Question

我想比较两个数据框并输出一个具有差异的数据框。但是，我可以忍受2天之内的日期差异，并且在5分之内获得分数。如果值在可接受范围内，我将保留df1中的值。

df1

id    group      date        score
10     A       2020-01-10     50
29     B       2020-01-01     80
39     C       2020-01-21     84
38     A       2020-02-02     29

df2

id    group      date        score
10     B       2020-01-11     56
29     B       2020-01-01     81
39     C       2020-01-22     85
38     A       2020-02-12     29

我的预期输出：

id    group           date                      score
10     A -> B       2020-01-10                50 -> 56
29     B            2020-01-01                   80
39     C            2020-01-21                   84
38     A            2020-02-02 -> 2020-02-12     29

因此，我想按单元格和某些列的条件比较数据框。

我从这里开始：

df1.set_index('id', inplace=True)
df2.set_index('id', inplace=True)
result = []
for col in df1.columns:
    for index, row in df1.iterrows():
        diff = []
        compare_item = row[col][index]
        for index, row in df2.iterrows():
            if col == 'date':
                # acceptable if it's within 2 days differences
            if col == 'score':
                # acceptable if it's within 5 points differences
            if compare_item == row[col][index]:
                diff.append(compare_item)
            else:
                diff.append('{} --> {}'.format(compare_item, row[col]))
    result.append(diff)
df = pd.DataFrame(result, columns = [df1.columns])

Answer 1

让我们尝试一下：

thresh = {'date':pd.to_timedelta('2D'),
          'score':5}

def update(col):
    name = col.name

    # if there is a threshold, we update only if threshold is surpassed
    if name in thresh:
        return col.where(col.sub(df2[name]).abs()<=thresh[name], df2[name])

    # there is no threshold for the column
    # return the corresponding column from df2
    return df2[name]

df1.apply(update)

输出：

   group       date  score
id                        
10     B 2020-01-10     56
29     B 2020-01-01     80
39     C 2020-01-21     84
38     A 2020-02-12     29

将两个数据框单元格与列条件进行比较

1 个答案: