entityMap|blocks|key|1r79n|text|我不确定这在速度上是否更好，但它更简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2a425|In+[68]:+np.array([list(astr)+for+astr+in+listOfStringOfINTs],int)+++++++++++
Out[68]:+
array([[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1]])|code-block|syntax|javascript|cjvl5|list(astr)将字符串拆分为一个包含1个字符串的列表。带有int数据类型的np.array负责转换所有字符串。|offset|length|style|CODE|amve6|或者，您可以将所有字符串连接到字符串中，生成列表，然后重塑数组：|gpvg|np.array(list(''.join(listOfStringOfINTs)),int).reshape(5,-1)|d5feu^0|0|0|0|A|W|3|14|8|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|S|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|T|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|U|9|@$J|V|K|W|L|M]|$J|X|K|Y|L|M]|$J|Z|K|10|L|M]]|A|@]|B|$]]|$2|N|4|O|6|7|8|11|9|@]|A|@]|B|$]]|$2|P|4|Q|6|E|8|12|9|@]|A|@]|B|$F|G]]|$2|R|4|-4|6|7|8|13|9|@]|A|@]|B|$]]]]

I'm not sure this is better in speed, but it's simpler:

<pre><code>In [68]: np.array([list(astr) for astr in listOfStringOfINTs],int) 
Out[68]: 
array([[1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1]])
</code></pre>

<code>list(astr)</code> split the string into a list of 1 character strings. The <code>np.array</code> with <code>int</code> dtype takes care of converting all strings.

Or you could join all strings into string, make the list, and then reshape the array:

<pre><code>np.array(list(''.join(listOfStringOfINTs)),int).reshape(5,-1)
</code></pre>

entityMap|blocks|key|6hrc4|text|为了好玩，这里有另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|behcj|>>>+np.vstack(np.frombuffer(a,dtype=np.uint8)-48+for+a+in+listOfStringOfINTs)
array([[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1]],+dtype=uint8)|code-block|syntax|javascript|4nbd9|此方法依赖于将ASCII字符作为无符号字符读取，然后依赖于数字1-9在ASCII表示中按顺序排列的事实。由于数字0表示为48，因此我们只需从所有值中减去48即可得到整数形式的值。|ct102|对于较小的字符串，它并不比@hpaulj的方法快，后者更具可读性：|fos24|In+[1]:+listOfStringOfINTs+=+["123231231231231"]*10000

In+[2]:+%25timeit+np.vstack(np.frombuffer(a,dtype=np.uint8)-48+for+a+in+listOfStringOfINTs)
10+loops,+best+of+3:+42.1+ms+per+loop

In+[3]:+%25timeit+np.array([list(astr)+for+astr+in+listOfStringOfINTs],int)
10+loops,+best+of+3:+36.3+ms+per+loop|340u8|但对于较大的字符串，它可以产生很大的不同：|r52p|In+[4]:+listOfStringOfINTs+=+["123231231231231"*1000]*10000

In+[5]:+%25timeit+np.vstack(np.frombuffer(a,dtype=np.uint8)-48+for+a+in+listOfStringOfINTs)
10+loops,+best+of+3:+115+ms+per+loop

In+[6]:+%25timeit+np.array([list(astr)+for+astr+in+listOfStringOfINTs],int)
1+loop,+best+of+3:+30.4+s+per+loop|9ijdj^0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|S|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|T|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|U|9|@]|A|@]|B|$]]|$2|J|4|K|6|7|8|V|9|@]|A|@]|B|$]]|$2|L|4|M|6|E|8|W|9|@]|A|@]|B|$F|G]]|$2|N|4|O|6|7|8|X|9|@]|A|@]|B|$]]|$2|P|4|Q|6|E|8|Y|9|@]|A|@]|B|$F|G]]|$2|R|4|-4|6|7|8|Z|9|@]|A|@]|B|$]]]]

Just for fun, here is another way to do it:

<pre><code>&gt;&gt;&gt; np.vstack(np.frombuffer(a,dtype=np.uint8)-48 for a in listOfStringOfINTs)
array([[1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1]], dtype=uint8)
</code></pre>

This method relies on reading in the ASCII characters as unsigned chars and then relying on the fact that the numerals 1-9 are in order in the ASCII representation. Since the numeral 0 is represented as 48, we just subtract 48 from all of the values to get their value as an integer.

For small strings it's not really faster than @hpaulj's method which is more readable:

<pre><code>In [1]: listOfStringOfINTs = ["123231231231231"]*10000

In [2]: %timeit np.vstack(np.frombuffer(a,dtype=np.uint8)-48 for a in listOfStringOfINTs)
10 loops, best of 3: 42.1 ms per loop

In [3]: %timeit np.array([list(astr) for astr in listOfStringOfINTs],int)
10 loops, best of 3: 36.3 ms per loop
</code></pre>

But for large strings it can make a big difference:

<pre><code>In [4]: listOfStringOfINTs = ["123231231231231"*1000]*10000

In [5]: %timeit np.vstack(np.frombuffer(a,dtype=np.uint8)-48 for a in listOfStringOfINTs)
10 loops, best of 3: 115 ms per loop

In [6]: %timeit np.array([list(astr) for astr in listOfStringOfINTs],int)
1 loop, best of 3: 30.4 s per loop
</code></pre>

entityMap|blocks|key|3mqf2|text|以上所有答案都是正确的，但直觉上，对我来说最容易理解的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|bovag|++++>>>+k+=+[list(x)+for+x+in+listOfStringOfINTs+]
++++>>>+print(np.array(k,+dtype=np.int64))
++++[[1+2+3+2+3+1+2+3+1+2+3+1+2+3+1]
+++++[1+2+3+2+3+1+2+3+1+2+3+1+2+3+1]
+++++[1+2+3+2+3+1+2+3+1+2+3+1+2+3+1]
+++++[1+2+3+2+3+1+2+3+1+2+3+1+2+3+1]
+++++[1+2+3+2+3+1+2+3+1+2+3+1+2+3+1]]|code-block|syntax|javascript|5pla3^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

All the above answers are correct, but intuitively, the easiest to understand for me is:

<pre><code> &gt;&gt;&gt; k = [list(x) for x in listOfStringOfINTs ]
 &gt;&gt;&gt; print(np.array(k, dtype=np.int64))
 [[1 2 3 2 3 1 2 3 1 2 3 1 2 3 1]
 [1 2 3 2 3 1 2 3 1 2 3 1 2 3 1]
 [1 2 3 2 3 1 2 3 1 2 3 1 2 3 1]
 [1 2 3 2 3 1 2 3 1 2 3 1 2 3 1]
 [1 2 3 2 3 1 2 3 1 2 3 1 2 3 1]]
</code></pre>

entityMap|blocks|key|3pai4|text|利用所有字符串都有相同字符数的事实，我们可以使用带有view的矢量化字符串-|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|54u4t|def+get_int_ar(a):
++++return+(np.array(a).view('u1')-48).reshape(len(a),-1)|code-block|syntax|javascript|621ao|示例运行-|8l6m0|In+[143]:+listOfStringOfINTs+=+["123231231231231"]*5

In+[144]:+get_int_ar(listOfStringOfINTs)
Out[144]:+
array([[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1],
+++++++[1,+2,+3,+2,+3,+1,+2,+3,+1,+2,+3,+1,+2,+3,+1]],+dtype=uint8)|6lhqc^0|Q|4|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|Q|9|@$A|R|B|S|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|T|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|U|9|@]|E|@]|F|$]]|$2|N|4|O|6|I|8|V|9|@]|E|@]|F|$J|K]]|$2|P|4|-4|6|7|8|W|9|@]|E|@]|F|$]]]]

Leveraging the fact that all strings have the same number of characters, we can use a vectorized one with <code>view</code> -

<pre><code>def get_int_ar(a):
 return (np.array(a).view('u1')-48).reshape(len(a),-1)
</code></pre>

Sample run -

<pre><code>In [143]: listOfStringOfINTs = ["123231231231231"]*5

In [144]: get_int_ar(listOfStringOfINTs)
Out[144]: 
array([[1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1],
 [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1]], dtype=uint8)
</code></pre>

entityMap|blocks|key|1a612|text|这是一个使用"".join的解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|ercbo|def+digit_ize(a):
++++r+=+np.fromstring(''.join(a),+'u1')
++++r+&=+0x0f
++++return+r.reshape(len(a),+-1)|code-block|syntax|javascript|9o1t6|或者(稍微快一点)：|d8kft|def+digit_ize(a):
++++r+=+np.frombuffer(''.join(a).encode(),+'u1')+&+0x0f
++++return+r.reshape(len(a),+-1)|2e4ps|计时：|9n8u8|small
pp1+4.314555088058114
pp2+2.933372976258397
div+3.740947926416993
usr+29.473979957401752
hpj+12.974489014595747
large
pp1+9.718517074361444
pp2+7.069707033224404
div+37.66830707900226
usr+2321.8201039126143
hpj+1103.1720889732242|e1tds|生成计时的脚本，包含必要时其他解决方案的Py3调整。|1ar96|import+numpy+as+np

def+digit_ize():
++++r+=+np.fromstring(''.join(a),+'u1')
++++r+&=+0x0f
++++return+r.reshape(len(a),+-1)

def+digit_ize_2():
++++r+=+np.frombuffer(''.join(a).encode(),+'u1')+&+0x0f
++++return+r.reshape(len(a),+-1)

def+get_int_ar():
++++return+(np.array(a,+'S').view('u1')-48).reshape(len(a),-1)

def+use_vstack():
++++np.vstack(np.frombuffer(b.encode(),+dtype=np.uint8)-48+for+b+in+a)

def+use_list():
++++return+np.array([list(astr)+for+astr+in+a],int)+++++++++++

from+timeit+import+timeit

listOfStringOfINTs+=+["123231231231231"]*5
a+=+listOfStringOfINTs
print("small")
print("pp1",+timeit(digit_ize,+number=1000)*1000)
print("pp2",+timeit(digit_ize_2,+number=1000)*1000)
print("div",+timeit(get_int_ar,+number=1000)*1000)
print("usr",+timeit(use_vstack,+number=1000)*1000)
print("hpj",+timeit(use_list,+number=1000)*1000)
a+=+a*100
print("large")
print("pp1",+timeit(digit_ize,+number=1000)*1000)
print("pp2",+timeit(digit_ize_2,+number=1000)*1000)
print("div",+timeit(get_int_ar,+number=1000)*1000)
print("usr",+timeit(use_vstack,+number=1000)*1000)
print("hpj",+timeit(use_list,+number=1000)*1000)|87pem^0|6|7|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|Y|9|@$A|Z|B|10|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|11|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|12|9|@]|E|@]|F|$]]|$2|N|4|O|6|I|8|13|9|@]|E|@]|F|$J|K]]|$2|P|4|Q|6|7|8|14|9|@]|E|@]|F|$]]|$2|R|4|S|6|I|8|15|9|@]|E|@]|F|$J|K]]|$2|T|4|U|6|7|8|16|9|@]|E|@]|F|$]]|$2|V|4|W|6|I|8|17|9|@]|E|@]|F|$J|K]]|$2|X|4|-4|6|7|8|18|9|@]|E|@]|F|$]]]]

Here is a soln using <code>"".join</code>:

<pre><code>def digit_ize(a):
 r = np.fromstring(''.join(a), 'u1')
 r &amp;= 0x0f
 return r.reshape(len(a), -1)
</code></pre>

or (slightly faster):

<pre><code>def digit_ize(a):
 r = np.frombuffer(''.join(a).encode(), 'u1') &amp; 0x0f
 return r.reshape(len(a), -1)
</code></pre>

Timings:

<pre><code>small
pp1 4.314555088058114
pp2 2.933372976258397
div 3.740947926416993
usr 29.473979957401752
hpj 12.974489014595747
large
pp1 9.718517074361444
pp2 7.069707033224404
div 37.66830707900226
usr 2321.8201039126143
hpj 1103.1720889732242
</code></pre>

Script to produce timings, contains Py3 adjustments of other solns where necessary.

<pre><code>import numpy as np

def digit_ize():
 r = np.fromstring(''.join(a), 'u1')
 r &amp;= 0x0f
 return r.reshape(len(a), -1)

def digit_ize_2():
 r = np.frombuffer(''.join(a).encode(), 'u1') &amp; 0x0f
 return r.reshape(len(a), -1)

def get_int_ar():
 return (np.array(a, 'S').view('u1')-48).reshape(len(a),-1)

def use_vstack():
 np.vstack(np.frombuffer(b.encode(), dtype=np.uint8)-48 for b in a)

def use_list():
 return np.array([list(astr) for astr in a],int) 

from timeit import timeit

listOfStringOfINTs = ["123231231231231"]*5
a = listOfStringOfINTs
print("small")
print("pp1", timeit(digit_ize, number=1000)*1000)
print("pp2", timeit(digit_ize_2, number=1000)*1000)
print("div", timeit(get_int_ar, number=1000)*1000)
print("usr", timeit(use_vstack, number=1000)*1000)
print("hpj", timeit(use_list, number=1000)*1000)
a = a*100
print("large")
print("pp1", timeit(digit_ize, number=1000)*1000)
print("pp2", timeit(digit_ize_2, number=1000)*1000)
print("div", timeit(get_int_ar, number=1000)*1000)
print("usr", timeit(use_vstack, number=1000)*1000)
print("hpj", timeit(use_list, number=1000)*1000)
</code></pre>

I've got a simple problem - I need to convert a string of integers to a list of integers and insert it into a numpy array.

I have code that works but I'm interested in a more efficient method if there is one. The starting condition is that I have a list of strings of integers (line 4) and the goal is to get a numpy array filled with those integers.

Here is an example of the code I use:

<pre><code>import numpy as np
print("Hello StackOverflow")

listOfStringOfINTs = ["123231231231231"]*5
print(listOfStringOfINTs)
numpyVectorOfInts = np.empty([len(listOfStringOfINTs),len(listOfStringOfINTs[0]) ], dtype='int')
for i, IntString in enumerate(listOfStringOfINTs):
 numpyVectorOfInts[i] = list(map(int, IntString))

print(numpyVectorOfInts)
</code></pre>

Most efficient way to turn a list of strings of integers to an array of integers

 我有一个简单的问题-我需要将一个整数字符串转换为一个整数列表，并将其插入到一个numpy数组中。 我有可以工作的代码，但我对更有效的方法感兴趣。开始条件是我有一个整数字符串列表(第4行)，目标是获得一个填充了这些整数的numpy数组。 下面是我使用的代码示例： import numpy as npprint("Hel...

问将整数字符串列表转换为整数数组的最有效方法
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