截断 pandas DataFrame 的行

Question

创建示例数据框的代码：

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)

可视化的示例数据框：

 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，以便截断任何形状大于 (2,x) 的行，从而导致以下 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

可以使用 .apply 函数结合 lambda 运算符轻松完成单元格级别的操作：

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

Answer 2

str 访问器专为字符串操作而设计，但对于像列表这样的可迭代对象，您也可以将其用于切片：

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 3

Pandas 不能很好地与系列列表一起播放，因此在处理它之前将其拉出可能会更好：

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

%timeit 结果以及 ayhan 和 Marjan 的出色回答：

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分成两列，您可以获得一种比上面的方法更好地扩展到更大数据帧的方法

>>>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

截断 pandas DataFrame 的行

Truncate rows of a pandas DataFrame

python-2.7

pandas

dataframe

truncate