我有两个dfs。一个只有1和0**（df_one_zero**）。另一个有不同的值df_value_total。这两个有一千行和列！
每个df的第一列是id，我们根本不想改变它。
我想移动与滑动窗口5通过列。所以，每个窗口，我想与这两个工作：df_one_zero_window，df_value_window。
在每个窗口中，1开始和结束的列很重要。
然后我想创建另一个与df_one_zero形状相同的df_out（最初设置为零），考虑在列col中，1开始并结束于col_end，
将值放入df_out（row，col-1）= df_value_window（row，col-1）-df_value_window（row，col），其他值为零。（如果1从索引0开始，或者在最后一列结束，那么它就可以了。它不需要为此设置值）此外，如果df_one_zero_window中的1在col_end结束，那么df_out（row，col_end+1）= df_value_window（row，col_end+1）-df_value_window（row，col_end）。在下面的dfs中，我想创建df_out= df 2。df_value_total中的值非常多样化，这里我只选择了我的df中的一些简单数字。
## only has zero and 1
df = pd.DataFrame()
df['id'] = ['a', 'b', 'c']
df['0'] = [0, 0, 0]
df['1'] = [1, 0, 1]
df['2'] = [1, 1, 1]
df['3'] = [0, 0, 0]
df['4'] = [0, 0, 0]
df['5'] = [0, 0, 0]
df['6'] = [0, 1, 1]
df['7'] = [0, 0, 1]
df['8'] = [0, 0, 0]
df['9'] = [0, 0, 0]
df['10'] = [0, 0, 0]
df['11'] = [0, 0, 1]
df['12'] = [1, 1, 1]
df['13'] = [1, 0, 0]
df['14'] = [0, 0, 0]
df['15'] = [0, 0, 0]
df['16'] = [0, 1, 1]
df['17'] = [1, 1, 0]
df['18'] = [0, 0, 0]
df['19'] = [0, 0, 0]
## this is that which has different values
df1 = pd.DataFrame()
df1['id'] = ['a', 'b', 'c']
df1['0'] = [4, 0, 9]
df1['1'] = [0, 0, 1]
df1['2'] = [1, 1, 3]
df1['3'] = [6, 2, 0]
df1['4'] = [0, 0, 0]
df1['5'] = [0, 5, 0]
df1['6'] = [0, 1, 2]
df1['7'] = [0, 0, 1]
df1['8'] = [0, 0, 3]
df1['9'] = [0, 0, 0]
df1['10'] = [0, 0, 0]
df1['11'] = [0, 0, 1]
df1['12'] = [1, 1, 1]
df1['13'] = [1, 3, 4]
df1['14'] = [9, 0, 0]
df1['15'] = [0, 0, 0]
df1['16'] = [2, 1, 1]
df1['17'] = [1, 1, 4]
df1['18'] = [0, 5, 0]
df1['19'] = [0, 0, 0]字符串
我试着做了一些部分，但我无法跟踪1是在哪里完成的，而且我认为这不是最佳的！你能帮我吗？
def generate_df_out(df_one_zero, df_value_total, window_size=5):
for col in range(1, len(df_one_zero.columns), window_size):
df1_window = df_one_zero.iloc[:, col:col + window_size]
df_value_window = df_value_total.iloc[:, col:col + window_size]
for row in range(df1_window.shape[0]):
start_idx = 0
for col in range(window_size):
if df1_window.iloc[row, col] == 1 and start_idx==0:
df_out.iloc[row, col-1] = df_value_window.iloc[row, col] - df_value_window.iloc[row, col-1]
start_idx += col
return df_out
df_out = generate_df_out(df, df1)型
我想要的输出是这样的：
df2 = pd.DataFrame()
df2['id'] = ['a', 'b', 'c']
df2['0'] = [4, 0, 8]
df2['1'] = [0, -1, 0]
df2['2'] = [0, 1, 0]
df2['3'] = [5, 1, -1]
df2['4'] = [0, 0, 0]
df2['5'] = [0, 4, -1]
df2['6'] = [0, 0, 0]
df2['7'] = [0, -1, 0]
df2['8'] = [0, 0, 2]
df2['9'] = [0, 0, 0]
df2['10'] = [0, 0, -1]
df2['11'] = [-1, -1, 0]
df2['12'] = [0, 0, 0]
df2['13'] = [0, 2, 3]
df2['14'] = [9, 0, 0]
df2['15'] = [0, -1, -1]
df2['16'] = [1, 0, 0]
df2['17'] = [0, 0, 3]
df2['18'] = [-1, 4, 0]
df2['19'] = [0, 0, 0]
df2
id 0 1 2 3 4 5 6 7 8 ... 10 11 12 13 14 15 16 17 18 19
0 a 4 0 0 5 0 0 0 0 0 ... 0 -1 1 1 9 0 1 0 -1 0
1 b 0 -1 1 1 0 4 0 -1 0 ... 0 -1 0 2 0 -1 0 0 4 0
2 c 8 0 0 -1 0 -1 0 0 2 ... -1 0 0 3 0 -1 0 3 0 0型
1条答案
按热度按时间yebdmbv41#
为了实现这一点，你需要一个函数来处理每个窗口并相应地更新
df_out。该函数应该遍历每个窗口，跟踪df_one_zero中1序列的开始和结束，并根据这些索引计算df_value_total中的差异。以下是更新后的函数：字符串
样品运行：
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输出量：
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