Python 如何根据同一数据帧的另一列中的值有条件地计算数据帧中的行数？

我有一个数据框，其中包含我希望有条件计数的行

     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

我想根据列“Prev_Time”上的条件计算行数，以便

在第一次迭代中，它开始计算行数，直到一行 在找到列中的“上一次”之前

在第二次迭代和其他迭代中，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

我还需要总计数，比如（len（df）），应该打印出来

Total Count: 5

---

这是可行的，你可以根据你的需要修改代码，但是基本的想法是可行的

#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)

---

这是可行的，你可以根据你的需要修改代码，但是基本的想法是可行的

#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)

---

这可能不是一个完美的答案，你会得到你想要的：

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)

---

这可能不是一个完美的答案，你会得到你想要的：

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)

找到好的台词：

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

中断：

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
"""

和计数：

len(row_counts)
"""
5
"""

找到好的台词：

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

中断：

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
"""

和计数：

len(row_counts)
"""
5
"""

---

Prev_Time

列是否已经存在，或者您是否询问如何创建该列，然后收集

Prev_Time

有值的行计数？@Grr Yes，“Prev_Time”列已经存在。

Prev_Time

列是否已经存在，或者您正在询问如何创建该列，然后收集

Prev_Time

有值的行数？@Grr是的，“Prev_Time”列已经存在。