如何在python / pandas中复制excel COUNTIFS?

时间:2016-10-26 20:44:51

标签: python pandas dataframe boolean countif

我想计算df ['A']中前5个值的#,它是< df中的当前值['A']&也是> = df2 ['A']。我试图避免循环遍历每一行和每列,因为我想将它应用于更大的数据集。

鉴于此......

list1 = [[21,101],[22,110],[25,113],[24,112],[21,109],[28,108],[30,102],[26,106],[25,111],[24,110]]
df = pd.DataFrame(list1,index=pd.date_range('2000-1-1',periods=10, freq='D'), columns=list('AB'))
df2 = pd.DataFrame(df * (1-.05))

我想将此返回(使用COUNTIFS在Excel中解决)...

enter image description here

下面这一行实现了第一部分(感谢亚历山大),而Divakar和DSM之前也称过(herehere)。

df3 = pd.DataFrame(df.rolling(center=False,window=6).apply(lambda rollwin: sum((rollwin[:-1] < rollwin[-1]))))

但我无法将比较添加到df2。请帮忙。

2016年10月27日跟进:

我如何将上面的lambda写为标准函数?

10/28/16:

见下文,从df和df2中取col'A',我试图计算df ['A']中前5个值中有多少落在当前df2 ['A']和df ['之间一个']。换句话说,每个橙色框中有多少落在黄色低 - 高范围之间?

enter image description here

更新:不同的list1数据产生错误的df3 ...

list1 = [[21,101],[22,110],[25,113],[24,112],[21,109],[26,108],[25,102],[26,106],[25,111],[22,110]]
df = pd.DataFrame(list1,index=pd.date_range('2000-1-1',periods=10, freq='D'), columns=list('AB'))
df2 = pd.DataFrame(df * (1-.05))

df3 = pd.DataFrame(
     df.rolling(center=False,window=6).apply(
          lambda rollwin: pd.Series(rollwin[:-1]).between(rollwin[-1]*0.95,rollwin[-1]).sum()))

df
Out[9]: 
             A    B
2000-01-01  21  101
2000-01-02  22  110
2000-01-03  25  113
2000-01-04  24  112
2000-01-05  21  109
2000-01-06  26  108
2000-01-07  25  102
2000-01-08  26  106
2000-01-09  25  111
2000-01-10  22  110


df3
Out[8]: 
              A    B
2000-01-01  NaN  NaN
2000-01-02  NaN  NaN
2000-01-03  NaN  NaN
2000-01-04  NaN  NaN
2000-01-05  NaN  NaN
2000-01-06  1.0  0.0
2000-01-07  2.0  0.0
2000-01-08  3.0  1.0
2000-01-09  2.0  3.0
2000-01-10  1.0  3.0
EXCEL EXAMPLES(11/14):见下图,试图计算蓝色框中有多少数字落在以橙色突出显示的范围之间。

enter image description here

2 个答案:

答案 0 :(得分:2)

list1 = [[21,101],[22,110],[25,113],[24,112],[21,109],[28,108],[30,102],[26,106],[25,111],[24,110]]
df = pd.DataFrame(list1,index=pd.date_range('2000-1-1',periods=10, freq='D'), columns=list('AB'))
df2 = pd.DataFrame(df * (1-.05))


window = 6
results = []
for i in range (len(df)-window+1):
    slice_df1 = df.iloc[i:i + window]
    slice_df2 = df2.iloc[i:i + window]
    compare1 = slice_df1['A'].iloc[-1]
    compare2 = slice_df2['A'].iloc[-1]
    a= slice_df1.iloc[:-1]['A'].between(compare2,compare1)  # series have a between metho
    results.append(a.sum())

df_res =  pd.DataFrame(data = results , index = df.index[window-1:] , columns = ['countifs'])
df_res = df_res.reindex(df.index,fill_value=0.0)
print df_res

which yields:

            countifs
2000-01-01    0.0000
2000-01-02    0.0000
2000-01-03    0.0000
2000-01-04    0.0000
2000-01-05    0.0000
2000-01-06    0.0000
2000-01-07    0.0000
2000-01-08    1.0000
2000-01-09    1.0000
2000-01-10    0.0000

BUT

看到你的上下界,价值和价值之间存在逻辑关系--5%。那么这也许就是你想要的。

    df3 = pd.DataFrame(
         df.rolling(center=False,window=6).apply(
            lambda rollwin: sum(np.logical_and(
                                    rollwin[-1]*0.95 <= rollwin[:-1]
                                   ,rollwin[:-1] < rollwin[-1]) 
                                )))

如果您更喜欢pd.Series.between()方法:

df3 = pd.DataFrame(
     df.rolling(center=False,window=6).apply(
          lambda rollwin: pd.Series(rollwin[:-1]).between(rollwin[-1]*0.95,rollwin[-1]).sum()))

答案 1 :(得分:1)

list1 = [[21,50,101],[22,52,110],[25,49,113],[24,49,112],[21,55,109],[28,54,108],[30,57,102],[26,56,106],[25,58,111],[24,60,110]]
df = pd.DataFrame(list1,index=pd.date_range('2000-1-1',periods=10, freq='D'), columns=list('ABC'))

print df

我认为这符合您的新屏幕截图&#34;鉴于数据&#34;。

             A   B    C
2000-01-01  21  50  101
2000-01-02  22  52  110
2000-01-03  25  49  113
2000-01-04  24  49  112
2000-01-05  21  55  109
2000-01-06  28  54  108
2000-01-07  30  57  102
2000-01-08  26  56  106
2000-01-09  25  58  111
2000-01-10  24  60  110

enter image description here

和相同的功能:

print pd.DataFrame(
           df.rolling(center=False,window=6).
              apply(lambda rollwin: pd.Series(rollwin[:-1]).
                   between(rollwin[-1]*0.95,rollwin[-1]).sum()))

提供您想要的输出&#34;期望的结果&#34;:

             A   B   C
2000-01-01 nan nan nan
2000-01-02 nan nan nan
2000-01-03 nan nan nan
2000-01-04 nan nan nan
2000-01-05 nan nan nan
2000-01-06   0   1   0
2000-01-07   0   1   0
2000-01-08   1   2   1
2000-01-09   1   2   3
2000-01-10   0   2   3

enter image description here