Question

我有一个数据框

menu.seePrice(item)

是否有某种自定义过滤方法，让Python知道B＆gt; M> K +

说我想过滤，0 1 2 3 Marketcap 0 1.707280 0.666952 0.638515 -0.061126 2.291747 1.71B 1 -1.017134 1.353627 0.618433 0.008279 0.148128 1.82B 2 -0.774057 -0.165566 -0.083345 0.741598 -0.139851 1.1M 3 -0.630724 0.250737 1.308556 -1.040799 1.064456 30.92M 4 2.029370 0.899612 0.261146 1.474148 -1.663970 476.74k 5 2.029370 0.899612 0.261146 1.474148 -1.663970 -1，是否有一种聪明或干净的方式来做到这一点？拥有M或B使得该值非常易读且易于区分。

谢谢。

编辑：重新打开线程作为Max U的答案，而优秀似乎产生了一个熊猫bug，我们在Github上打开了一个问题。

Answer 1

这不是超级干净，但它可以解决问题并且不会使用任何python迭代：

<强>代码：

# Create a separate column (which you can omit later) that converts 'Marketcap' strings to numbers
df['cap'] = df.loc[df['Marketcap'].str.contains('B'), 'Marketcap'].str.replace('B','').astype(float) * 1000
df['cap'].fillna(df.loc[df['Marketcap'].str.contains('M'), 'Marketcap'].str.replace('M',''), inplace = True)

# For pandas pre-0.20.0 (<May 2017)
print df.ix[df['cap'].astype(float) > 35, :-1]

# For pandas 0.20.0+ (.ix[] deprecated)
print df.iloc[df[df['cap'].astype(float) > 35].index, :-1]

# Or, alternate pandas 0.20.0+ option (thanks @Psidom)
print df[df['cap'].astype(float) > 35].iloc[:,:-1]

<强>输出：

          0         1         2         3         4 Marketcap
0  1.707280  0.666952  0.638515 -0.061126  2.291747     1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128     1.82B
4  2.029370  0.899612  0.261146  1.474148 -1.663970    100.9M

Answer 2

<强>更新

In [44]: df
Out[44]:
          0         1         2         3         4 Marketcap
0  1.707280  0.666952  0.638515 -0.061126  2.291747     1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128     1.82B
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851      1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456    30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970   476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970        -1

In [45]: df[pd.eval(df.Marketcap.replace(['[Kk]','[Mm]','[Bb]'],
                                         ['*10**3','*10**6','*10**9'], regex=True) \
                      .add(' < 35*10**6'))]
Out[45]:
          0         1         2         3         4 Marketcap
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851      1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456    30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970   476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970        -1

我这样做：

In [13]: df[pd.eval(df.Marketcap.replace(['M','B'],['','*1000'], regex=True).add(' > 35'))]
Out[13]:
          0         1         2         3         4 Marketcap
0  1.707280  0.666952  0.638515 -0.061126  2.291747     1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128     1.82B
4  2.029370  0.899612  0.261146  1.474148 -1.663970    100.9M

说明：

In [14]: df.Marketcap.replace(['M','B'],['','*1000'], regex=True)
Out[14]:
0    1.71*1000
1    1.82*1000
2          1.1
3        30.92
4        100.9
Name: Marketcap, dtype: object

In [15]: df.Marketcap.replace(['M','B'],['','*1000'], regex=True).add(' > 35')
Out[15]:
0    1.71*1000 > 35
1    1.82*1000 > 35
2          1.1 > 35
3        30.92 > 35
4        100.9 > 35
Name: Marketcap, dtype: object

In [16]: pd.eval(df.Marketcap.replace(['M','B'],['','*1000'], regex=True).add(' > 35'))
Out[16]: array([True, True, False, False, True], dtype=object)

Answer 3

来源DF：

In [176]: df
Out[176]:
                    0         1         2         3 Market Cap
0  1.707280  0.666952  0.638515 -0.061126  2.291747      1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128      1.82B
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851       1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456     30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970    476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970         -1

<强>解决方案：

to_replace = ['\d+\s*[Kk]','\d+\s*[Mm]','\d+\s*[Bb]', '-1', 'N/A']
value = [1000,1000000,1000000000, 1, 1]

mask = df.assign(
    f=df['Market Cap'].replace(to_replace, value, regex=True),
    Marketcap=pd.to_numeric(df['Market Cap'].str.replace(r'[^\d\.]', ''), errors='coerce')
).eval("Marketcap * f < 35000000")

df[mask]

<强>结果：

In [178]: df[mask]
Out[178]:
                    0         1         2         3 Market Cap
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851       1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456     30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970    476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970         -1

PS如果要在结果数据集更改中保留非数字值（如N/A）：

pd.to_numeric(df['Market Cap'].str.replace(r'[^\d\.]', ''), errors='coerce')

到

pd.to_numeric(df['Market Cap'].str.replace(r'[^\d\.]', ''), errors='coerce').fillna('0')

Pandas中的自定义布尔过滤？

3 个答案: