根据循环内另一列的值将列的值更改为nan
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【中文标题】根据循环内另一列的值将列的值更改为nan【英文标题】:Change the value of a column into a nan based on the value of another column inside a loop 【发布时间】:2020-10-12 10:03:58 【问题描述】:我有大量带有后缀“mean”或“sum”的列。有时带有“平均”后缀的是NaN。发生这种情况时,我也想将带有“sum”后缀的那个也变成 NaN。我有大量变量,所以我需要 (?) 使用循环。我创建了一个假数据框,并添加了基于 SO 中类似帖子尝试过的 3 件事。不幸的是,没有任何效果
original_data_set = (pd.DataFrame
(
'customerId':[1,2]
,'usage_1_sum':[100, 200]
,'usage_1_mean':[np.nan,100]
,'usage_2_sum':[420,330]
,'usage_2_mean':[45,np.nan]
)
)
print('original dataset')
original_data_set
desired_data_set = (pd.DataFrame
(
'customerId':[1,2]
,'usage_1_sum':[np.nan, 200]
,'usage_1_mean':[np.nan,100]
,'usage_2_sum':[420,np.nan]
,'usage_2_mean':[45,np.nan]
)
)
print('desired dataset')
desired_data_set
holder_set = original_data_set.copy()
for number in range(1,3):
holder_set['usage__sum'.format(number)] = (
holder_set['usage__sum'.format(number)]
.where(holder_set['usage__mean'.format(number)] == np.nan, np.nan
)
)
print('using an np.where statement changed all sum variables into NaN with no discretion')
holder_set
holder_set = original_data_set.copy()
for number in range(1,3):
conditions = [holder_set['usage__mean'.format(number)]==np.nan]
outcome = [np.nan]
holder_set['usage__sum'.format(number)] = np.select(conditions, outcome, default=holder_set['usage__sum'.format(number)])
print('using an np.select did not have any effect on the dataframe')
holder_set
holder_set = original_data_set.copy()
for number in range(1,3):
holder_set.loc[holder_set['usage__mean'.format(number)]==np.nan, 'usage__sum'.format(number)] = 12
print('using a loc did not have any effect on the dataframe')
holder_set
【问题讨论】:
也许可以尝试查看DataFrame.where()
功能。您应该能够直接索引到问题区域,而无需自己编写 for 循环。
【参考方案1】:
假设original
数据框为df
:
df = pd.DataFrame('customerId': [1, 2], 'usage_1_sum': [100, 200], 'usage_1_mean': [
np.nan, 100], 'usage_2_sum': [420, 330], 'usage_2_mean': [45, np.nan])
使用Series.str.endswith
过滤以_mean
结尾的列,然后对于以_mean
结尾的列中的每一列,将_sum
列中的相应值更改为NaN
,其中均值列中的值为@ 987654330@:
for col in df.columns[df.columns.str.endswith('_mean')]:
df.loc[df[col].isna(), col.rstrip('_mean') + '_sum'] = np.nan
结果:
# print(df)
customerId usage_1_sum usage_1_mean usage_2_sum usage_2_mean
0 1 NaN NaN 420.0 45.0
1 2 200.0 100.0 NaN NaN
【讨论】:
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