Question

我有一个数据框，其中包含有条件的行

     TIME  VALUE Prev_Time
0   23:01      0       NaN
1   23:02      0       NaN
2   23:03      1     23:02
3   23:04      0       NaN
4   23:05      0       NaN
5   23:06      1     23:05
6   23:07      0       NaN
7   23:08      0       NaN
8   23:09      0       NaN
9   23:10      0       NaN
10  23:11      1     23:10
11  23:12      0       NaN
12  23:13      0       NaN
13  23:14      0       NaN
14  23:15      0       NaN
15  23:16      1     23:15
16  23:17      0       NaN

我想根据Column'Prev_Time'上的条件计算行数，以便......

在第一次迭代中，它开始计算行直到一行在它找到列中的'Prev_Time'之前。
第二次和其余的迭代，它开始计数，包括打印时间的行。

所需的输出应为

   ROW_COUNT
0          2
1          3
2          5
3          5
4          2

我也想要Total Counts，有点像（len（df）），应该打印

Total Count: 5

Answer 1

找到好的方法：

notnull=df[df.VALUE>0]
"""
     TIME  VALUE Prev_Time
2   23:03      1     23:02
5   23:06      1     23:05
10  23:11      1     23:10
15  23:16      1     23:15
"""

使用np.split来打破：

row_counts=pd.DataFrame({'ROW_COUNT':[len(x) for x in np.split(df,notnull.index)]})
"""
   ROW_COUNT
0          2
1          3
2          5
3          5
4          2
"""

并计算：

len(row_counts)
"""
5
"""

Answer 2

这有点奏效，你可以根据自己的需要调整代码，但有点基本的想法！

#Dummy data set
df1 = pd.DataFrame({'TIME': np.arange(17), 'VALUE': np.arange(-17,0), 'Prev_time': [np.nan, np.nan,1, np.nan, np.nan,2, np.nan, np.nan, np.nan, np.nan,4, np.nan, np.nan, np.nan, np.nan,5, np.nan]})
#gets the rows that are not null and extracts their index number
df=df1[df1['Prev_time'].notnull()].reset_index()
#Checking for the case where the last row might be null, 
#need to add it manually to the index
if df.loc[len(df)-1]['index'] != (len(df1)-1):
   df.loc[len(df)]=[len(df1),0,0,0]
count=df['index']-df['index'].shift(1).fillna(0)
len(count)

Answer 3

这可能不是一个完美的答案，应该得到你想要的东西：

import pandas as pd

#read the data 
d = pd.read_csv('stackdata.txt')

#we need the last row to be identified, so give it a value
d['Prev_Time'][len(d)-1]=1

#get all the rows where Prev_Time is not null
ds = d[d.Prev_Time.notnull()]

#reset the index, you shall get an additional column named index
ds = ds.reset_index()
#get only the newly added index column
dst = ds[ds.columns[0]]

#get the diff of the series
dstr = dst.diff()

#Get the first value from the previous series and assign it. 
dstr[0] = dst[0]

#Addd +1 to the last item -- result required.
dstr[len(dstr)-1] +=1
len(dstr)

如何有条理地计算Pandas中数据帧中的行与同一数据帧的另一列中的值？

3 个答案: