Question: The array is generated following way:

```
for(var i = start; i <= end;i+=step) {
    arr[i] = i + 'is the number';
}

```

Array structure looks like this

```
arr = [
1100: "1100 is the number",
1101: "1101 is the number",
1102: "1102 is the number",
1103: "1103 is the number",
1104: "1104 is the number",
1105: "1105 is the number",
]; 

```

and so on.

This array will be mapped into a list element in React.

```
render() {
    const list = this.arr;
    const timeList = list.map((element) =>
        <li>{element}</li>
        );

    return (
      <div className="App" >
        {timeList}
      </div>
    ); 

```

It works fine with the regular arrays, but using this one gives me an infinite loop.

Why is this happening and what are the possibilities to solve this ?

Thank you

UPDATE:
Found out that
"If you use named indexes, JavaScript will redefine the array to a standard object.
After that, some array methods and properties will produce incorrect results."

But still no solution to the problem

Answer: 
It is happening like that because `tweet_text` is a copy of a column `df.ix[:, 2]` for starters. Secondly, this is not pandas way to iterate over `Series` - you should use `apply()`.

To update your code, everything that goes into the loop, change into function:

```
def parse_tweet(tweet):
    ## everything from loop goes here
    return tweet

```

Then, instead of:

```
tweet_text = df.ix[:, 2]

```

do:

```
df.iloc[:, 2] = df.iloc[:, 2].apply(parse_tweet)

```

BTW, do not use `ix` indexer as it is depreciated and going to be removed in the future versions of pandas.
Python strings are immutable. You are just changing the value attributed to variable `tweet`, but never actually updating the dataframe.

You just have to reinsert the updated value back to your dataframe. Example of a simple fix:

```
for i, tweet in enumerate(tweet_text):
    tweet = re.sub('https://t.co/[a-zA-Z0-9]*', "", tweet)
    tweet = re.sub('@[a-zA-Z0-9]*', '', tweet)

    # ...

    # update dataframe
    df.ix[i, 2] = tweet

```