Python将递增地使用上述单元格的单元格值填充NaN
我有一个数据框架如下Python将递增地使用上述单元格的单元格值填充NaN,python,pandas,dataframe,nan,Python,Pandas,Dataframe,Nan,我有一个数据框架如下 Word Count Team Sex Code 0 develop 9 1 M P45.01 1 Effective 7 NaN M NaN 2 professional 8 NaN M NaN 3 approach 5 NaN M NaN 4 raster 34 NaN M
Word Count Team Sex Code
0 develop 9 1 M P45.01
1 Effective 7 NaN M NaN
2 professional 8 NaN M NaN
3 approach 5 NaN M NaN
4 raster 34 NaN M NaN
5 Sad 55 NaN M NaN
6 water 2 NaN M NaN
7 soil 7 NaN M NaN
8 farming 9 NaN M NaN
9 deep 12 NaN M NaN
我相信您可以创建范围并将其添加到第一个值中，以供性使用： 更一般的广播：
cols = ['Team','Code']
r = np.arange(len(df))
df[cols] = r[:, None] + df.loc[0, cols].values
df['Sex'] = df['Sex'].ffill()
print (df)
Word Count Team Sex Code
0 develop 9 1.0 M 45.01
1 Effective 7 2.0 M 46.01
2 professional 8 3.0 M 47.01
3 approach 5 4.0 M 48.01
4 raster 34 5.0 M 49.01
5 Sad 55 6.0 M 50.01
6 water 2 7.0 M 51.01
7 soil 7 8.0 M 52.01
8 farming 9 9.0 M 53.01
9 deep 12 10.0 M 54.01
r = np.arange(len(df))
df['Team'] = (df.loc[0, 'Team'] + r).astype(int)
df1 = df.loc[[0], 'Code'].str.extract('(\D+)(\d+\.\d+)', expand=False)
print (df1)
0 1
0 P 45.01
df['Code'] = float(df1.loc[0, 1]) + r
df['Code'] = df1.loc[0, 0] + df['Code'].astype(str)
df['Sex'] = df['Sex'].ffill()
print (df)
Word Count Team Sex Code
0 develop 9 1 M P45.01
1 Effective 7 2 M P46.01
2 professional 8 3 M P47.01
3 approach 5 4 M P48.01
4 raster 34 5 M P49.01
5 Sad 55 6 M P50.01
6 water 2 7 M P51.01
7 soil 7 8 M P52.01
8 farming 9 9 M P53.01
9 deep 12 10 M P54.01
r = np.arange(len(df))
df['Team'] = (df.iloc[0, df.columns.get_loc('Team')] + r).astype(int)
df1 = df.iloc[[0], df.columns.get_loc('Code')].str.extract('(\D+)(\d+\.\d+)', expand=False)
df['Code'] = float(df1.loc[0, 1]) + r
df['Code'] = df1.loc[0, 0] + df['Code'].astype(str)
df['Sex'] = df['Sex'].ffill()
print (df)
Word Count Team Sex Code
0 develop 9 1 M P45.01
1 Effective 7 2 M P46.01
2 professional 8 3 M P47.01
3 approach 5 4 M P48.01
4 raster 34 5 M P49.01
5 Sad 55 6 M P50.01
6 water 2 7 M P51.01
7 soil 7 8 M P52.01
8 farming 9 9 M P53.01
9 deep 12 10 M P54.01
我相信您可以创建范围并将其添加到第一个值中，以供性使用： 更一般的广播：
cols = ['Team','Code']
r = np.arange(len(df))
df[cols] = r[:, None] + df.loc[0, cols].values
df['Sex'] = df['Sex'].ffill()
print (df)
Word Count Team Sex Code
0 develop 9 1.0 M 45.01
1 Effective 7 2.0 M 46.01
2 professional 8 3.0 M 47.01
3 approach 5 4.0 M 48.01
4 raster 34 5.0 M 49.01
5 Sad 55 6.0 M 50.01
6 water 2 7.0 M 51.01
7 soil 7 8.0 M 52.01
8 farming 9 9.0 M 53.01
9 deep 12 10.0 M 54.01
r = np.arange(len(df))
df['Team'] = (df.loc[0, 'Team'] + r).astype(int)
df1 = df.loc[[0], 'Code'].str.extract('(\D+)(\d+\.\d+)', expand=False)
print (df1)
0 1
0 P 45.01
df['Code'] = float(df1.loc[0, 1]) + r
df['Code'] = df1.loc[0, 0] + df['Code'].astype(str)
df['Sex'] = df['Sex'].ffill()
print (df)
Word Count Team Sex Code
0 develop 9 1 M P45.01
1 Effective 7 2 M P46.01
2 professional 8 3 M P47.01
3 approach 5 4 M P48.01
4 raster 34 5 M P49.01
5 Sad 55 6 M P50.01
6 water 2 7 M P51.01
7 soil 7 8 M P52.01
8 farming 9 9 M P53.01
9 deep 12 10 M P54.01
r = np.arange(len(df))
df['Team'] = (df.iloc[0, df.columns.get_loc('Team')] + r).astype(int)
df1 = df.iloc[[0], df.columns.get_loc('Code')].str.extract('(\D+)(\d+\.\d+)', expand=False)
df['Code'] = float(df1.loc[0, 1]) + r
df['Code'] = df1.loc[0, 0] + df['Code'].astype(str)
df['Sex'] = df['Sex'].ffill()
print (df)
Word Count Team Sex Code
0 develop 9 1 M P45.01
1 Effective 7 2 M P46.01
2 professional 8 3 M P47.01
3 approach 5 4 M P48.01
4 raster 34 5 M P49.01
5 Sad 55 6 M P50.01
6 water 2 7 M P51.01
7 soil 7 8 M P52.01
8 farming 9 9 M P53.01
9 deep 12 10 M P54.01
df=df.reset_index()
df['Team']=df.index+1
df=df.reset_index()
df['Team']=df.index+1
我还没有测试过它，但它应该可以工作我意识到，考虑到这里更新的数据帧，您还可以执行以下操作之一： 安装程序
我们需要注意的一点是，发电机正在动态生成数字，只能使用一次；但它确实起到了作用。我意识到，考虑到这里更新的数据帧，您还可以执行以下操作之一： 安装程序
我们需要注意的一点是，发电机正在动态生成数字，只能使用一次；但是它确实起到了作用。啊，我只是用np.arange写的，很好的一个。我认为这不能很好地推广到多个列。而且，从第一行开始添加不是一个好主意，因为你不能保证它从第一行开始是连续的。此外，并非所有列都需要相同数量的NAN。@cᴏʟᴅsᴘᴇᴇᴅ - 是的，这取决于数据。如果需要其他内容，我会等待OP。@UmarYusuf-那么如果在浮点代码检查编辑的答案之前只差一些字符串。啊，我只是在用np.arange编写，很好。我认为这不会很好地推广到多个列。而且，从第一行添加不是一个好主意，因为您不能保证它从第一行开始是连续的。此外，并非所有列都需要相同数量的NAN。@cᴏʟᴅsᴘᴇᴇᴅ - 是的，这取决于数据。如果需要其他东西，我在等待OP。@UmarYusuf-那么如果在浮点代码检查编辑的答案之前只差一些字符串。总是有步骤1吗？或者应该是不同的？还有，在数字列中总是会有相同数量的N吗？@jezrael只是想知道你在这里把什么称为步骤1？@stucash-我想如果有时有不同的步骤，例如2，那么解决方案应该是不同的。@jezrael啊，我现在明白你的意思了，谢谢。总是有步骤1吗？或者应该是不同的？还有，在数字列中总是有相同数量的“N”吗？@jezrael只是想知道这里的第一步是什么？@stucash-我想如果有时有不同的步骤，例如2，那么解决方案应该是不同的。@jezrael啊，我现在明白你的意思了，谢谢。
import re
def count(n):
x = 0
while x < n:
yield x
x += 1
def populate(s, step):
chars = re.split('(\d*)', s)
number = int(chars[1]) + step
chars[1] = str(number)
return ''.join(chars)
number_generator = count(10)
number_generator2 = count(10)
# option 1 for `team` column
df['Team'] = df.index + 1 # just use your indices.
# option 2 for 'team' column
df.loc[:, 'Team'] = df.Team.apply(lambda _: next(number_generator)) # use generator.
# output
Word Count Team Sex Code
0 develop 9 1 M P45.01
1 Effective 7 2 M NaN
2 professional 8 3 M NaN
3 approach 5 4 M NaN
4 raster 34 5 M NaN
5 Sad 55 6 M NaN
6 water 2 7 M NaN
7 soil 7 8 M NaN
8 farming 9 9 M NaN
9 deep 12 10 M NaN
# we can use generator again for `Code` column
df.loc[:, 'Code']df.Code.apply(lambda _: populate(df.loc[0, 'Code'], next(number_generator2)))
# output
Word Count Team Sex Code
0 develop 9 1 M P45.01
1 Effective 7 2 M P46.01
2 professional 8 3 M P47.01
3 approach 5 4 M P48.01
4 raster 34 5 M P49.01
5 Sad 55 6 M P50.01
6 water 2 7 M P51.01
7 soil 7 8 M P52.01
8 farming 9 9 M P53.01
9 deep 12 10 M P54.01