Python：如果条件仅返回最后一个值，则在内部返回语句

Question

我有一个带有列的fruits 数据框：(Name, Color, ID) 和一个带有列的sentence 数据框：(Sentence, ID)。我需要将水果数据框的每条记录与句子数据框进行比较，如果水果名称在句子中完全如此，请将其颜色连接到句子中水果名称之前。

这就是我所做的：

import pandas as pd
import regex as re

# create fruit dataframe 
fruit_data = [['Apple', 'Red', 1], ['Mango', 'Yellow', 2], ['Grapes', 'Green', 3]] 
fruit_df = pd.DataFrame(fruit_data, columns = ['Name', 'Color', 'ID']) 
print(fruit_df)

# create sentence dataframe 
sentence = [['I like Apple', 1], ['I like ripe Mango', 2], ['Grapes are juicy', 3]] 
sentence_df = pd.DataFrame(sentence, columns = ['Sentence', 'ID']) 
print(sentence_df)


def search(desc, name, color, id):
    flag = 0
    if re.findall(r"\b" + name + r"\b", desc):
        desc_id = (sentence_df[sentence_df['Sentence'] == desc]['ID'].values[0])
        if desc_id == id:
            flag = 1
        
        if flag == 1:
            # for loop is used because fruit can appear more than once in sentence
            all_indexes = []
            for match in re.finditer(r"\b" + name + r"\b", desc):
                     all_indexes.append(match.start())
            
            arr = list(desc)
            for idx in sorted(all_indexes, reverse=True):
                       arr.insert(idx, color + " ")

            new_desc = ''.join(arr)
           
            print("modified sentence: ", new_desc)
            return new_desc 

def compare(name, color, id):
    sentence_df['Result'] = sentence_df['Sentence'].apply(lambda x: search(x, name, color, id))
    

fruit_df.apply(lambda x: compare(x['Name'], x['Color'], x['ID']), axis=1)
print ("The final result is: ")
print(sentence_df['Result'])

代码的结果是：

     Name    Color  ID
0   Apple     Red    1
1   Mango  Yellow    2
2  Grapes   Green    3

            Sentence  ID
0       I like Apple   1
1  I like ripe Mango   2
2   Grapes are juicy   3


modified sentence:  I like Red Apple
modified sentence:  I like ripe Yellow Mango
modified sentence:  Green Grapes are juicy


The final result is: 
0                      None
1                      None
2    Green Grapes are juicy
Name: Result, dtype: object

句子被正确修改，但问题是，前两个句子没有存储在句子数据框的Result 列中，只有最后一个句子被存储。这是正确的做法还是我错过了什么？

Answer 1

经过一些修改：

import pandas as pd
import re
# create fruit dataframe 
fruit_data = [['Apple', 'Red', 1], ['Mango', 'Yellow', 2], ['Grapes', 'Green', 3]] 
fruit_df = pd.DataFrame(fruit_data, columns = ['Name', 'Color', 'ID']) 
print(fruit_df)

# create sentence dataframe 
sentence = [['I like Apple', 1], ['I like ripe Mango', 2], ['Grapes are juicy', 3]] 
sentence_df = pd.DataFrame(sentence, columns = ['Sentence', 'ID']) 
print(sentence_df)


def search(ids):
    name = fruit_df[fruit_df['ID']==ids]['Name'].values[0]
    desc = sentence_df[sentence_df['ID']==ids]['Sentence'].values[0]
    color = fruit_df[fruit_df['ID']==ids]['Color'].values[0]
    if True:# kept to maintain this indentation
            # for loop is used because fruit can appear more than once in sentence
            all_indexes = []
            for match in re.finditer(r"\b" + name + r"\b", desc):
                     all_indexes.append(match.start())
            
            arr = list(desc)
            for idx in sorted(all_indexes, reverse=True):
                       arr.insert(idx, color + " ")

            new_desc = ''.join(arr)
           
            print("modified sentence: ", new_desc)
            return new_desc


sentence_df['Result'] = sentence_df['ID'].apply(lambda x: search(x))
    

print("The final result is: ")
print(sentence_df['Result'])

---

变化：

主要问题在这里fruit_df.apply。这会为fruits_df 中的每个项目调用比较函数，这意味着在示例中提供了 3 次。

然后compare根据fruit_df.apply传递的当前水果修改了Result列中的所有条目。

所以，第一步是只调用一次。

需要进行的另一项更改是使用 foreign-key : ID。

ID 存在于两个数据帧中，因此足以识别search 函数中的name、desc 和color。

---

输出：

      Name   Color  ID
0   Apple     Red   1
1   Mango  Yellow   2
2  Grapes   Green   3


            Sentence  ID
0       I like Apple   1
1  I like ripe Mango   2
2   Grapes are juicy   3


modified sentence:  I like Red Apple
modified sentence:  I like ripe Yellow Mango
modified sentence:  Green Grapes are juicy


The final result is: 
0            I like Red Apple
1    I like ripe Yellow Mango
2      Green Grapes are juicy
Name: Result, dtype: object

编辑：根据 OP 的要求，解决方案的快速修复版本。

只要把你原来的代码底部改成如下，

def compare(name, color, id):
    sentence_df['Result'] = sentence_df['Result'].apply(lambda x: search(x, name, color, id) or x)
    
sentence_df['Result'] = sentence_df['Sentence']

fruit_df.apply(lambda x: compare(x['Name'], x['Color'], x['ID']), axis=1)
print ("The final result is: ")
print(sentence_df['Result'])

注意：在此修复中，上述问题并未在技术上得到解决。只是引入了一个小旁路来达到所需的输出。

Answer 2

您在这两个地方都使用 Dataframe.apply 方法是代码中的问题。 apply 方法可帮助您在数据框的任何轴上应用方法，而不是添加新列并为其分配值。 如果您要处理相同的数据帧并想要执行上述操作，您可以使用 .assign 方法。这使您可以分配一个新列，其值是根据其他列的值计算的。 对于您的代码，如果您想保留相同的代码而不是按照上面的建议重构它，您需要的只是一个循环。

for idx, row in fruit_df.iterrows():
result = search(sentence_df.loc[idx,"Sentence"], row["Name"], row["Color"], row["ID"])
sentence_df.loc[idx,"Result"] = result