python将编号相同的一列按固定值拆分

Question

问题遇到的现象和发生背景

希望将v这一列按照buck这一列进行拆分，同一个buck的量应该为326881602，拆分结果应该为标黄的vol这一列，如果buck的值相同，比如都为1，可以看到D2位置应该填326881602，如果buck都为3，那么对应的v这一列的值的和为326881602，也即D4+D5=326881602。

另外需要注意，对于新的一天，比如2020/1/3，9：31分的buck和2020/1/2的buck同为56，这就需要参考2020/1/2，14：57-15：00的v这一列，标黄的vol这一列是正确的结果，怎么才能得到

Answer 1

代码写长了一不小心，题主先试试行不行，数据格式没问题是前提，个别细节不懂可以留言

import pandas as pd

df = pd.read_csv('test.csv')
v_idx = list(df.columns).index('v')
df.insert(v_idx+1, column = 'vol', value = '')
b_idx = list(df.columns).index('buck')
df.loc[df['buck'].duplicated(keep = False) == False, 'vol'] = 326881602
df.loc[df['buck'].duplicated(keep = False) == True, 'vol'] = 0
v_set = list(set(df['v']))
b_set = list(set(df['buck']))


def v_checker():
    flag = 0
    for v in v_set:
        part = df.loc[df.v == v]
        check = 0
        for row in range(part.shape[0]):
            check += int(part.iloc[row, v_idx + 1])
        if check != int(v):
            flag += 1
        else:
            flag = flag
    if flag == 0:
        return True
    else:
        return False


def b_checker():
    flag = 0
    for b in b_set:
        part = df.loc[df.buck == b]
        check = 0
        for row in range(part.shape[0]):
            check += int(part.iloc[row, v_idx + 1])
        if check != 326881602:
            flag += 1
        else:
            flag = flag
    if flag == 0:
        return True
    else:
        return False


def v_correct():
    global df
    dfs = []
    for v in v_set:
        part = df.loc[df.v == v]
        check = 0
        for row in range(part.shape[0]):
            check += int(part.iloc[row, v_idx+1])
        if check != int(v):
            if check > int(v):
                over = check - int(v)
                max_row = part.shape[0]-1
                part.iloc[max_row, v_idx+1] -= over
                dfs.append(part)
            elif check < int(v):
                loss = int(v) - check
                min_row = part.shape[0]-1
                part.iloc[min_row, v_idx+1] += loss
                dfs.append(part)
        else:
            dfs.append(part)
    df = dfs[0]
    for index in range(1, len(dfs)):
        add = dfs[index]
        df = pd.concat([df, add])
        df.sort_index(ascending=True, inplace = True)


def b_correct():
    global df
    dfs = []
    for b in b_set:
        part = df.loc[df.buck == b]
        check = 0
        for row in range(part.shape[0]):
            check += int(part.iloc[row, v_idx+1])
        if check != 326881602:
            if check > 326881602:
                over = check - 326881602
                max_row = part.shape[0]-1
                part.iloc[max_row, v_idx+1] -= over
                dfs.append(part)
            elif check < 326881602:
                loss = 326881602 - check
                min_row = part.shape[0]-1
                part.iloc[min_row, v_idx+1] += loss
                dfs.append(part)
        else:
            dfs.append(part)
    df = dfs[0]
    for index in range(1, len(dfs)):
        add = dfs[index]
        df = pd.concat([df, add])
        df.sort_index(ascending=True, inplace = True)


def go():
    try:
        if v_checker() is False or b_checker() is False:
            v_correct()
            b_correct()
            go()
        else:
            df.to_csv('res.csv', index = False)  # 保存结果文件
    except:
        print('出问题了')


go()
print(df)
"""
实现的效果如下：
原数据
    date  time          v  buck
0      1    11  768944800     1
1      1    22  768944800     2
2      1    22  768944800     3
3      1    33  297902500     3
4      1    33  297902500     4
5      1    33  286462900     4
6      1    33  286462900     5
7      1    33  275848000     5
8      1    33  203289200     5
9      1    33  203289200     6
10     1    33  128842212     6

结果数据
    date  time          v        vol  buck
0      1    11  768944800  326881602     1
1      1    22  768944800  326881602     2
2      1    22  768944800  115181596     3
3      1    33  297902500  211700006     3
4      1    33  297902500   86202494     4
5      1    33  286462900  240679108     4
6      1    33  286462900   45783792     5
7      1    33  275848000  275848000     5
8      1    33  203289200    5249810     5
9      1    33  203289200  198039390     6
10     1    33  128842212  128842212     6
"""
"""