我正在尝试迭代Python Pandas数据帧的行。在数据帧的每一行中,我试图通过列名引用行中的每个值。
这就是我所拥有的:
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.rand(10,4),columns=list('ABCD'))
print df
A B C D
0 0.351741 0.186022 0.238705 0.081457
1 0.950817 0.665594 0.671151 0.730102
2 0.727996 0.442725 0.658816 0.003515
3 0.155604 0.567044 0.943466 0.666576
4 0.056922 0.751562 0.135624 0.597252
5 0.577770 0.995546 0.984923 0.123392
6 0.121061 0.490894 0.134702 0.358296
7 0.895856 0.617628 0.722529 0.794110
8 0.611006 0.328815 0.395859 0.507364
9 0.616169 0.527488 0.186614 0.278792
我使用this approach进行迭代,但它只给了我解决方案的一部分 - 在每次迭代中选择一行后,如何按列名访问行元素?
以下是我要做的事情:
for row in df.iterrows():
print row.loc[0,'A']
print row.A
print row.index()
我的理解是该行是一只熊猫series。但是我无法索引该系列。
是否可以在同时迭代行时使用列名?
答案 0 :(得分:25)
我也喜欢itertuples()
for row in df.itertuples():
print(row.A)
print(row.Index)
因为row是一个命名元组,如果你想访问每一行的值,那么这应该是 MUCH 更快
速度跑:
df = pd.DataFrame([x for x in range(1000*1000)], columns=['A'])
st=time.time()
for index, row in df.iterrows():
row.A
print(time.time()-st)
45.05799984931946
st=time.time()
for row in df.itertuples():
row.A
print(time.time() - st)
0.48400020599365234
答案 1 :(得分:18)
iterrows()
中的项目不是系列,而是(索引,系列)的元组,因此您可以像这样解压缩for循环中的元组:
for (idx, row) in df.iterrows():
print(row.loc['A'])
print(row.A)
print(row.index)
#0.890618586836
#0.890618586836
#Index(['A', 'B', 'C', 'D'], dtype='object')
答案 2 :(得分:0)
for i in range(1,len(na_rm.columns)):
print ("column name:", na_rm.columns[i])
输出:
column name: seretide_price
column name: symbicort_mkt_shr
column name: symbicort_price
答案 3 :(得分:0)
我试图弄清楚如何通过熊猫数据框有效地进行迭代。有不同的方法,通常的iterrows()远非最佳。 itertuples()可以快100倍。
df = pd.DataFrame(np.random.randint(0, 100, size=(1000000, 4)), columns=list('ABCD'))
print(df)
1)iterrows()
很方便,但是该死的很慢
start_time = time.clock()
result = 0
for _, row in df.iterrows():
result += max(row['B'], row['C'])
total_elapsed_time = round(time.clock() - start_time, 2)
print("1. Iterrows done in {} seconds, result = {}".format(total_elapsed_time, result))
2)默认的itertuples()
已经快了
start_time = time.clock()
result = 0
for row in df.itertuples(index=False):
result += max(row.B, row.C)
total_elapsed_time = round(time.clock() - start_time, 2)
print("2. Named Itertuples done in {} seconds, result = {}".format(total_elapsed_time, result))
3)使用另一种访问方法的默认itertuples()
start_time = time.clock()
result = 0
for row in df.itertuples():
result += max(getattr(row, "B"), getattr(row, "C"))
total_elapsed_time = round(time.clock() - start_time, 2)
print("3. Named Itertuples using getattr done in {} seconds, result = {}".format(total_elapsed_time, result))
4)使用name = None的默认itertuples()
更快
start_time = time.clock()
result = 0
for(_, col1, col2, col3, col4) in df.itertuples(name=None):
result += max(col2, col3)
total_elapsed_time = round(time.clock() - start_time, 2)
print("4. Itertuples done in {} seconds, result = {}".format(total_elapsed_time, result))
5)命名为itertuples()
的列名,例如My Col-Name is very Strange
start_time = time.clock()
result = 0
for row in df.itertuples(index=False):
result += max(row[df.columns.get_loc('B')], row[df.columns.get_loc('C')])
total_elapsed_time = round(time.clock() - start_time, 2)
print("5. Polyvalent Itertuples working even with special characters in the column name done in {} seconds, result = {}".format(total_elapsed_time, result))
输出:
A B C D
0 41 63 42 23
1 54 9 24 65
2 15 34 10 9
3 39 94 82 97
4 4 88 79 54
... .. .. .. ..
999995 48 27 4 25
999996 16 51 34 28
999997 1 39 61 14
999998 66 51 27 70
999999 51 53 47 99
[1000000 rows x 4 columns]
1. Iterrows done in 104.96 seconds, result = 66151519
2. Named Itertuples done in 1.26 seconds, result = 66151519
3. Named Itertuples using getattr done in 1.56 seconds, result = 66151519
4. Itertuples done in 0.94 seconds, result = 66151519
5. Polyvalent Itertuples working even with special characters in the column name done in 2.94 seconds, result = 66151519
This article is a very interesting comparison between iterrows and itertuples
总结:
df.itertuples(name=None)
df.itertuples()
否则,除非您的列中有特殊字符(''或'-',)itertuples()
。iterrows()
答案 4 :(得分:0)
这并不像我希望的那么简单。您需要使用 enumerate 来跟踪您有多少列。然后使用该计数器查找列的名称。接受的答案没有向您展示如何动态访问列名。
for row in df.itertuples(index=False, name=None):
for k,v in enumerate(row):
print("column: {0}".format(df.columns.values[k]))
print("value: {0}".format(v)