Python如何获取2维列表中的每个第一个元素

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 (EVERY first element in the list) :

我想要一个像这样的结果(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

我刚刚开始学习Python几周前。 Python版本= 2.7.9

4 个解决方案

#1

You can get the index [0] from each element in a list comprehension

您可以从列表推导中的每个元素获取索引[0]

>>> [i[0] for i in a]
[4.0, 3.0, 3.5]

Also just to be pedantic, you don't have a list of list, you have a tuple of tuple.

也只是为了迂腐,你没有列表列表,你有一个元组元组。

#2

use zip

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]

#3

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

它将返回a中每个列表的第一个元素的列表

#4

Try using

for i in a :
  print(i[0])

i represents individual row in a.So,i[0] represnts the 1st element of each row.

我代表a.So中的单个行,i [0]重新表示每行的第一个元素。

#1