我正在遍历一个有数千行的数据框。理想情况下，我想知道我的循环进度——即已完成多少行，已完成总行数的百分比等。
是否有一种方法可以打印出行号，甚至更好的是遍历的行数百分比？
目前我的代码如下。目前，打印它看起来显示了一些元组/列表，但我只需要行号。这可能很简单。
for row in testDF.iterrows():
print("Currently on row: "+str(row))
理想的打印响应：
Currently on row 1; Currently iterated 1% of rows
Currently on row 2; Currently iterated 2% of rows
Currently on row 3; Currently iterated 3% of rows
Currently on row 4; Currently iterated 4% of rows
Currently on row 5; Currently iterated 5% of rows
首先，iterrows会给出包含索引和行的元组。因此，正确的代码如下：
for index, row in testDF.iterrows():
通常情况下，索引不是行号，而是一些标识符（这是pandas的优势，但它会导致一些混淆，因为它的行为与python中普通的list不同，那里的索引是行号）。这就是为什么我们需要独立计算行数的原因。我们可以引入 line_number = 0 并在每个循环中递增它 line_number += 1。但是Python为我们提供了一个现成的工具：enumerate，它返回元组 (line_number, value) 而不仅仅是 value。所以我们得到了以下代码：
for line_number, (index, row) in enumerate(testDF.iterrows()):
print("Currently on row: {}; Currently iterated {}% of rows".format(
line_number, 100*(line_number + 1)/len(testDF)))
顺便提一句，当你在python2中除以整数时，会返回整数，这就是为什么999/1000 == 0，这是你不期望的。因此，你可以强制转换成浮点数或者在开头添加100*来获得百分比。
使用format的一种可能的解决方案，如果索引是唯一且单调递增(0,1,2,...)：
for i, row in testDF.iterrows():
print("Currently on row: {}; Currently iterrated {}% of rows".format(i, (i + 1)/len(testDF.index) * 100))
示例：
np.random.seed(1332)
testDF = pd.DataFrame(np.random.randint(10, size=(10, 3)))
print (testDF)
0 1 2
0 8 1 9
1 4 3 5
2 0 1 3
3 1 8 6
4 7 4 7
5 7 5 3
6 7 9 9
7 0 1 2
8 1 3 4
9 0 0 3
for i, row in testDF.iterrows():
print("Currently on row: {}; Currently iterrated {}% of rows".format(i, (i + 1)/len(testDF.index) * 100))
Currently on row: 0; Currently iterrated 10.0% of rows
Currently on row: 1; Currently iterrated 20.0% of rows
Currently on row: 2; Currently iterrated 30.0% of rows
Currently on row: 3; Currently iterrated 40.0% of rows
Currently on row: 4; Currently iterrated 50.0% of rows
Currently on row: 5; Currently iterrated 60.0% of rows
Currently on row: 6; Currently iterrated 70.0% of rows
Currently on row: 7; Currently iterrated 80.0% of rows
Currently on row: 8; Currently iterrated 90.0% of rows
Currently on row: 9; Currently iterrated 100.0% of rows
编辑：
如果有一些自定义的索引值，可以使用zip和numpy.arange的解决方案，其中索引的长度与数据框的长度相同。
np.random.seed(1332)
testDF = pd.DataFrame(np.random.randint(10, size=(10, 3)), index=[2,4,5,6,7,8,2,1,3,5])
print (testDF)
0 1 2
2 8 1 9
4 4 3 5
5 0 1 3
6 1 8 6
7 7 4 7
8 7 5 3
2 7 9 9
1 0 1 2
3 1 3 4
5 0 0 3
for i, (idx, row) in zip(np.arange(len(testDF.index)), testDF.iterrows()):
print("Currently on row: {}; Currently iterrated {}% of rows".format(idx, (i + 1)/len(testDF.index) * 100))
Currently on row: 2; Currently iterrated 10.0% of rows
Currently on row: 4; Currently iterrated 20.0% of rows
Currently on row: 5; Currently iterrated 30.0% of rows
Currently on row: 6; Currently iterrated 40.0% of rows
Currently on row: 7; Currently iterrated 50.0% of rows
Currently on row: 8; Currently iterrated 60.0% of rows
Currently on row: 2; Currently iterrated 70.0% of rows
Currently on row: 1; Currently iterrated 80.0% of rows
Currently on row: 3; Currently iterrated 90.0% of rows
Currently on row: 5; Currently iterrated 100.0% of rows
print('currently at row',i,'. iterated through ',100 * i / testDF.shape[0],'%') 为什么？感谢你的回答。 - Rayhane Mamalen(df.index)，因为它是最快的方式。 - jezraeli 是每一行的索引。它适用于索引包含从0到len(df)-1的整数的情况，但如果testDF使用自定义索引值，则不适用。 - Adrien Matissart对于大型数据框，最好限制打印输出，因为这是一个耗时的任务。以下是一种解决方法：
dftest=pd.DataFrame(np.random.rand(10**5,5))
percent=0
n=len(dftest)//100
for i,row in dftest.iterrows():
if (i+1)//n>percent :
percent +=1
print (percent, "% realized")
dftest.iloc[i] = 2*row #a job
enumerate来轻松计算进度，它返回当前行的索引（以及行本身），可以将其除以总行数。for index, row in enumerate(testDF.iterrows()): ... progress = index / len(testDF)- DeepSpace