如何修改此python代码以使其运行更快?
如何修改下面的代码以使其运行更快?当前从i=0运行到i=75125需要30秒如何修改此python代码以使其运行更快?,python,performance,dataframe,Python,Performance,Dataframe,如何修改下面的代码以使其运行更快?当前从i=0运行到i=75125需要30秒 for i in range(0, len(df.index)): if i % 25 == 0: df.iloc[i, 15] = df.iloc[i, 27] df.iloc[i + 1, 15] = df.iloc[i, 26] df.iloc[i + 2, 15] = df.iloc[i, 25] df.ilo
for i in range(0, len(df.index)):
if i % 25 == 0:
df.iloc[i, 15] = df.iloc[i, 27]
df.iloc[i + 1, 15] = df.iloc[i, 26]
df.iloc[i + 2, 15] = df.iloc[i, 25]
df.iloc[i + 3, 15] = df.iloc[i, 24]
df.iloc[i + 4, 15] = df.iloc[i, 23]
df.iloc[i + 5, 15] = df.iloc[i, 22]
df.iloc[i + 6, 15] = df.iloc[i, 21]
df.iloc[i + 7, 15] = df.iloc[i, 20]
df.iloc[i + 8, 15] = df.iloc[i, 19]
df.iloc[i + 9, 15] = df.iloc[i, 18]
df.iloc[i + 10, 15] = df.iloc[i, 17]
df.iloc[i + 11, 15] = df.iloc[i, 16]
df.iloc[i + 12, 15] = df.iloc[i, 15]
df.iloc[i + 13, 15] = df.iloc[i, 28]
df.iloc[i + 14, 15] = df.iloc[i, 29]
df.iloc[i + 15, 15] = df.iloc[i, 30]
df.iloc[i + 16, 15] = df.iloc[i, 31]
df.iloc[i + 17, 15] = df.iloc[i, 32]
df.iloc[i + 18, 15] = df.iloc[i, 33]
df.iloc[i + 19, 15] = df.iloc[i, 34]
df.iloc[i + 20, 15] = df.iloc[i, 35]
df.iloc[i + 21, 15] = df.iloc[i, 36]
df.iloc[i + 22, 15] = df.iloc[i, 37]
df.iloc[i + 23, 15] = df.iloc[i, 38]
df.iloc[i + 24, 15] = df.iloc[i, 39]
尝试执行
range(0,len(df.index),25)
而不执行if
语句。如果不了解更多关于数据帧中的数据或算法试图实现的内容,则很难推荐任何内容。