填充 DataFrame 的 Python 熊猫

Question

我在填充数据框时遇到问题。 这是初始情况（图1）

我的代码是这样运行的（图 2）：

但我想要这个（图 3）：

因此，如果从 -1 到 4 的行为空，则它应该为空。但如果有数字则应填写“0”

我的代码看起来像这样...

import pandas as pd
df = pd.read_csv('/Users/Hanna/Code/ZERO.csv')


indx = df[df['D'] == -1].index.values

for i, j in zip(indx[:-1], indx[1:]):
df.loc[i:j-1, 'E'] = df.loc[i:j-1, 'E'].fillna(0)

    if j == indx[-1]:
    df.loc[j:, 'E'] = df.loc[j:, 'E'].fillna(0)

那是我的代码，但我不确定'NaN'

d = { 'A'：[4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074，4000074]，“B “：[SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746，SP000796746]， 'C'：[-1,0， 1, 2, 3, 4, -1, 0, 1, 2, 3, 4, -1, 0, 1, 2, 3, 4, -1, 0, 1, 2, 3, 4], 'D ': [0, 1000, 1000, 0, 0, 0, 'NaN', 'NaN', 'NaN', 'NaN', 'NaN', 'NaN', 0, 0, 0, 3000, 3000, 0 ], 'E': [2000, 2000, 2000, 2000, 2000, 2000, 'NaN', 'NaN', 'NaN', 'NaN', 'NaN', 'NaN', 4000, 4000, 4000, 4000 , 4000, 4000]}

谢谢你，汉娜

也许它不起作用，因为我以前这样做过，另一列 F：

indx = df[df['Diff Load Due Week'] == -1].index.values


for i, j in zip(indx[:-1], indx[1:]):
    df.loc[i:j-1, 'F'] = df.loc[i:j-1, 'F'].max()

if j == indx[-1]:
        df.loc[j:, 'F'] = df.loc[j:, 'F'].max()

可以，我必须先删除索引吗？

这是我最后的输出：

base_list =[-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26]
df_c = pd.MultiIndex.from_product([
[4000074],
["SP000796746","SP001811642"],
[201824, 201828, 201832, 201835, 201837, 201839, 201845, 201850, 201910, 201918, 201922, 201926, 201909, 201916, 201918, 201920],
base_list],

    names=["A", "B", "C", "D"]).to_frame(index=False)

# Verbinden der neuen Liste und der kleinen Rohdatenliste

df_3 = pd.merge(df_c, df_1, how='outer')

# Zusammengefügte Daten in Excel und csv speichern für Überprüfung und Weiterarbeit

df_3.to_csv('GROß.csv')
df_3.to_excel('GROß.xlsx')

Einlesen der neu erstellten csv

df = pd.read_csv('/Users/Hanna/Desktop/Daten Projektseminar/Coding/GROß.csv')

#Index setzen für -1, damit Spalten und Reihen aufgefüllt werden können

indx = df[df['D'] == -1].index.values

#Aufüllen der Billings mit maximalen Wert

for i, j in zip(indx[:-1], indx[1:]):
    df.loc[i:j-1, 'F'] = df.loc[i:j-1, 'F'].max()

if j == indx[-1]:
        df.loc[j:, 'F'] = df.loc[j:, 'F'].max()

Answer 1

好的，现在应该可以解决问题了：

import pandas as pd
import numpy as np

#your data
d = {'A': [4000074, 4000074, 4000074, 4000074, 4000074, 4000074, 4000074, 4000074, 4000074,
   4000074, 4000074, 4000074, 4000074, 4000074, 4000074, 4000074, 4000074, 4000074],
 'B': ['SP000796746', 'SP000796746', 'SP000796746', 'SP000796746', 'SP000796746', 'SP000796746',
   'SP000796746', 'SP000796746', 'SP000796746', 'SP000796746', 'SP000796746', 'SP000796746', 
   'SP000796746', 'SP000796746', 'SP000796746', 'SP000796746', 'SP000796746', 'SP000796746'],
 'C': [201926, 201926, 201926, 201926, 201926, 201926, 201909,201909, 201909, 201909, 201909, 
   201909, 201933, 201933, 201933, 201933, 201933, 201933],
 'D': [-1, 0, 1, 2, 3, 4, -1, 0, 1, 2, 3, 4, -1, 0, 1, 2, 3, 4], 
 'E': [np.nan, 1000, 1000, 0, 0, 0, np.nan, np.nan, np.nan, np.nan, np.nan, np.nan, np.nan, np.nan, np.nan, 3000, 3000, np.nan]}

#create data frame
df = pd.DataFrame(data = d)

#group sum column E by ID columns
groupedSum = df.groupby(['A', 'B', 'C'])['E'].sum().reset_index()

#loop over unique IDs
for i, row in groupedSum.iterrows():
    #define values
    idValue = groupedSum.at[i,'C']
    sumValue = groupedSum.at[i,'E']
    #if sum is not zero
    if (sumValue != 0):
        #change values to zero if greater than zero
        df['E'].loc[df['C'] == idValue] = df['E'].apply(lambda x: x if x > 0 else 0)

print(df)

          A            B       C  D       E
0   4000074  SP000796746  201926 -1     0.0
1   4000074  SP000796746  201926  0  1000.0
2   4000074  SP000796746  201926  1  1000.0
3   4000074  SP000796746  201926  2     0.0
4   4000074  SP000796746  201926  3     0.0
5   4000074  SP000796746  201926  4     0.0
6   4000074  SP000796746  201909 -1     NaN
7   4000074  SP000796746  201909  0     NaN
8   4000074  SP000796746  201909  1     NaN
9   4000074  SP000796746  201909  2     NaN
10  4000074  SP000796746  201909  3     NaN
11  4000074  SP000796746  201909  4     NaN
12  4000074  SP000796746  201933 -1     0.0
13  4000074  SP000796746  201933  0     0.0
14  4000074  SP000796746  201933  1     0.0
15  4000074  SP000796746  201933  2  3000.0
16  4000074  SP000796746  201933  3  3000.0
17  4000074  SP000796746  201933  4     0.0