尝试使用满足特定条件的值创建新列。下面我列出了一些代码，它在某种程度上解释了逻辑，但没有产生正确的输出:
import pandas as pd
import numpy as np
df = pd.DataFrame({'date': ['2019-08-06 09:00:00', '2019-08-06 12:00:00', '2019-08-06 18:00:00', '2019-08-06 21:00:00', '2019-08-07 09:00:00', '2019-08-07 16:00:00', '2019-08-08 17:00:00' ,'2019-08-09 16:00:00'],
'type': [0, 1, np.nan, 1, np.nan, np.nan, 0 ,0],
'colour': ['blue', 'red', np.nan, 'blue', np.nan, np.nan, 'blue', 'red'],
'maxPixel': [255, 7346, 32, 5184, 600, 322, 72, 6000],
'minPixel': [86, 96, 14, 3540, 528, 300, 12, 4009],
'colourDate': ['2019-08-06 12:00:00', '2019-08-08 16:00:00', '2019-08-06 23:00:00', '2019-08-06 22:00:00', '2019-08-08 09:00:00', '2019-08-09 16:00:00', '2019-08-08 23:00:00' ,'2019-08-11 16:00:00'] })
max_conditions = [(df['type'] == 1) & (df['colour'] == 'blue'),
(df['type'] == 1) & (df['colour'] == 'red')]
max_choices = [np.where(df['date'] <= df['colourDate'], max(df['maxPixel']), np.nan),
np.where(df['date'] <= df['colourDate'], min(df['minPixel']), np.nan)]
df['pixelLimit'] = np.select(max_conditions, max_choices, default=np.nan)
date type colour maxPixel minPixel colourDate pixelLimit
0 2019-08-06 09:00:00 0.0 blue 255 86 2019-08-06 12:00:00 NaN
1 2019-08-06 12:00:00 1.0 red 7346 96 2019-08-08 16:00:00 12.0
2 2019-08-06 18:00:00 NaN NaN 32 14 2019-08-06 23:00:00 NaN
3 2019-08-06 21:00:00 1.0 blue 5184 3540 2019-08-06 22:00:00 6000.0
4 2019-08-07 09:00:00 NaN NaN 600 528 2019-08-08 09:00:00 NaN
5 2019-08-07 16:00:00 NaN NaN 322 300 2019-08-09 16:00:00 NaN
6 2019-08-08 17:00:00 0.0 blue 72 12 2019-08-08 23:00:00 NaN
7 2019-08-09 16:00:00 0.0 red 6000 4009 2019-08-11 16:00:00 NaN
值
12.0在列 df['pixelLimit'] 的索引行 1 中不正确，因为该值来自 df['minPixel']索引行 6 具有 df['date'] 2019-08-08 17:00:00 的日期时间大于 2019-08-08 16:00:00 df['date']索引行 1 中包含的日期时间。值
6000.0在列 df['pixelLimit'] 的索引行 3 中不正确，因为该值来自 df['maxPixel']索引行 7 具有 df['date'] 2019-08-09 16:00:00 的日期时间大于 2019-08-06 22:00:00 df['date']索引行中包含的日期时间。正确的输出:
date type colour maxPixel minPixel colourDate pixelLimit
0 2019-08-06 09:00:00 0.0 blue 255 86 2019-08-06 12:00:00 NaN
1 2019-08-06 12:00:00 1.0 red 7346 96 2019-08-08 16:00:00 14.0
2 2019-08-06 18:00:00 NaN NaN 32 14 2019-08-06 23:00:00 NaN
3 2019-08-06 21:00:00 1.0 blue 5184 3540 2019-08-06 22:00:00 5184.0
4 2019-08-07 09:00:00 NaN NaN 600 528 2019-08-08 09:00:00 NaN
5 2019-08-07 16:00:00 NaN NaN 322 300 2019-08-09 16:00:00 NaN
6 2019-08-08 17:00:00 0.0 blue 72 12 2019-08-08 23:00:00 NaN
7 2019-08-09 16:00:00 0.0 red 6000 4009 2019-08-11 16:00:00 NaN
值
14.0在列 df['pixelLimit'] 的索引行 1 中是正确的，因为我们正在寻找列 df['minPixel'] 中的最小值在 df['date'] 列中有一个日期时间小于 df['colourDate'] 列的索引行 1 中的日期时间并且大于或等于索引行 1 中列 df['date'] 的日期时间值
5184.0在列 df['pixelLimit'] 的索引行 3 中是正确的，因为我们正在寻找列 df['maxPixel'] 中的最大值在 df['date'] 列中有一个日期时间小于 df['colourDate'] 列的索引第 3 行中的日期时间并且大于或等于索引行 3 中列 df['date'] 的日期时间注意事项:
也许
np.select不是最适合这项任务，某种功能可能会更好地为这项任务服务？另外，也许我需要创建某种动态
len用作每一行的起点？请求
请任何人都可以帮助我修改我的代码以实现正确的输出
最佳答案
对于像这样的匹配问题，一种可能性是使用 bool 系列对所有满足条件的行(对于该行)进行完整合并，然后进行子集化，然后找到 max或 min在所有可能的匹配项中。由于这需要稍微不同的列和不同的函数，我将这些操作分成两段非常相似的代码，一段处理 1/blue，另一段处理 1/red。
首先一些家务，使事情日期时间
import pandas as pd
df['date'] = pd.to_datetime(df['date'])
df['colourDate'] = pd.to_datetime(df['colourDate'])
计算每行时间之间 1/红色的最小像素
# Subset of rows we need to do this for
dfmin = df[df.type.eq(1) & df.colour.eq('red')].reset_index()
# To each row merge all rows from the original DataFrame
dfmin = dfmin.merge(df[['date', 'minPixel']], how='cross')
# If pd.version < 1.2 instead use:
#dfmin = dfmin.assign(t=1).merge(df[['date', 'minPixel']].assign(t=1), on='t')
# Only keep rows between the dates, then among those find the min minPixel
smin = (dfmin[dfmin.date_y.between(dfmin.date_x, dfmin.colourDate)]
.groupby('index')['minPixel_y'].min()
.rename('pixel_limit'))
#index
#1 14
#Name: pixel_limit, dtype: int64
# Max is basically a mirror
dfmax = df[df.type.eq(1) & df.colour.eq('blue')].reset_index()
dfmax = dfmax.merge(df[['date', 'maxPixel']], how='cross')
#dfmax = dfmax.assign(t=1).merge(df[['date', 'maxPixel']].assign(t=1), on='t')
smax = (dfmax[dfmax.date_y.between(dfmax.date_x, dfmax.colourDate)]
.groupby('index')['maxPixel_y'].max()
.rename('pixel_limit'))
最后，因为以上对原始索引(即
'index' )的分组，我们可以简单地分配回以与原始 DataFrame 对齐。df['pixel_limit'] = pd.concat([smin, smax])
date type colour maxPixel minPixel colourDate pixel_limit
0 2019-08-06 09:00:00 0.0 blue 255 86 2019-08-06 12:00:00 NaN
1 2019-08-06 12:00:00 1.0 red 7346 96 2019-08-08 16:00:00 14.0
2 2019-08-06 18:00:00 NaN NaN 32 14 2019-08-06 23:00:00 NaN
3 2019-08-06 21:00:00 1.0 blue 5184 3540 2019-08-06 22:00:00 5184.0
4 2019-08-07 09:00:00 NaN NaN 600 528 2019-08-08 09:00:00 NaN
5 2019-08-07 16:00:00 NaN NaN 322 300 2019-08-09 16:00:00 NaN
6 2019-08-08 17:00:00 0.0 blue 72 12 2019-08-08 23:00:00 NaN
7 2019-08-09 16:00:00 0.0 red 6000 4009 2019-08-11 16:00:00 NaN
如果您需要为具有最小/最大像素的行带来许多不同的信息，那么代替
groupby min/max我们将 sort_values 然后 gropuby + head或 tail获取最小或最大像素。对于 min 这看起来像(后缀的轻微重命名):# Subset of rows we need to do this for
dfmin = df[df.type.eq(1) & df.colour.eq('red')].reset_index()
# To each row merge all rows from the original DataFrame
dfmin = dfmin.merge(df[['date', 'minPixel']].reset_index(), how='cross',
suffixes=['', '_match'])
# For older pandas < 1.2
#dfmin = (dfmin.assign(t=1)
# .merge(df[['date', 'minPixel']].reset_index().assign(t=1),
# on='t', suffixes=['', '_match']))
# Only keep rows between the dates, then among those find the min minPixel row.
# A bunch of renaming.
smin = (dfmin[dfmin.date_match.between(dfmin.date, dfmin.colourDate)]
.sort_values('minPixel_match', ascending=True)
.groupby('index').head(1)
.set_index('index')
.filter(like='_match')
.rename(columns={'minPixel_match': 'pixel_limit'}))
.tail 类似dfmax = df[df.type.eq(1) & df.colour.eq('blue')].reset_index()
dfmax = dfmax.merge(df[['date', 'maxPixel']].reset_index(), how='cross',
suffixes=['', '_match'])
smax = (dfmax[dfmax.date_match.between(dfmax.date, dfmin.colourDate)]
.sort_values('maxPixel_match', ascending=True)
.groupby('index').tail(1)
.set_index('index')
.filter(like='_match')
.rename(columns={'maxPixel_match': 'pixel_limit'}))
axis=1现在我们需要将多个列连接到原始列:result = pd.concat([df, pd.concat([smin, smax])], axis=1)
date type colour maxPixel minPixel colourDate index_match date_match pixel_limit
0 2019-08-06 09:00:00 0.0 blue 255 86 2019-08-06 12:00:00 NaN NaN NaN
1 2019-08-06 12:00:00 1.0 red 7346 96 2019-08-08 16:00:00 2.0 2019-08-06 18:00:00 14.0
2 2019-08-06 18:00:00 NaN NaN 32 14 2019-08-06 23:00:00 NaN NaN NaN
3 2019-08-06 21:00:00 1.0 blue 5184 3540 2019-08-06 22:00:00 3.0 2019-08-06 21:00:00 5184.0
4 2019-08-07 09:00:00 NaN NaN 600 528 2019-08-08 09:00:00 NaN NaN NaN
5 2019-08-07 16:00:00 NaN NaN 322 300 2019-08-09 16:00:00 NaN NaN NaN
6 2019-08-08 17:00:00 0.0 blue 72 12 2019-08-08 23:00:00 NaN NaN NaN
7 2019-08-09 16:00:00 0.0 red 6000 4009 2019-08-11 16:00:00 NaN NaN NaN
关于python - 找到满足条件的特定值 - python，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/66165400/