Ale*_*rdt 8 python performance json pandas
假设我有以下 DataFrame,其中该data列包含一个嵌套的 JSON 字符串,我想将其解析为单独的列:
import pandas as pd
df = pd.DataFrame({
'bank_account': [101, 102, 201, 301],
'data': [
'{"uid": 100, "account_type": 1, "account_data": {"currency": {"current": 1000, "minimum": -500}, "fees": {"monthly": 13.5}}, "user_name": "Alice"}',
'{"uid": 100, "account_type": 2, "account_data": {"currency": {"current": 2000, "minimum": 0}, "fees": {"monthly": 0}}, "user_name": "Alice"}',
'{"uid": 200, "account_type": 1, "account_data": {"currency": {"current": 3000, "minimum": 0}, "fees": {"monthly": 13.5}}, "user_name": "Bob"}',
'{"uid": 300, "account_type": 1, "account_data": {"currency": {"current": 4000, "minimum": 0}, "fees": {"monthly": 13.5}}, "user_name": "Carol"}'
]},
index = ['Alice', 'Alice', 'Bob', 'Carol']
)
df
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我找到了这个json_normalize函数,目前正在列表理解中解析 JSON;结果是正确的,但这需要很长时间。1000 行需要 1-2 秒,而我在实际运行中大约有一百万行:
import json
from pandas.io.json import json_normalize
parsed_df = pd.concat([json_normalize(json.loads(js)) for js in df['data']])
parsed_df['bank_account'] = df['bank_account'].values
parsed_df.index = parsed_df['user_id']
parsed_df
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有没有更快的方法将这些数据解析为漂亮的 DataFrame?
我发现绕过pandas.concat.
否则,重写/优化json_normalize似乎并不简单。
def original(df):
parsed_df = pd.concat([json_normalize(json.loads(js)) for js in df['data']])
parsed_df['bank_account'] = df['bank_account'].values
parsed_df.index = parsed_df['uid']
return parsed_df
def jp(df):
cols = ['account_data.currency.current', 'account_data.currency.minimum',
'account_data.fees.monthly', 'account_type', 'uid', 'user_name']
parsed_df = pd.DataFrame([json_normalize(json.loads(js)).values[0] for js in df['data']],
columns=cols)
parsed_df['bank_account'] = df['bank_account'].values
parsed_df.index = parsed_df['uid']
return parsed_df
df = pd.concat([df]*100, ignore_index=True)
%timeit original(df) # 675 ms per loop
%timeit jp(df) # 526 ms per loop
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