Question

我有一个数组和一个输入，如果我输入一些我想用<html> <head> <title>Typehead</title> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> <link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" /> <script src="https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.5.0-rc.2/angular.js"></script> <script src="https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.5.0-rc.2/angular-animate.js"></script> <script src="https://cdnjs.cloudflare.com/ajax/libs/angular-ui-bootstrap/1.1.2/ui-bootstrap-tpls.min.js"></script> </head> <body ng-app="ui.bootstrap.demo" > <style> .full button span { background-color: limegreen; border-radius: 32px; color: black; } .partially button span { background-color: orange; border-radius: 32px; color: black; } .appointment>button { color: white; background-color: red; } </style> <div ng-controller="DatepickerDemoCtrl"> <pre>Selected date is: <em>{{dt | date:'fullDate' }}</em></pre> <h4>Popup</h4> <div class="row"> <div class="col-md-6"> <p class="input-group"> <input type="text" class="form-control" uib-datepicker-popup="{{format}}" ng-model="dt" ng-change="dateSelected()"is-open="popup1.opened" datepicker-options="dateOptions" ng-required="true" close-text="Close" alt-input-formats="altInputFormats" date-disabled="disabled(date, mode)" custom-class="dayClass(date, mode)" /> <span class="input-group-btn"> <button type="button" class="btn btn-default" ng-click="open1()"><i class="glyphicon glyphicon-calendar"></i></button> </span> </p> </div> </div> </div> </body> </html>和我的数组的东西，例如，如果我有这个：

.startswith()

如果我输入Array = ['foo','bar']，我希望它与"fo"匹配，然后返回索引，在本例中为"foo"。我该怎么做？

Answer 1

MaryPython的答案一般都很好。或者，在O（n）而不是O（n ^ 2）中，您可以使用

for index, item in enumerate(my_list):
       if item.startswith('fo'):
            print(index)

我已使用enumerate使用项目

来遍历索引

请注意，Marky的实现在此阵列上失败

 ['fo','fo','fobar','fobar','hi']

因为.index总是返回重复出现的first instance（但是否则他的解决方案很好而直观）

Answer 2

这是一个解决方案。我遍历列表并检查每个项目是否以字符串'fo'开头（或者您想要检查的任何内容）。如果它以该字符串开头，则打印该项的索引。我希望这有帮助！

Array = ['foo', 'bar']

for item in Array:
    if item.startswith('fo'):
        print(Array.index(item))

Answer 3

#!/usr/bin/python
# -*- coding: ascii -*-

Data = ['bleem', 'flargh', 'foo', 'bar' 'beep']

def array_startswith(search, array):
    """Search an iterable object in order to find the index of the first
    .startswith(search) match of the items."""
    for index, item in enumerate(array):
        if item.startswith(search):
            return(index)
    raise(KeyError)
    ## Give some sort of error. You probably want to raise an error rather
    ##     than returning None, as this might cause a logic error in the
    ##     later program. I think KeyError is correct, based on the question.
    ## See Effective Python by Brett Slatkin, Page 29...

if __name__ == '__main__':
    lookfor='fo'
    try:
        result=array_startswith(lookfor, Data)
    except KeyError as e:
        print("'{0}' not found in Data, Sorry...".format(lookfor))
    else:
        print("Index where '{0}' is found is: {1}. Found:{2}".format(lookfor, 
            result, Data[result]))
    finally:
        print("Try/Except/Else/Finally Cleanup happens here...")
    print("Program done.")

返回存储在数组中的字符串的索引

3 个答案: