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python - 通过将其他列与字典匹配来更改数据框的列

问题描述

我有一个情况：

postStr = """{
                     "zoneId":"0",
                     "id":["a","b","c","d","f","g"],
                     "currencycode":["USD"],

                }"""


postData = json.loads(postStr, object_pairs_hook=OrderedDict)

我有一个数据框：

df = {
'id':['a','b','c','d','f','g','h','i','j','k'],
'B':['c','d','e','d','d','c','s','e','s','q'],
'S':['f','g','h','j','e','j','t','r','p','p']
}
df1 = pd.DataFrame(df)

现在我想要一个数据框架，如果 id 在字典中，则 B 对应列变为 XX

输出：

    df = {
'id':['a','b','c','d','f','g','h','i','j','k'],
'B' :['XX','XX','XX','XX','XX','c','s','e','s','q'],
'S' :['f','g','h','j','e','j','t','r','p','p']
}
df1 = pd.DataFrame(df)

请帮忙

标签： pythonpandasdataframeanacondadata-science

解决方案

我认为isin需要loc：

df1.loc[df1['id'].isin(postData['id']), 'id'] = 'XX'
print (df1)
   id  B  S
0  XX  c  f
1  XX  d  g
2  XX  e  h
3  XX  d  j
4  XX  d  e
5  XX  c  j
6   h  s  t
7   i  e  r
8   j  s  p
9   k  q  p

如果想要更多动态解决方案 -intersection用于 DataFrame 和字典中的列名并在循环中设置值：

postStr = """{
                     "S":["f","h"],
                     "id":["a","b","c","d","f","g"],
                     "currencycode":["USD"]

                }"""

postData = json.loads(postStr, object_pairs_hook=OrderedDict)
print (postData)
OrderedDict([('S', ['f', 'h']), 
             ('id', ['a', 'b', 'c', 'd', 'f', 'g']), 
             ('currencycode', ['USD'])])

df = {
'id':['a','b','c','d','f','g','h','i','j','k'],
'B':['c','d','e','d','d','c','s','e','s','q'],
'S':['f','g','h','j','e','j','t','r','p','p']
}
df1 = pd.DataFrame(df)

for col in df1.columns.intersection(postData.keys()):
    df1.loc[df1[col].isin(postData[col]), col] = 'XX'
print (df1)
   id  B   S
0  XX  c  XX
1  XX  d   g
2  XX  e  XX
3  XX  d   j
4  XX  d   e
5  XX  c   j
6   h  s   t
7   i  e   r
8   j  s   p
9   k  q   p

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