英文:
Set variable column values to nan based on row condition
问题
我想要能够根据第一列的值可变地更改一列的值。
假设我有以下数据框：
col_ind col_1 col_2 col_3
3 a b c
2 d e f
1 g h i
我实际上想要执行以下操作：
df.loc[:, df.columns[-df['col_ind']]:] = np.nan
这将导致：
col_ind col_1 col_2 col_3
3 nan nan nan
2 d nan nan
1 g h nan
英文:
I want to be able to variably change a column value based on the value of the first column.
Say I have a dataframe as follows:
col_ind col_1 col_2 col_3
3 a b c
2 d e f
1 g h i
I effectively want to do
df.loc[:, df.columns[-df['col_ind']:]] = np.nan
Which would result in:
col_ind col_1 col_2 col_3
3 nan nan nan
2 d nan nan
1 g h nan
答案1
得分: 5
让我们使用广播来检查可以被掩盖的索引。
c = df.columns[1:]
m = range(len(c), 0, -1) <= df['col_ind'].values[:, None]
df[c] = df[c].mask(m)
结果：
col_ind col_1 col_2 col_3
0 3 NaN NaN NaN
1 2 d NaN NaN
2 1 g h NaN
英文:
Lets use broadcasting to check the indices which can be masked
c = df.columns[1:]
m = range(len(c), 0, -1) <= df['col_ind'].values[:, None]
df[c] = df[c].mask(m)
Result
col_ind col_1 col_2 col_3
0 3 NaN NaN NaN
1 2 d NaN NaN
2 1 g h NaN
答案2
得分: 1
你可以获取 df["col_ind"] 的 values，对它们进行迭代并将 slice 设置为 np.nan：
vals = df["col_ind"].values
for i, v in enumerate(vals):
df.iloc[i, -v:] = np.nan
英文:
You can get the values of df["col_ind"], iterate through them and set the slice to np.nan:
vals = df["col_ind"].values
for i, v in enumerate(vals):
df.iloc[i, -v:] = np.nan
答案3
得分: 1
你可以使用apply并指定result_type='broadcast'。 (编辑：借用 @marcelo-paco 的代码)
def make_nan(row):
row[-row[0]:] = np.nan
return row
df = pd.DataFrame({'col_ind': [3, 2, 1], 'col_1': ['a', 'd', 'g'], 'col_2': ['b', 'e', 'h'], 'col_3': ['c', 'f', 'i']})
df[:] = df.apply(make_nan, axis=1, result_type='broadcast')
df
这将得到：
col_ind col_1 col_2 col_3
0 3 NaN NaN NaN
1 2 d NaN NaN
2 1 g h NaN
英文:
You an use apply with result_type='broadcast'. (Edit: borrowing @marcelo-paco's code)
def make_nan(row):
row[-row[0]:] = np.nan
return row
df = pd.DataFrame({'col_ind': [3, 2, 1], 'col_1': ['a', 'd', 'g'], 'col_2': ['b', 'e', 'h'], 'col_3': ['c', 'f', 'i']})
df[:] = df.apply(make_nan, axis=1, result_type='broadcast')
df
This will give:
col_ind col_1 col_2 col_3
3 NaN NaN NaN
2 d NaN NaN
1 g h NaN
答案4
得分: 1
你可以使用当前列的切片创建新列，然后替换原始列的内容。
for i, cn in enumerate(df.columns, 1):
df[cn] = [*[np.nan]*i, *df[cn].loc[i:]]
> 
英文:
You could create new columns with slices of the current columns and then replace
for i, cn in enumerate(df.columns,1):
df[cn] = [*[np.nan]*i, *df[cn].loc[i:]]
>
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