如何迭代pandas dataframe的行

Posted 2021-01-09 bonelee

from:https://blog.csdn.net/tanzuozhev/article/details/76713387

How to iterate over rows in a DataFrame in Pandas-DataFrame按行迭代

https://stackoverflow.com/questions/16476924/how-to-iterate-over-rows-in-a-dataframe-in-pandas

http://stackoverflow.com/questions/7837722/what-is-the-most-efficient-way-to-loop-through-dataframes-with-pandas

在对DataFrame进行操作时，我们不可避免的需要逐行查看或操作数据，那么有什么高效、快捷的方法呢？

index序号索引

import pandas as pd
inp = [{‘c1‘:10, ‘c2‘:100}, {‘c1‘:11,‘c2‘:110}, {‘c1‘:12,‘c2‘:120}]
df = pd.DataFrame(inp)
for x in xrange(len(df.index)):
    print df[‘c1‘].iloc[x]

这似乎是最常规的办法，而且可以在迭代的过程中对DataFrame进行操作。

enumerate

for i, row in enumerate(df.values):
    index= df.index[i]
    print row

df.values 是 numpy.ndarray 类型
这里 i 是index的序号， row是numpy.ndarray类型。

iterrows

https://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.iterrows.html

import pandas as pd
inp = [{‘c1‘:10, ‘c2‘:100}, {‘c1‘:11,‘c2‘:110}, {‘c1‘:12,‘c2‘:120}]
df = pd.DataFrame(inp)

for index, row in df.iterrows():
    print row[‘c1‘], row[‘c2‘]

#10 100
#11 110
#12 120

df.iterrows() 的每次迭代都是一个tuple类型,包含了index和每行的数据。

1. 采用iterrows的方法，得到的 row 是一个Series，DataFrame的dtypes不会被保留。
2. 返回的Series只是一个原始DataFrame的复制，不可以对原始DataFrame进行修改；

itertuples

http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.itertuples.html

import pandas as pd
inp = [{‘c1‘:10, ‘c2‘:100}, {‘c1‘:11,‘c2‘:110}, {‘c1‘:12,‘c2‘:120}]
df = pd.DataFrame(inp)

for row in df.itertuples():
    # print row[0], row[1], row[2] 等同于
    print row.Index, row.c1, row.c2

itertuples 返回的是一个 pandas.core.frame.Pandas 类型。

普遍认为itertuples 比 iterrows的速度要快。

zip / itertools.izip

zip 和 itertools.izip的用法是相似的， 但是zip返回一个list，而izip返回一个迭代器。 如果数据量很大，zip的性能不及izip

from itertools import izip
import pandas as pd
inp = [{‘c1‘:10, ‘c2‘:100}, {‘c1‘:11,‘c2‘:110}, {‘c1‘:12,‘c2‘:120}]
df = pd.DataFrame(inp)

for row in izip(df.index, df[‘c1‘], df[‘c2‘]):
    print row

时间测评

import time
from numpy.random import randn

df = pd.DataFrame({‘a‘: randn(100000), ‘b‘: randn(100000)})

time_stat = []

# range(index)
test_list = []
t = time.time()
for r in xrange(len(df)):
    test_list.append((df.index[r], df.iloc[r,0], df.iloc[r,1]))
time_stat.append(time.time()-t)

# enumerate
test_list = []
t = time.time()
for i, r in enumerate(df.values):
    test_list.append((df.index[i], r[0], r[1]))
time_stat.append(time.time()-t)

# iterrows
test_list = []
t = time.time()
for i,r in df.iterrows():
    test_list.append((df.index[i], r[‘a‘], r[‘b‘]))
time_stat.append(time.time()-t)

#itertuples
test_list = []
t = time.time()
for ir in df.itertuples():
    test_list.append((ir[0], ir[1], ir[2]))    
time_stat.append(time.time()-t)

# zip
test_list = []
t = time.time()
for r in zip(df.index, df[‘a‘], df[‘b‘]):
    test_list.append((r[0], r[1], r[2]))
time_stat.append(time.time()-t)

# izip
test_list = []
t = time.time()
from itertools import izip
for r in izip(df.index, df[‘a‘], df[‘b‘]):
    test_list.append((r[0], r[1], r[2]))
time_stat.append(time.time()-t)

time_df = pd.DataFrame({‘items‘:[‘range(index)‘, ‘enumerate‘,  ‘iterrows‘, ‘itertuples‘ , ‘zip‘, ‘izip‘], ‘time‘:time_stat})

time_df.sort_values(‘time‘)


items   time
5   izip    0.034869
4   zip 0.040440
3   itertuples  0.072604
1   enumerate   0.174094
2   iterrows    4.026293
0   range(index)    21.921407

可以发现在时间花销上， izip > zip > itertuples > enumerate > iterrows > range(index)