我试图找出每行DataFrame中第一个有效值和最后一个有效值之间的差异。在
我有一个工作代码与for循环,并寻找更快的东西。 下面是我目前正在做的一个例子:
import pandas as pd
import numpy as np
df = pd.DataFrame(
np.arange(16).astype(np.float).reshape(4, 4),
columns=['a', 'b', 'c', 'd'])
# Fill some NaN
df.loc[0, ['a', 'd']] = np.nan
df.loc[1, ['c', 'd']] = np.nan
df.loc[2, 'b'] = np.nan
df.loc[3, :] = np.nan
print(df)
# a b c d
# 0 NaN 1.0 2.0 NaN
# 1 4.0 5.0 NaN NaN
# 2 8.0 NaN 10.0 11.0
# 3 NaN NaN NaN NaN
diffs = pd.Series(index=df.index)
for i in df.index:
row = df.loc[i]
min_i = row.first_valid_index()
max_i = row.last_valid_index()
if min_i is None or min_i == max_i: # 0 or 1 valid values
continue
diffs[i] = df.loc[i, max_i] - df.loc[i, min_i]
df['diff'] = diffs
print(df)
# a b c d diff
# 0 NaN 1.0 2.0 NaN 1.0
# 1 4.0 5.0 NaN NaN 1.0
# 2 8.0 NaN 10.0 11.0 3.0
# 3 NaN NaN NaN NaN NaN
熊猫让你的生活更轻松,一次一个方法(first_valid_values())。请注意,您必须删除所有具有allNaN值的行(无论如何,没有必要使用这些值):
对于第一个有效值:
对于最后一个有效值:
^{pr2}$减去得到最终结果:
一种方法是back and forward fill丢失的值,然后比较第一行和最后一行。在
如果您想在一行中完成,而不创建新的数据帧:
^{pr2}$相关问题 更多 >
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