Python 如何获取二维列表中的每个第一个元素
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Python How to get every first element in 2 Dimensional List
提问by CodingBeginner
I have a list like this :
我有一个这样的列表：
a = ((4.0, 4, 4.0), (3.0, 3, 3.6), (3.5, 6, 4.8))
I want an outcome like this (EVERYfirst element in the list) :
我想要这样的结果（列表中的每个第一个元素）：
4.0, 3.0, 3.5
I tried a[::1][0], but it doesn't work
我尝试了 [::1][0]，但它不起作用
I'm just start learning Python weeks ago. Python version = 2.7.9
我几周前才开始学习 Python。Python 版本 = 2.7.9
采纳答案by Cory Kramer
You can get the index [0]from each element in a list comprehension
您可以[0]从列表理解中的每个元素获取索引
>>> [i[0] for i in a]
[4.0, 3.0, 3.5]
Also just to be pedantic, you don't have a listof list, you have a tupleof tuple.
也正好是迂腐，你没有list的list，你有一个tuple的tuple。
回答by Eric Renouf
You can get it like
你可以得到它
[ x[0] for x in a]
which will return a list of the first element of each list in a
这将返回每个列表的第一个元素的列表 a
回答by Joran Beasley
use zip
使用邮编
columns = zip(*rows) #transpose rows to columns
print columns[0] #print the first column
#you can also do more with the columns
print columns[1] # or print the second column
columns.append([7,7,7]) #add a new column to the end
backToRows = zip(*columns) # now we are back to rows with a new column
print backToRows
you can also use numpy
你也可以使用 numpy
a = numpy.array(a)
print a[:,0]
回答by Akash Srivastava
Try using
尝试使用
for i in a :
print(i[0])
i represents individual row in a.So,i[0] represnts the 1st element of each row.
i 代表 a.So 中的单个行，i[0] 代表每行的第一个元素。
回答by notilas
Compared the 3 methods
3种方法比较
- 2D list: 5.323603868484497 seconds
- Numpy library : 0.3201274871826172 seconds
- Zip (Thanks to Joran Beasley) : 0.12395167350769043 seconds
- 2D 列表：5.323603868484497 秒
- Numpy 库：0.3201274871826172 秒
- Zip（感谢 Joran Beasley）：0.12395167350769043 秒
D2_list=[list(range(100))]*100
t1=time.time()
for i in range(10**5):
for j in range(10):
b=[k[j] for k in D2_list]
D2_list_time=time.time()-t1
array=np.array(D2_list)
t1=time.time()
for i in range(10**5):
for j in range(10):
b=array[:,j]
Numpy_time=time.time()-t1
D2_trans = list(zip(*D2_list))
t1=time.time()
for i in range(10**5):
for j in range(10):
b=D2_trans[j]
Zip_time=time.time()-t1
print ('2D List:',D2_list_time)
print ('Numpy:',Numpy_time)
print ('Zip:',Zip_time)
The Zip method works best. It was quite useful when I had to do some column wise processes for mapreduce jobs in the cluster servers where numpy was not installed.
Zip 方法效果最好。当我必须为未安装 numpy 的集群服务器中的 mapreduce 作业执行一些列明智的过程时，它非常有用。
回答by Angel Lira
You could use this:
你可以用这个：
a = ((4.0, 4, 4.0), (3.0, 3, 3.6), (3.5, 6, 4.8))
a = np.array(a)
a[:,0]
returns >>> array([4. , 3. , 3.5])
