英文:
Finding the index of an element is list in a Pandas Dataframe column
问题
我需要在 pandas 中执行以下操作：
import pandas as pd# 创建示例数据帧data = {'col_a': ['hello'],'col_b': [['goodbye', 'how are you', 'hello']],'col_c': [['farewell', 'question', 'greeting']]}df = pd.DataFrame(data)# 使用 apply 方法获取所需的值df['col_d'] = df.apply(lambda row: row['col_c'][row['col_b'].index(row['col_a'][0])], axis=1)# col_d 中包含所需的值
通过这种方式，你可以在 Pandas 中找到所需的值并将其放入一个新的列（在此示例中是 'col_d'）。
英文:
I have a pandas dataframe:
col_a col_b col_chello [goodbye, how are you, hello] [farewell, question, greeting]
I need to find the value in the list in col_c which has the same index as the value of col_a in the list col_b. So in this case: "hello" has index 2 in [goodbye, how are you, hello]. I need to get the value with index 2 from col_c - in this case "greeting".
How can I do this in pandas?
答案1
得分: 2
df['out'] = [next((c2 for b2, c2 in zip(b, c) if a in b2), None)for a, b, c in zip(df['col_a'], df['col_b'], df['col_c'])]
输出：
col_a col_b col_c out0 hello [goodbye, how are you, hello] [farewell, question, greeting] greeting
英文:
You can use a list comprehension with zip and next:
df['out'] = [next((c2 for b2, c2 in zip(b, c) if a in b2), None)for a, b, c in zip(df['col_a'], df['col_b'], df['col_c'])]
Output:
col_a col_b col_c out0 hello [goodbye, how are you, hello] [farewell, question, greeting] greeting
答案2
得分: 2
已添加get_result_value(row)函数并实现了逻辑，该函数会根据列b中与列a对应的索引获取结果，并返回列c中的结果。
已附上结果。
data_set = {'col_a': ['hello'],'col_b': [['goodbye', 'how are you', 'hello']],'col_c': [['farewell', 'question', 'greeting']]}df = pd.DataFrame(data_set)def get_result_value(row):index = row['col_b'].index(row['col_a'])return row['col_c'][index]print(df.apply(get_result_value, axis=1))
结果：
英文:
Just added the function for get_result_value(row) and implemented logic as the index will contain column b corresponding to column a then will get a result from the index and column c.
I have attached the result as well.
data_set = {'col_a': ['hello'],'col_b': [['goodbye', 'how are you', 'hello']],'col_c': [['farewell', 'question', 'greeting']]}df = pd.DataFrame(data_set)def get_result_value(row):index = row['col_b'].index(row['col_a'])return row['col_c'][index]print(df.apply(get_result_value, axis=1))
Result:
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