用最快的解决方案:
for col in myList:
df[col] = 0
print(df)
a b c d
0 computer 0 0 0
1 printer 0 0 0
另一种解决方案是使用concat
与DataFrame
构造:
pd.concat([df3,pd.DataFrame(columns=myList, index=df.index, data=0)], axis=1)
时序:
[20000行×300个colu MNS]:
In [286]: %timeit pd.concat([df,pd.DataFrame(columns=myList)], axis=1).fillna(0)
1 loop, best of 3: 1.17 s per loop
In [287]: %timeit pd.concat([df3,pd.DataFrame(columns=myList, index=df.index,data=0)],axis=1)
10 loops, best of 3: 81.7 ms per loop
In [288]: %timeit (orig(df4))
10 loops, best of 3: 59.2 ms per loop
代码计时:
myList=["b","c","d"] * 100
df = pd.DataFrame({"a":["computer", "printer"]})
print(df)
df = pd.concat([df]*10000).reset_index(drop=True)
df3 = df.copy()
df4 = df.copy()
df1= pd.concat([df,pd.DataFrame(columns=myList)], axis=1).fillna(0)
df2 = pd.concat([df3,pd.DataFrame(columns=myList, index=df.index, data=0)], axis=1)
print(df1)
print(df2)
def orig(df):
for col in range(300):
df[col] = 0
return df
print (orig(df4))
嗯,我认为'concat'解决方案更快,但不是。很有意思。 – jezrael
@jezrael也是令人惊讶的,我期望在这里逐渐增加df,使其不具有高性能 – EdChum