Question

我有以下数据框：

load = pd.DataFrame({'A':list('abcdef'),
                   'B':[4,5,4,5,5,4],
                   'C':[7,8,9,4,2,0],
                   'D':[1,3,5,4,2,0],
                   'E':[5,3,6,9,2,4],
                   'F':list('aaabbb')})

如果条件满足，我需要检查b>c和c=d，其中c,d!=0和False False False True True False都必须为True，否则为False。

输出

var parser, xmlDoc;
    var text = "<InventoryResponse><Inventories><a:Inventory><a:ID>12345</a:ID><a:Cost>20.00</a:Cost></a:Inventory></Inventories></InventoryResponse>";

    parser = new DOMParser();
    xmlDoc = parser.parseFromString(text, "text/xml");


    alert(xmlDoc.getElementsByTagName("InventoryResponse")[0].childNodes[1].childNodes[0].childNodes[1].innerHTML);

Answer 1

我认为需要compilation-speed为&链接布尔掩码：

bitwise AND

如果要m = load.B.gt(load.C) & load.C.eq(load.D) & load[['C','D']].ne(0).any(axis=1) #alternative m = load.B.gt(load.C) & load.C.eq(load.D) & load['C'].ne(0) & load['D'].ne(0) print (m) 0 False 1 False 2 False 3 True 4 True 5 False dtype: bool获得相同的输出：

numpy.where

如果需要按如下条件设置2个值，则使用它：

load['new'] = np.where(m, True, False)
print (load)
   A  B  C  D  E  F    new
0  a  4  7  1  5  a  False
1  b  5  8  3  3  a  False
2  c  4  9  5  6  a  False
3  d  5  4  4  9  b   True
4  e  5  2  2  2  b   True
5  f  4  0  0  4  b  False

性能：

load['new'] = np.where(m, 10, 20)
print (load)
   A  B  C  D  E  F  new
0  a  4  7  1  5  a   20
1  b  5  8  3  3  a   20
2  c  4  9  5  6  a   20
3  d  5  4  4  9  b   10
4  e  5  2  2  2  b   10
5  f  4  0  0  4  b   20

Answer 2

尝试：

load['print'] = load.apply(lambda x:(x.B>x.C)&(x.C==x.D)&(x.C!=0)&(x.D!=0),axis=1)
   A  B  C  D  E  F  print
0  a  4  7  1  5  a  False
1  b  5  8  3  3  a  False
2  c  4  9  5  6  a  False
3  d  5  4  4  9  b   True
4  e  5  2  2  2  b   True
5  f  4  0  0  4  b  False

或

load['print'] = np.where((load.B>load.C)&(load.C==load.D)&(load.C!=0)&(load.D!=0),True, False)

Answer 3

您可以将np.logical_and.reduce与布尔系列元组一起使用：

m1 = load['B'] > load['C']
m2 = load['C'] == load['D']
m3 = load['C'] != 0
m4 = load['D'] != 0

res = load[np.logical_and.reduce((m1, m2, m3, m4))]

print(res)

   A  B  C  D  E  F
3  d  5  4  4  9  b
4  e  5  2  2  2  b

在python pandas上使用np.where放置多个条件？

3 个答案: