Question

我有两个列表，其中包含字符串格式的术语。这些术语分为两类：水果和车辆。我正在尝试显示仅包含来自冲突类别的术语对的数据框。这样做的最佳方法是什么？以下是我的列表和数据框的示例。任何帮助将不胜感激！

  dataframe:

         col 1                 
  ['apple', 'truck' ]
  ['truck', 'orange']
  ['pear',  'motorcycle']
  ['pear', 'orange' ]
  ['apple', 'pear'  ]
  ['truck', 'car'   ]


  vehicles = ['car', 'truck', 'motorcycle']
  fruits = ['apple', 'orange', 'pear']


  desired output:

        col 2

  ['apple', 'truck' ]
  ['pear', 'motorcycle']
  ['truck', 'orange']

Answer 1

从列表列创建DataFrame，通过DataFrame.isin测试成员资格，然后通过~反转掩码，使用DataFrame.any检查每行至少一个True列表和最后一个链条条件都按位与-&并按boolean indexing进行过滤：

df1 = pd.DataFrame(df['col 1'].values.tolist())
df = df[(~df1.isin(vehicles)).any(axis=1) & (~df1.isin(fruits)).any(axis=1)]
print (df)
                col 1
0      [apple, truck]
1     [truck, orange]
2  [pear, motorcycle]

另一种解决方案，其中set由and链接（由于标量）而交集并转换为bool-空集将转换为False：

def func(x):
    s = set(x)
    v = set(vehicles)
    f = set(fruits)
    return bool((s & v) and (s & f))

df = df[df['col 1'].apply(func)]
print (df)
                col 1
0      [apple, truck]
1     [truck, orange]
2  [pear, motorcycle]

Answer 2

np.isin可能对您有用！

super_set = np.array([vehicles,fruits])

def f(x):
    return all(np.isin(super_set,x).sum(axis=1))

df[df.col1.apply(f)]

#
col1
0   [apple, truck]
1   [truck, orange]
2   [pear, motorcycle]

检查两个列表是否在一个熊猫列中

2 个答案: