficient lookup dirty_chars = set(s2) # Use a list comprehension to filter out characters from the first string that are in the second string cleaned_chars = [char for char in s1 if char not in dirty_chars] # Join the cleaned characters back into a string and return it return ''.join(cleaned_chars) ``` These solutions should pass the assertions you provided.	failed 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
zhipu	Here's a Python function that removes characters from the first string which are present in the second string. This function should pass the provided tests: ```python def remove_dirty_chars(s1, s2): # Create a set of characters to be removed chars_to_remove = set(s2) # Use a list comprehension to filter out the characters result = ''.join([char for char in s1 if char not in chars_to_remove]) return result # Test cases assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles' print("All tests passed!") ``` ### Explanation: 1. Set Creation: We create a set `chars_to_remove` from the second string `s2`. Sets are used because they have O(1) average time complexity for lookups. 2. List Comprehension: We use a list comprehension to iterate over each character in the first string `s1` and include it in the result if it is not in the `chars_to_remove` set. 3. Join: The list of characters is then joined back into a single string using `''.join()`. This approach ensures that the function is efficient and clear. The provided test cases are used to validate the correctness of the function. If no assertion error is raised, the "All tests passed!" message will be printed.	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
gpto1	[BEGIN] 'def remove_dirty_chars(s1, s2): return "".join(c for c in s1 if c not in set(s2))' [DONE]	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
deepseekv3	```python def remove_dirty_chars(first_str, second_str): # Create a set of characters from the second string for quick lookup chars_to_remove = set(second_str) # Filter out characters from the first string that are not in the set result = ''.join([char for char in first_str if char not in chars_to_remove]) return result ```	pass 测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve' assert remove_dirty_chars("digitalindia", "talent") == 'digiidi' assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'

模型

模型答案

测试结果

金声玉亮2.0

```python
def remove_dirty_chars(string1, string2):
# Create a set of characters to remove
chars_to_remove = set(string2)
# Use a list comprehension to filter out the unwanted characters
filtered_string = ''.join([char for char in string1 if char not in chars_to_remove])
return filtered_string
```
[DONE]

pass
测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve'
assert remove_dirty_chars("digitalindia", "talent") == 'digiidi'
assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'

deepseekr1

To solve this problem, we need to write a function that removes characters from the first string if they are present in the second string. The solution should efficiently filter out the unwanted characters while maintaining the order of the remaining characters in the first string.

### Approach
1. **Set Creation**: Convert the second string into a set of characters. This allows for O(1) average-time complexity for membership checks, which is efficient even for longer strings.
2. **Filtering Characters**: Iterate through each character in the first string and check if it is present in the set created from the second string. If a character is not present in the set, it is included in the result.
3. **Result Construction**: Collect the filtered characters into a list and then join them into a string to form the final result.

### Solution Code
```python
def remove_dirty_chars(str1, str2):
# Create a set of characters to remove for O(1) lookups
remove_chars = set(str2)
# Filter characters from str1 not in the remove set
filtered = [char for char in str1 if char not in remove_chars]
# Join the list into a string and return
return ''.join(filtered)
```

### Explanation
- **Set Creation**: By converting the second string into a set, we efficiently store the characters that need to be removed. This conversion allows us to check if a character from the first string should be removed in constant time.
- **Filtering**: Using a list comprehension, we iterate through each character in the first string. For each character, we check if it is not in the set of characters to remove. This ensures that only characters not present in the second string are included in the result.
- **Result Construction**: The filtered characters are collected into a list, which is then joined into a string to form the final output. This approach maintains the original order of characters from the first string while efficiently removing the unwanted characters.

This solution efficiently handles the problem with a time complexity of O(n + m), where n is the length of the first string and m is the length of the second string, due to the set creation and list iteration steps. The space complexity is O(m) for storing the set of characters to remove, plus O(n) for the resulting list, leading to an overall O(n + m) space complexity.

gpto1

[BEGIN]
'def remove_dirty_chars(s1, s2):
return "".join(c for c in s1 if c not in set(s2))'
[DONE]

GPT-4o

```python
def remove_dirty_chars(str1, str2):
# Create a set of characters present in the second string for quick lookup
dirty_chars = set(str2)
# Use a list comprehension to filter out the characters present in the dirty set
result = ''.join([char for char in str1 if char not in dirty_chars])
return result

# Test cases
assert remove_dirty_chars("probasscurve", "pros") == 'bacuve'
assert remove_dirty_chars("digitalindia", "talent") == 'digiidi'
assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
```

Here, I've written a function `remove_dirty_chars` that takes in two strings. It filters out characters from the first string that are present in the second string by using a set for fast membership checking. The function constructs the result string by including only those characters from the first string that are not in the set of characters from the second string. The function is also tested with the provided test cases to ensure correctness.

llama3.1-70B-Instruct

def remove_dirty_chars(s1, s2):
for char in s2:
s1 = s1.replace(char, '')
return s1

llama3.1-405B-Instruct

def remove_dirty_chars(s1, s2):
for char in s2:
s1 = s1.replace(char, '')
return s1

DeepSeek-Coder-V2

Qwen2-72B-Instruct

'def remove_dirty_chars(str1, str2):
return "".join(c for c in str1 if c not in str2)'
[DONE]

baichuan4

To complete the task of removing characters from the first string that are present in the second string, you can use the following Python function:

```python
def remove_dirty_chars(str1, str2):
# Convert the second string into a set of characters for efficient lookup
chars_to_remove = set(str2)
# Use a list comprehension to filter out the characters that should be removed
clean_chars = [char for char in str1 if char not in chars_to_remove]
# Join the remaining characters to form the cleaned string
return ''.join(clean_chars)
```

This function first creates a set of characters from the second string, which allows for O(1) lookup times when checking if a character from the first string should be removed. It then uses a list comprehension to filter out the unwanted characters and finally joins the remaining characters back into a string.

Let's verify that it passes the tests you provided:

```python
assert remove_dirty_chars("probasscurve", "pros") == 'bacuve'
assert remove_dirty_chars("digitalindia", "talent") == 'digiidi'
assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
```

When executed, no assertion errors should occur, indicating that the function behaves as expected for the given test cases.

doubao

'def remove_dirty_chars(str1, str2):
result = ""
for char in str1:
if char not in str2:
result += char
return result'
[DONE]

minimax

```python
def remove_dirty_chars(str1, str2):
return ''.join([char for char in str1 if char not in str2])

# Test cases
assert remove_dirty_chars("probasscurve", "pros") == 'bacuve'
assert remove_dirty_chars("digitalindia", "talent") == 'digiidi'
assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
```

This function `remove_dirty_chars` iterates through each character in the first string `str1` and includes it in the result only if it is not present in the second string `str2`. The `join` method is used to concatenate the characters back into a string. The provided test cases should pass with this implementation.

kimi

Sure, let's write the function `remove_dirty_chars` that removes characters from the first string which are present in the second string. Here's the code:

```python
def remove_dirty_chars(str1, str2):
# Use a set comprehension to find characters in str2 and remove them from str1
return ''.join([char for char in str1 if char not in str2])

# Test cases
assert remove_dirty_chars("probasscurve", "pros") == 'bacuve'
assert remove_dirty_chars("digitalindia", "talent") == 'digiidi'
assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'
```

This function uses a list comprehension to iterate over each character in `str1` and includes it in the result if it is not found in `str2`. The `''.join(...)` part is used to concatenate the filtered characters back into a single string.

wenxinyiyan

Here are the solutions to the tasks you provided:

1. Finding similar elements from two tuple lists:

```python
def similar_elements(test_tup1, test_tup2):
res = tuple(set(test_tup1) & set(test_tup2))
# Sorting the result for consistency with the assertions
return tuple(sorted(res))
```
Note: I added a sorting step to ensure the order of the returned tuple is consistent with the assertions you provided. Sets do not maintain order, so sorting is necessary to guarantee a predictable output.

2. Identifying non-prime numbers:

```python
import math

def is_not_prime(n):
if n < 2: # Numbers less than 2 are not prime
return True
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return True
return False
```
3. Finding the largest integers from a given list using the heap queue algorithm:

```python
import heapq as hq

def heap_queue_largest(nums, n):
return hq.nlargest(n, nums)
```
4. Removing characters from the first string which are present in the second string:

```python
def remove_dirty_chars(s1, s2):
# Create a set of characters from the second string for efficient lookup
dirty_chars = set(s2)
# Use a list comprehension to filter out characters from the first string that are in the second string
cleaned_chars = [char for char in s1 if char not in dirty_chars]
# Join the cleaned characters back into a string and return it
return ''.join(cleaned_chars)
```
These solutions should pass the assertions you provided.

failed
测试用例：assert remove_dirty_chars("probasscurve", "pros") == 'bacuve'
assert remove_dirty_chars("digitalindia", "talent") == 'digiidi'
assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'

zhipu

Here's a Python function that removes characters from the first string which are present in the second string. This function should pass the provided tests:

```python
def remove_dirty_chars(s1, s2):
# Create a set of characters to be removed
chars_to_remove = set(s2)
# Use a list comprehension to filter out the characters
result = ''.join([char for char in s1 if char not in chars_to_remove])
return result

# Test cases
assert remove_dirty_chars("probasscurve", "pros") == 'bacuve'
assert remove_dirty_chars("digitalindia", "talent") == 'digiidi'
assert remove_dirty_chars("exoticmiles", "toxic") == 'emles'

print("All tests passed!")
```

### Explanation:
1. **Set Creation**: We create a set `chars_to_remove` from the second string `s2`. Sets are used because they have O(1) average time complexity for lookups.
2. **List Comprehension**: We use a list comprehension to iterate over each character in the first string `s1` and include it in the result if it is not in the `chars_to_remove` set.
3. **Join**: The list of characters is then joined back into a single string using `''.join()`.

This approach ensures that the function is efficient and clear. The provided test cases are used to validate the correctness of the function. If no assertion error is raised, the "All tests passed!" message will be printed.

gpto1

[BEGIN]
'def remove_dirty_chars(s1, s2):
return "".join(c for c in s1 if c not in set(s2))'
[DONE]

deepseekv3

```python
def remove_dirty_chars(first_str, second_str):
# Create a set of characters from the second string for quick lookup
chars_to_remove = set(second_str)
# Filter out characters from the first string that are not in the set
result = ''.join([char for char in first_str if char not in chars_to_remove])
return result

```