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遍歷字典列表值我有一個熊貓數據幀有效地跳過缺失值的Python 3
import pandas as pd
df=pd.DataFrame({'Location': [ 'NY', 'SF', 'NY', 'NY', 'SF', 'SF', 'TX', 'TX', 'TX', 'DC'],
'Class': ['H','L','H','L','L','H', 'H','L','L','M'],
'Address': ['12 Silver','10 Fak','12 Silver','1 North','10 Fak','2 Fake', '1 Red','1 Dog','2 Fake','1 White'],
'Score':['4','5','3','2','1','5','4','3','2','1',]})
而且我要補充,我存儲在字典2個標籤。需要注意的是,第二字典不包括關鍵的「A」
df['Tag1'] =''
df['Tag2'] =''
tagset1 = {'A':['NY|SF'],
'B':['DC'],
'C':['TX'],
}
for key in tagset1:
df.loc[df.Location.str.contains(tagset1[key][0]) & (df.Tag1 == ''),'Tag1'] = key
tagset2= {'B':['H|M'],
'C':['L'],
}
for key in tagset2:
df.loc[df.Class.str.contains(tagset2[key][0]) & (df.Tag2 == ''),'Tag2'] = key
print (df)
如果我想將兩個字典相結合,使代碼更易讀,高效的,我應該填寫在newtagset['A'][1]
當場A和''
還是有在迭代列表中的位置時,使迭代器忽略或跳過位置newtagset['A'][1]
的另一種方法是什麼?
newtagset = {'A':['NY|SF', '',],
'B':['DC','H|M',],
'C':['TX','L',],
}
for key in newtagset:
df.loc[df.Location.str.contains(newtagset[key][0]) & (df.Tag1 == ''),'Tag1'] = key
for key in newtagset:
df.loc[df.Class.str.contains(newtagset[key][1]) & (df.Tag2 == ''),'Tag2'] = key
print (df)
我找到的大多數解決方案都使用itertools Skip multiple iterations in loop python這是唯一的方法嗎?