Question

我有2个字符串列表，需要加入列表，因此列表2中的行被附加到列表1的每一行的末尾。排序是正确的，每个列表中的记录数相同。以下示例：

list1 =  ["A", "B" , "C"]

list2 = ["E", "F", "G"]

newlist = ["A,E", "B,F", "C,G"]

Answer 1

只需简单地压缩列表然后加入元组：

>>> list1 =  ["A", "B" , "C"]
>>> list2 = ["E", "F", "G"]
>>> [','.join(i) for i in zip(list1, list2)]
['A,E', 'B,F', 'C,G']

Answer 2

轻松完成任务，请使用renderer : function (oRM, oControl) { oRM.write("<div"); oRM.writeControlData(oControl); oRM.addClass("myAppDemoWTProductRating"); oRM.writeClasses(); oRM.write(">"); oRM.renderControl(oControl.getAggregation("_rating")); oRM.renderControl(oControl.getAggregation("_label")); oRM.renderControl(oControl.getAggregation("_button")); oRM.write("</div>"); oControl.someFunction(); }, someFunction: function(){ //Do something }：

compile 'com.android.support:appcompat-v7:23.4.0'
compile 'com.google.android.gms:play-services:9.0.0'
compile 'com.google.firebase:firebase-database:9.0.0'

你甚至可以为一般长度的purpouse生成它：

map

Answer 3

你可以结合列表推导和内置data.frame(...,stringsAsFactors=FALSE)功能的强大功能：

     stringsAsFactors=TRUE;

Answer 4

如果您有多个列表，则会执行以下操作：

list_of_list = [list1,list2]
[",".join(n) for n in  zip( *list_of_list )]

Answer 5

这应该可以正常工作

list1 = ["A", "B" , "C"]
list2 = ["E", "F", "G"]
newlist = []
for i in range(0, len(list1)):
    newlist.append(list1[i]+","+list2[i])
print(newlist)

bash@bash:~$ python a.py 
['A,E', 'B,F', 'C,G']

python join list 2逐行列出1

5 个答案: