難以理解如何使用Lambda對dict進行排序

Question

我正在嘗試對這本字典進行排序。

d = {'FNP': ['0.02', '0.02', '0.02', '0.02'],
     'TestName': ['Test1205', 'Test1206', 'Test1207', 'Test1208'],
     'eno': ['0', '0', '0', '0'],
     'GRE': ['0.00', '0.00', '0.00', '0.00'],
     'TPS': ['78.00', '45.00', '73400', '34.00'],
     'id': ['1', '1', '1', '1']}

我想做的是基於TPS排序並輸出名稱wrt排序值。 在此，TPS [0]對應於TesTName [0]。 我想對TPS排序時對Testname排序。 預期產量：

d = {'TestName': ['Test1208','Test1206','Test1205','Test1207'],
     'TPS':['34.00', '45.00', '78.00', '73400']

這是我到目前為止所擁有的： sorted(d['TPS'],key=lambda d:d)

在對TPS排序時，如何確保Testname也可以互換？

Answer 1

這是一個功能解決方案。 訣竅是使用enumerate來提取索引，記住要轉換為float以進行排序，並使用帶有多個參數的itemgetter來提取正確的順序。

from operator import itemgetter

idx, _ = zip(*sorted(enumerate(d['TPS']), key=lambda x: float(x[1])))

res = {k: itemgetter(*idx)(d[k]) for k in ('TestName', 'TPS')}

{'TestName': ('Test1208', 'Test1206', 'Test1205', 'Test1207'),
 'TPS': ('34.00', '45.00', '78.00', '73400')}

Answer 2

假設您希望TPS列表按數字排序，而不是按詞法排序（按字母順序按其中每個字符串元素的值排序）：

from pprint import pprint

d = {'FNP': ['0.02', '0.02', '0.02', '0.02'],
     'TestName': ['Test1205', 'Test1206', 'Test1207', 'Test1208'],
     'eno': ['0', '0', '0', '0'],
     'GRE': ['0.00', '0.00', '0.00', '0.00'],
     'TPS': ['78.00', '45.00', '73400', '34.00'],
     'id': ['1', '1', '1', '1']}

d['TestName'] = sorted(d['TestName'])
d['TPS'] = sorted(d['TPS'], key=float)

pprint(d)

輸出：

{'FNP': ['0.02', '0.02', '0.02', '0.02'],
 'GRE': ['0.00', '0.00', '0.00', '0.00'],
 'TPS': ['34.00', '45.00', '78.00', '73400'],
 'TestName': ['Test1205', 'Test1206', 'Test1207', 'Test1208'],
 'eno': ['0', '0', '0', '0'],
 'id': ['1', '1', '1', '1']}

請注意， TestName列表已經排序。

Answer 3

除了@jpp的解決方案之外，如果您想要新dict中原始字典中所有列表的排序版本，則可以使用原始TPS壓縮每個數據列表，對元組進行排序，並從排序后的數據中創建新數據列表：

original_tps = [float(t) for t in d['TPS']]
sorted_d = {}

for k, v in d.items():
    sorted_d[k] = [x for tps, x in sorted([(t, e) for t, e in zip(original_tps, v)])]


print(sorted_d['TPS'])
# ['34.00', '45.00', '78.00', '73400']                                                                                                    
print(sorted_d['TestName'])
# ['Test1208', 'Test1206', 'Test1205', 'Test1207']
# ... similarly other lists are sorted

Answer 4

我將創建一個Record namedtuple來保存每個單獨的記錄，然后輕松進行排序，並在需要時強制返回到列表的字典。

from collections import namedtuple

Record = namedtuple("Record", "FNP TestName eno GRE TPS id")

def extract(d, keys=None):
    if keys is None:
        keys = ["FNP", "TestName", "eno", "GRE", "TPS", "id"]
    records = [Record(*vals) for vals in zip(*map(d.get, keys))]
    return records

def transform(records):
    records.sort(key=lambda r: float(r.TPS))

def load(records, keys=None):
    if keys is None:
        keys = ["FNP", "TestName", "eno", "GRE", "TPS", "id"]
    d = {k: [getattr(record, k) for record in records] for k in keys}
    return d

d = {'FNP': ['0.02', '0.02', '0.02', '0.02'],
     'TestName': ['Test1205', 'Test1206', 'Test1207', 'Test1208'],
     'eno': ['0', '0', '0', '0'],
     'GRE': ['0.00', '0.00', '0.00', '0.00'],
     'TPS': ['78.00', '45.00', '73400', '34.00'],
     'id': ['1', '1', '1', '1']}
records = extract(d)
transform(records)
new_d = load(records)