[英]Python Pandas - Iterate over unique columns
我正在嘗試遍歷唯一列值的列表,以在字典中創建帶有字典的三個不同鍵。 這是我現在擁有的代碼:
import pandas as pd
dataDict = {}
metrics = frontendFrame['METRIC'].unique()
for metric in metrics:
dataDict[metric] = frontendFrame[frontendFrame['METRIC'] == metric].to_dict('records')
print(dataDict)
這適用於少量數據,但是隨着數據量的增加最快可能需要一秒鍾(!!!!)。
我已經嘗試了在熊貓中使用groupby,它甚至更慢,而且還嘗試了map,但是我不想將其返回到列表中。 我該如何迭代並以更快的方式創建我想要的東西? 我正在使用Python 3.6
更新:
輸入:
DATETIME METRIC ANOMALY VALUE
0 2018-02-27 17:30:32 SCORE 2.0 -1.0
1 2018-02-27 17:30:32 VALUE NaN 0.0
2 2018-02-27 17:30:32 INDEX NaN 6.6613381477499995E-16
3 2018-02-27 17:31:30 SCORE 2.0 -1.0
4 2018-02-27 17:31:30 VALUE NaN 0.0
5 2018-02-27 17:31:30 INDEX NaN 6.6613381477499995E-16
6 2018-02-27 17:32:30 SCORE 2.0 -1.0
7 2018-02-27 17:32:30 VALUE NaN 0.0
8 2018-02-27 17:32:30 INDEX NaN 6.6613381477499995E-16
輸出:
{
"INDEX": [
{
"DATETIME": 1519759710000,
"METRIC": "INDEX",
"ANOMALY": null,
"VALUE": "6.6613381477499995E-16"
},
{
"DATETIME": 1519759770000,
"METRIC": "INDEX",
"ANOMALY": null,
"VALUE": "6.6613381477499995E-16"
}],
"SCORE": [
{
"DATETIME": 1519759710000,
"METRIC": "SCORE",
"ANOMALY": 2,
"VALUE": "-1.0"
},
{
"DATETIME": 1519759770000,
"METRIC": "SCORE",
"ANOMALY": 2,
"VALUE": "-1.0"
}],
"VALUE": [
{
"DATETIME": 1519759710000,
"METRIC": "VALUE",
"ANOMALY": null,
"VALUE": "0.0"
},
{
"DATETIME": 1519759770000,
"METRIC": "VALUE",
"ANOMALY": null,
"VALUE": "0.0"
}]
}
一種可能的解決方案:
a = defaultdict( list )
_ = {x['METRIC']: a[x['METRIC']].append(x) for x in frontendFrame.to_dict('records')}
a = dict(a)
from collections import defaultdict
a = defaultdict( list )
for x in frontendFrame.to_dict('records'):
a[x['METRIC']].append(x)
a = dict(a)
慢:
dataDict = frontendFrame.groupby('METRIC').apply(lambda x: x.to_dict('records')).to_dict()
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