我有一個如下的字典，如果發(fā)現(xiàn)相同的值 more tahn once 則必須使用增量編號創(chuàng)建字典鍵。TT = { "TYPE_1" : ['ERROR'], "TYPE_2": ['FATAL'], "TYPE_3" : ["TIME_OUT"], "TYPE_4" : ['SYNTAX'], "TYPE_5" : ['COMPILE'], } m1 = "ERROR the input is not proper" m2 = "This should have not occured FATAL" m3 = "Sorry TIME_OUT" m4 = "SYNTAX not proper" m5 = "u r late its TIME_OUT"值“TIME_OUT”在搜索中出現(xiàn)兩次。 count = 0 for key in TT.keys(): print(key) Key_1 = key while key_1 in TT: count = count+1 key_1 = key + "_{}".format(count)上面的代碼給出錯誤 Key_1 未定義。預(yù)期輸出：如果相同的值出現(xiàn)不止一次，則應(yīng)使用增量編號“TYPE_3_1”創(chuàng)建字典鍵：[“TIME_OUT”]， TT = { "TYPE_1" : ['ERROR'], "TYPE_2": ['FATAL'], "TYPE_3" : ["TIME_OUT"], "TYPE_3_1" : ["TIME_OUT"], "TYPE_4" : ['SYNTAX'], "TYPE_5" : ['COMPILE'], }請對此提出建議。
1 回答
叮當(dāng)貓咪
TA貢獻1776條經(jīng)驗 獲得超12個贊
如果您可以重新考慮一些數(shù)據(jù)結(jié)構(gòu)，那么可以有一種更有效的方法來解決這個問題，但如果那不是一個選項，您可以嘗試這個。
inputs = ["ERROR the input is not proper",
"This should have not occured FATAL",
"Sorry TIME_OUT",
"SYNTAX not proper",
"u r late its TIME_OUT"]
basic_types = {
"TYPE_1" : ['ERROR'],
"TYPE_2": ['FATAL'],
"TYPE_3" : ["TIME_OUT"],
"TYPE_4" : ['SYNTAX'],
"TYPE_5" : ['COMPILE'],
}
type_counts = {}
results = {}
for sentence in inputs:
for basic_type in basic_types:
if basic_types.get(basic_type)[0] in sentence:
type_counts[basic_type] = type_counts.get(basic_type, 0) + 1
if type_counts[basic_type] == 1:
results[basic_type] = [basic_types.get(basic_type)[0]]
else:
results[basic_type+"_{}".format(type_counts[basic_type] - 1)] = [basic_types.get(basic_type)[0]]
print(results)
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