基于列条件筛选出行的问题

Question

我有以下数据帧：

df = 
VD_1   VD_2   VD_3    VD_4   VD_5  TYPE  VAL
NaN    XX     VV      DD     NaN   ABC   5
NaN    XX     MM     VV      NaN   ABC   6
XX     MM     NaN     NaN    NaN   ABC   6
TT     XX     MM      NaN    NaN   ABC   5

我只想保留那些第一个非NaN值等于XX的行，以及至少两个不等于NaN和XX的后续值。你知道吗

问题是return x返回None, None, None。。。它只在我使用return row时有效，但是结果不包含与df相同的列数。代码既不从分析中排除列TYPE和VAL。你知道吗

    def customFilter(x):
        row = x.dropna()
        if (row[0] == 'XX') & (('XX' not in row[1:]) & (len(row[1:]) >= 2)):
            return row
        return np.nan
    df = df.apply(customFilter, axis=1).dropna(how='all', axis=0)

Is there any trick to solve the mentioned issues?

更新：

# Delete rows that do not start from AG
def calculate_correct_rows(df):
    # Create drop rows
    drop_rows = []
    i = 0
    for index, x in df.iterrows():
        row = x.dropna()
        if (row[0] == 'XX') & (('XX' not in row[1:]) & (len(row[1:]) >= 2)):
            drop_rows.append(i)
        i = i + 1
    return drop_rows

# Drop the rows in list
subset2 = df.filter(like='VD_')
correct_rows = calculate_correct_rows(subset2)
final2 = df.loc[correct_rows,:]

Answer 1

也许有一种更漂亮的方法可以做到这一点，但是您可以简单地分两步而不是一步来执行过滤器。首先，创建一个不符合上述标准的所有行的列表。第二，使用df.drop(rows)删除步骤1中创建的列表中的行。你知道吗

这是到drop：drop的链接

例如

def calculate_drop_rows(df):
    # Create drop rows
    drop_rows = []
    i = 0
    for row in df:
        if [condition]:
            drop_rows.append(i)
        i = i + 1
    return drop_rows

# Drop the rows in list
drop_rows = calculate_drop_rows(df)
df = df.drop(drop_rows)