英文:
How to extract dict values of pandas DataFrame in new columns?
问题
I would like to extract the values of a dictionary inside a Pandas DataFrame df into new columns of that DataFrame. All keys in the referring dict are the same across all rows.
import pandas as pddf = pd.DataFrame({'a': [1, 2, 3], 'b': [{'x':[101], 'y': [102], 'z': [103]}, {'x':[201], 'y': [202], 'z': [203]}, {'x':[301], 'y': [302], 'z': [303]}]})
dfResult = pd.DataFrame({'a': [1, 2, 3], 'x':[101, 201, 301], 'y': [102, 202, 302], 'z': [103, 203, 303]})
I am as far as I can get the keys and values out of the dict, but I do not know how to make new columns out of them:
df.b.apply(lambda x: [x[y] for y in x.keys()])0 [[101], [102], [103]]1 [[201], [202], [203]]2 [[301], [302], [303]]df.b.apply(lambda x: [y for y in x.keys()])0 [x, y, z]1 [x, y, z]2 [x, y, z]
英文:
I would like to extract the values of a dictionary inside a Pandas DataFrame df into new columns of that DataFrame. All keys in the referring dict are the same across all rows.
import pandas as pddf = pd.DataFrame({'a': [1, 2, 3], 'b': [{'x':[101], 'y': [102], 'z': [103]}, {'x':[201], 'y': [202], 'z': [203]}, {'x':[301], 'y': [302], 'z': [303]}]})
dfResult = pd.DataFrame({'a': [1, 2, 3], 'x':[101, 201, 301], 'y': [102, 202, 302], 'z': [103, 203, 303]})
I am as far as I can get the keys and values out of the dict, but I do not know how to make new columns out of them:
df.b.apply(lambda x: [x[y] for y in x.keys()])0 [[101], [102], [103]]1 [[201], [202], [203]]2 [[301], [302], [303]]df.b.apply(lambda x: [y for y in x.keys()])0 [x, y, z]1 [x, y, z]2 [x, y, z]
答案1
得分: 3
如果始终存在只有一个元素的列表，可以使用嵌套列表与字典推导，然后传递给 DataFrame 构造函数：
df = df.join(pd.DataFrame([{k: v[0] for k, v in x.items()} for x in df.pop('b')],index=df.index))print (df)a x y z0 1 101 102 1031 2 201 202 2032 3 301 302 303
另一种方法是为字典推导中的每一行创建一个 DataFrame，并使用 concat 进行连接：
df = df.join(pd.concat({k: pd.DataFrame(v) for k, v in df.pop('b').items()}).droplevel(1))print (df)a x y z0 1 101 102 1031 2 201 202 2032 3 301 302 303
英文:
If there are always one element lists is possible use nested list with dictionary comprehension and pass to DataFrame constructor:
df = df.join(pd.DataFrame([{k: v[0] for k, v in x.items()} for x in df.pop('b')],index=df.index))print (df)a x y z0 1 101 102 1031 2 201 202 2032 3 301 302 303
Another idea is create DataFrame for each row in dictionary comprehension and join by concat:
df = df.join(pd.concat({k: pd.DataFrame(v) for k, v in df.pop('b').items()}).droplevel(1))print (df)a x y z0 1 101 102 1031 2 201 202 2032 3 301 302 303
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