截断熊猫数据帧的行

Question

创建示例数据帧的代码：

Sample = [{'account': 'Jones LLC', 'Jan': 150, 'Feb': 200, 'Mar': [[.332, .326], [.058, .138]]},
     {'account': 'Alpha Co',  'Jan': 200, 'Feb': 210, 'Mar': [[.234, .246], [.234, .395], [.013, .592]]},
     {'account': 'Blue Inc',  'Jan': 50,  'Feb': 90,  'Mar': [[.084, .23], [.745, .923], [.925, .843]]}]
df = pd.DataFrame(Sample)

可视化的示例数据帧：

 df:
  account        Jan      Feb          Mar
Jones LLC  |     150   |   200    | [.332, .326], [.058, .138]
Alpha Co   |     200   |   210    | [[.234, .246], [.234, .395], [.013, .592]]
Blue Inc   |     50    |   90     | [[.084, .23], [.745, .923], [.925, .843]]

我正在寻找一个公式来截断"Mar"列，以便截断形状大于 (2，x( 的任何行，从而产生以下 df

 df:
  account        Jan      Feb          Mar
Jones LLC  |     150   |   200    | [.332, .326], [.058, .138]
Alpha Co   |     200   |   210    | [[.234, .246], [.234, .395] 
Blue Inc   |     50    |   90     | [[.084, .23], [.745, .923]

Answer 1

str 访问器是为字符串操作而设计的，但对于列表等可迭代对象，您也可以使用它进行切片：

df['Mar'] = df['Mar'].str[:2]
df
Out: 
   Feb  Jan                               Mar    account
0  200  150  [[0.332, 0.326], [0.058, 0.138]]  Jones LLC
1  210  200  [[0.234, 0.246], [0.234, 0.395]]   Alpha Co
2   90   50   [[0.084, 0.23], [0.745, 0.923]]   Blue Inc

Answer 2

可以使用.apply函数结合lambda运算符轻松完成单元级别的操作：

df["Mar"] = df["Mar"].apply(lambda x: x[:2])

Answer 3

Pandas 在处理系列列表时不是特别好，所以在处理它之前最好把它拉出来：

df['Mar'] = [row[:2] for row in df['Mar'].tolist()]

%timeit Ayhan 和 Marjan 的出色答案的结果：

3 行：

%timeit df['Mar'].str[:2]
10000 loops, best of 3: 154 µs per loop
%timeit df['Mar'].apply(lambda x: x[:2])
10000 loops, best of 3: 133 µs per loop
%timeit [row[:2] for row in df['Mar'].tolist()]
The slowest run took 5.51 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 12 µs per loop
>>>5.51*12
66.12

3,000,000 行：

%timeit df['Mar'].str[:2]
1 loop, best of 3: 1.23 s per loop
%timeit df['Mar'].apply(lambda x: x[:2])
1 loop, best of 3: 1.25 s per loop
%timeit [row[:2] for row in df['Mar'].tolist()]
1 loop, best of 3: 940 ms per loop

如果您愿意将列表对拆分为两列，则可以获得一种方法，该方法可以更好地扩展到比上述更大的数据帧

>>>pd.DataFrame(df['Mar'].tolist()).iloc[:, :2]]
                0               1
0  [0.332, 0.326]  [0.058, 0.138]
1  [0.234, 0.246]  [0.234, 0.395]
2   [0.084, 0.23]  [0.745, 0.923]

在 3,000,000 行上：

%timeit pd.DataFrame(df['Mar'].tolist()).iloc[:, :2]
1 loop, best of 3: 276 ms per loop