所以我的數(shù)據(jù)看起來像這樣：data = {"technology1": [ [ 20, 0.02, u'10.00,106.10,107.00,107.00,0.45', u'24.00,-47.15,-49.50,-51.00,0.12', u'11.00,0.35,0.00,0.00,0.92', u'0.00',0.04,0.16, u'0.223196881092', u'f',0.02, ], [ 100, 0.02, u'10.00,106.10,107.00,107.00,0.45', u'24.00,-47.15,-49.50,-51.00,0.12', u'11.00,0.35,0.00,0.00,0.92', u'0.00', 0.04, 0.16, u'0.223196881092', u'f', 0.01 ] ... ], "technology2": ...}如您所見，它是一個字典，每個鍵都訪問一個列表列表，所有列表都具有相同的格式。每個“內(nèi)部”列表都包含整數(shù)，浮點數(shù)。還有unicode字符串，其中一些帶有單個值，某些在unicode字符串中帶有一組5個數(shù)字。我想要的是：為每種技術(shù)制作一個陣列。在每個數(shù)組中，行將是上面的“外部”列表，列將是“內(nèi)部列表”的不同元素。理想情況下，需要將unicode轉(zhuǎn)換為字符串（因為我知道如何更好地使用它們），并且unicode字符串中的5個數(shù)字的集合需要擴(kuò)展為每個元素。即技術(shù)陣列20, 0.02, 10.00, 106.10, ... "f", 0.02100, 0.02, ... "f", 0.01到目前為止的嘗試：for tech in data: features = data[tech] # i.e. grab technologyn for row in features: for i in row[2:5]: # 2 til 5 defines the instance which are sets of 5 #print i,"\n" i = str(i) i = i.split(',')這是行不通的，當(dāng)我在代碼執(zhí)行后查看功能時，它是完全一樣的！這并不是一個完整解決方案的嘗試，因為它顯然不會將所有unicode類型都轉(zhuǎn)換為字符串，但這是一個墊腳石。我還嘗試使用列表理解為：for row in features: [i.split(',') for i in row if (type(i)==unicode and "," in i)]
2 回答
Smart貓小萌
TA貢獻(xiàn)1911條經(jīng)驗 獲得超7個贊
您需要為每行創(chuàng)建一個新的列表對象，然后替換原始列表值：
def row_to_values(row):
values = []
for col in row:
if isinstance(col, unicode) and col != u'f':
# split and convert all entries to float
values += (float(v) for v in col.split(','))
else:
values.append(col)
return values
for value in data.values():
value[:] = [row_to_values(row) for row in value]
該value[:] = ...分配告訴Python來代替包含的所有指標(biāo)中的列表對象具有一組新的對象。由于每個value列表都是data詞典中的外部列表，因此將所有子列表替換為擴(kuò)展的行。
演示部分樣本數(shù)據(jù)：
>>> data = {"technology1": [
... [
... 20, 0.02,
... u'10.00,106.10,107.00,107.00,0.45',
... u'24.00,-47.15,-49.50,-51.00,0.12',
... u'11.00,0.35,0.00,0.00,0.92',
... u'0.00',0.04,0.16, u'0.223196881092', u'f',0.02,
... ],
... [
... 100, 0.02,
... u'10.00,106.10,107.00,107.00,0.45',
... u'24.00,-47.15,-49.50,-51.00,0.12',
... u'11.00,0.35,0.00,0.00,0.92', u'0.00', 0.04,
... 0.16, u'0.223196881092', u'f', 0.01
... ]],
... }
>>> from pprint import pprint
>>> pprint(data["technology1"][0])
[20,
0.02,
u'10.00,106.10,107.00,107.00,0.45',
u'24.00,-47.15,-49.50,-51.00,0.12',
u'11.00,0.35,0.00,0.00,0.92',
u'0.00',
0.04,
0.16,
u'0.223196881092',
u'f',
0.02]
>>> pprint(row_to_values(data["technology1"][0]))
[20,
0.02,
10.0,
106.1,
107.0,
107.0,
0.45,
24.0,
-47.15,
-49.5,
-51.0,
0.12,
11.0,
0.35,
0.0,
0.0,
0.92,
0.0,
0.04,
0.16,
0.223196881092,
u'f',
0.02]
因此，可以通過返回新列表對象的函數(shù)調(diào)用來擴(kuò)展一行，使其包含字符串中的所有浮點值。
使用該函數(shù)替換所有字典值中的所有行：
>>> for value in data.values():
... value[:] = [row_to_values(row) for row in value]
...
我們可以看到之前查看的第一行已更新：
>>> pprint(data["technology1"][0])
[20,
0.02,
10.0,
106.1,
107.0,
107.0,
0.45,
24.0,
-47.15,
-49.5,
-51.0,
0.12,
11.0,
0.35,
0.0,
0.0,
0.92,
0.0,
0.04,
0.16,
0.223196881092,
u'f',
0.02]
字典的其余部分也是如此：
>>> pprint(data)
{'technology1': [[20,
0.02,
10.0,
106.1,
107.0,
107.0,
0.45,
24.0,
-47.15,
-49.5,
-51.0,
0.12,
11.0,
0.35,
0.0,
0.0,
0.92,
0.0,
0.04,
0.16,
0.223196881092,
u'f',
0.02],
[100,
0.02,
10.0,
106.1,
107.0,
107.0,
0.45,
24.0,
-47.15,
-49.5,
-51.0,
0.12,
11.0,
0.35,
0.0,
0.0,
0.92,
0.0,
0.04,
0.16,
0.223196881092,
u'f',
0.01]]}
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